हेलो स्टूडेंट्स! MA/MSc Mathematics Entrance Exam 2026 की तैयारी कर रहे गणित के सभी छात्रों का स्वागत है। यदि आप IIT JAM, CUET PG, DU, BHU, JNU, CMI या किसी अन्य प्रतिष्ठित विश्वविद्यालय से मैथमेटिक्स में मास्टर्स करने का लक्ष्य रख रहे हैं, तो केवल थ्योरी पढ़ना काफी नहीं है; आपको परीक्षा के स्तर के अनुसार वस्तुनिष्ठ प्रश्नों (MCQs) की निरंतर प्रैक्टिस भी करनी होगी।

आपकी इसी आवश्यकता को पूरा करने के लिए, हम 'Most Important Questions' की एक स्पेशल सीरीज़ शुरू कर रहे हैं। आज Part 1 में, हमने 'Real Analysis' (वास्तविक विश्लेषण) और 'Linear Algebra' (रैखिक बीजगणित) के वो बहुविकल्पीय प्रश्न शामिल किए हैं जो गणित की प्रवेश परीक्षाओं में सबसे अधिक बार पूछे जाते हैं।

💡 प्रो टिप (Pro Tip): गणित की प्रवेश परीक्षाओं में Linear Algebra के तहत Eigenvalues, Eigenvectors और Matrices के गुणों (Properties) से, जबकि Real Analysis में Series की Convergence, Limits और Continuity से हमेशा प्रश्न बनते हैं। महत्वपूर्ण Theorems और Formulas की एक लिस्ट बनाकर अपने स्टडी रूम में चिपका लें।

MA/MSc Mathematics Entrance Important Questions: Part 1

नीचे दिए गए प्रश्नों का ध्यानपूर्वक अभ्यास करें। आपकी सुविधा और त्वरित मूल्यांकन के लिए सही उत्तर को हरे रंग (Green) से हाईलाइट कर दिया गया है:

ALGEBRA • EASY

Question 1

If \(G\) is a finite group of order \(17\), then which one of the following statements is always true?

(A) \(G\) is non-Abelian.
(B) \(G\) is cyclic.
(C) \(G\) has exactly two generators.
(D) \(G\) has a proper non-trivial subgroup of order \(8\).


Correct Answer

(B) \(G\) is cyclic.


Detailed Explanation

Since the order of the group is \(17\), which is a prime number, an important theorem of Group Theory applies.

\[ \boxed{\textbf{Every finite group of prime order is cyclic.}} \]

Because \(17\) is prime, every finite group of order \(17\) must be cyclic. Every cyclic group is also Abelian. Moreover, according to Lagrange's Theorem, the order of every subgroup must divide the order of the group. Since \(17\) is prime, the only possible subgroup orders are

\[ \boxed{1 \text{ and } 17} \]

Therefore, the group cannot have a proper non-trivial subgroup of order \(8\), and there is no reason for it to have exactly two generators. Hence, Option (B) is the correct answer.

LINEAR ALGEBRA • EASY

Question 2

If \[ A=\begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}, \] then the eigenvalues of the matrix \(A\) are

(A) \(1,\;3\)
(B) \(2,\;2\)
(C) \(0,\;4\)
(D) \(-1,\;5\)


Correct Answer

(A) \(1,\;3\)


Detailed Explanation

The eigenvalues of a matrix are obtained by solving its characteristic equation.

\[ \boxed{\det(A-\lambda I)=0} \]

Therefore, \[ \begin{aligned} \left| \begin{matrix} 2-\lambda & 1\\ 1 & 2-\lambda \end{matrix} \right| &=0\\[2mm] (2-\lambda)^2-1&=0\\[2mm] \lambda^2-4\lambda+3&=0\\[2mm] (\lambda-1)(\lambda-3)&=0 \end{aligned} \] Hence,

\[ \boxed{\lambda=1,\;3} \]

Thus, the eigenvalues of the matrix are \(1\) and \(3\). Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • EASY

Question 3

If \[ f(x)=x^3-3x^2+2, \] then the local maximum value of the function is

(A) \(2\)
(B) \(0\)
(C) \(-2\)
(D) \(1\)


Correct Answer

(A) \(2\)


Detailed Explanation

To determine the local maximum, first compute the first derivative.

\[ \boxed{f'(x)=0} \]

Differentiate the function: \[ \begin{aligned} f'(x) &=3x^2-6x\\ &=3x(x-2) \end{aligned} \] Therefore, the critical points are \[ x=0,\;2. \] Now calculate the second derivative: \[ f''(x)=6x-6. \] At \(x=0\), \[ f''(0)=-6<0, \] which indicates a local maximum. Now evaluate the function: \[ f(0)=0-0+2=2. \]

\[ \boxed{f(0)=2} \]

Hence, the local maximum value of the function is \(2\). Therefore, the correct answer is Option (A).

REAL ANALYSIS • EASY

Question 4

Evaluate the limit \[ \lim_{x\to 0}\frac{\sin x}{x}. \]

(A) \(0\)
(B) \(1\)
(C) \(\infty\)
(D) Does not exist


Correct Answer

(B) \(1\)


Detailed Explanation

This is one of the most fundamental limits in Differential Calculus and Real Analysis.

\[ \boxed{\lim_{x\to0}\frac{\sin x}{x}=1} \]

The result follows from the Squeeze Theorem (Sandwich Theorem) or from the geometric interpretation of the unit circle. Since the given limit is exactly the standard limit, \[ \lim_{x\to0}\frac{\sin x}{x}=1. \] Therefore, the correct answer is Option (B).

DIFFERENTIAL EQUATIONS • EASY

Question 5

The differential equation \[ \frac{dy}{dx}+2y=e^{-2x} \] is

(A) Linear differential equation
(B) Bernoulli differential equation
(C) Exact differential equation
(D) Homogeneous differential equation


Correct Answer

(A) Linear differential equation


Detailed Explanation

A first-order differential equation is said to be linear if it can be written in the standard form

\[ \boxed{\frac{dy}{dx}+P(x)y=Q(x)} \]

Comparing the given equation, \[ \frac{dy}{dx}+2y=e^{-2x}, \] we obtain \[ P(x)=2,\qquad Q(x)=e^{-2x}, \] where \(P(x)\) and \(Q(x)\) depend only on \(x\). Hence, the equation satisfies the standard form of a first-order linear differential equation. It is neither Bernoulli nor Exact nor Homogeneous. Therefore, the correct answer is Option (A).

INTEGRAL CALCULUS • EASY

Question 6

Evaluate the integral \[ \int_{0}^{1}(3x^2+2x+1)\,dx. \]

(A) \(2\)
(B) \(3\)
(C) \(4\)
(D) \(5\)


Correct Answer

(B) \(3\)


Detailed Explanation

To evaluate a definite integral, first determine the antiderivative of the integrand.

\[ \boxed{\int_a^b f(x)\,dx=F(b)-F(a)} \]

The antiderivative of the given function is \[ \int(3x^2+2x+1)\,dx=x^3+x^2+x. \] Now apply the limits: \[ \begin{aligned} \int_{0}^{1}(3x^2+2x+1)\,dx &=\left[x^3+x^2+x\right]_0^1\\[2mm] &=(1+1+1)-0\\[2mm] &=3. \end{aligned} \]

\[ \boxed{\int_{0}^{1}(3x^2+2x+1)\,dx=3} \]

Hence, the correct answer is Option (B).

COMPLEX ANALYSIS • EASY

Question 7

If \[ z=3+4i, \] then the modulus of \(z\) is

(A) \(4\)
(B) \(5\)
(C) \(6\)
(D) \(7\)


Correct Answer

(B) \(5\)


Detailed Explanation

The modulus of a complex number \(z=a+bi\) is defined by

\[ \boxed{|z|=\sqrt{a^2+b^2}} \]

For the given complex number, \[ a=3,\qquad b=4. \] Therefore, \[ \begin{aligned} |z| &=\sqrt{3^2+4^2}\\[2mm] &=\sqrt{9+16}\\[2mm] &=\sqrt{25}\\[2mm] &=5. \end{aligned} \]

\[ \boxed{|3+4i|=5} \]

Hence, the modulus of the given complex number is \(5\). Therefore, the correct answer is Option (B).

VECTOR CALCULUS • EASY

Question 8

If \[ \vec{F}(x,y,z)=x\hat{i}+y\hat{j}+z\hat{k}, \] then the divergence of the vector field \(\vec{F}\) is

(A) \(0\)
(B) \(1\)
(C) \(2\)
(D) \(3\)


Correct Answer

(D) \(3\)


Detailed Explanation

The divergence of a vector field measures the net outward flow at a point and is defined as

\[ \boxed{\nabla\cdot\vec{F}= \frac{\partial F_x}{\partial x} +\frac{\partial F_y}{\partial y} +\frac{\partial F_z}{\partial z}} \]

Here, \[ F_x=x,\qquad F_y=y,\qquad F_z=z. \] Therefore, \[ \begin{aligned} \nabla\cdot\vec{F} &=\frac{\partial x}{\partial x} +\frac{\partial y}{\partial y} +\frac{\partial z}{\partial z}\\[2mm] &=1+1+1\\[2mm] &=3. \end{aligned} \]

\[ \boxed{\nabla\cdot(x\hat{i}+y\hat{j}+z\hat{k})=3} \]

Hence, the divergence of the given vector field is \(3\). Therefore, the correct answer is Option (D).

NUMERICAL ANALYSIS • EASY

Question 9

The Newton-Raphson method is primarily used for

(A) Numerical differentiation
(B) Solving linear systems
(C) Finding the approximate root of a nonlinear equation
(D) Numerical integration


Correct Answer

(C) Finding the approximate root of a nonlinear equation


Detailed Explanation

The Newton-Raphson method is one of the fastest iterative techniques for approximating the root of a nonlinear equation. The iterative formula is

\[ \boxed{ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} } \]

Starting from an initial approximation \(x_0\), successive approximations are obtained until the desired accuracy is achieved. The method requires that the derivative \(f'(x)\) exists and is not equal to zero near the root. It is not used for numerical differentiation, numerical integration, or directly solving systems of linear equations.

\[ \boxed{ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} } \]

Hence, the correct answer is Option (C).

PROBABILITY & STATISTICS • EASY

Question 10

A fair die is rolled once. What is the probability of obtaining a prime number?

(A) \(\frac{1}{6}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{2}{3}\)


Correct Answer

(C) \(\frac{1}{2}\)


Detailed Explanation

The sample space for a single throw of a fair die is \[ S=\{1,2,3,4,5,6\}. \] The prime numbers among these outcomes are \[ 2,\;3,\;5. \] Thus, the number of favorable outcomes is \(3\), while the total number of possible outcomes is \(6\).

\[ \boxed{ P(E)=\frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} } \]

Therefore, \[ P(\text{Prime Number}) = \frac{3}{6} = \frac{1}{2}. \]

\[ \boxed{ P(\text{Prime Number})=\frac12 } \]

Hence, the probability of obtaining a prime number is \(\frac12\). Therefore, the correct answer is Option (C).

DISCRETE MATHEMATICS • EASY

Question 11

How many subsets does a set containing \(5\) elements have?

(A) \(16\)
(B) \(25\)
(C) \(32\)
(D) \(64\)


Correct Answer

(C) \(32\)


Detailed Explanation

For a finite set containing \(n\) elements, every element has exactly two choices while forming a subset:

  • Include the element.
  • Exclude the element.

Hence, the total number of subsets is given by

\[ \boxed{ \text{Number of Subsets}=2^n } \]

Since \(n=5\), \[ 2^5=32. \] Therefore, the given set contains exactly \(32\) subsets, including the empty set and the set itself.

\[ \boxed{ 2^5=32 } \]

Hence, the correct answer is Option (C).

ALGEBRA • EASY

Question 12

According to Lagrange's Theorem, the order of every subgroup of a finite group \(G\) must

(A) be greater than the order of \(G\)
(B) divide the order of \(G\)
(C) always be a prime number
(D) always be equal to the order of \(G\)


Correct Answer

(B) divide the order of \(G\)


Detailed Explanation

Lagrange's Theorem is one of the most important results in Group Theory. It establishes a relationship between the order of a finite group and the order of its subgroups.

\[ \boxed{|H|\mid |G|} \]

Here, \[ |H|=\text{Order of the subgroup},\qquad |G|=\text{Order of the group}. \] The notation \[ |H|\mid |G| \] means that the order of the subgroup exactly divides the order of the group. For example, if \[ |G|=24, \] then the possible subgroup orders are \[ 1,\;2,\;3,\;4,\;6,\;8,\;12,\;24. \] Numbers such as \(5,\;7,\;10\) cannot be subgroup orders because they do not divide \(24\). Therefore, the correct answer is Option (B).

LINEAR ALGEBRA • EASY-MODERATE

Question 13

If \[ A= \begin{bmatrix} 4 & 2\\ 1 & 3 \end{bmatrix}, \] then the determinant of the matrix \(A\) is

(A) \(8\)
(B) \(10\)
(C) \(12\)
(D) \(14\)


Correct Answer

(B) \(10\)


Detailed Explanation

The determinant of a \(2\times2\) matrix is calculated using the following formula.

\[ \boxed{ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc } \]

For the given matrix, \[ a=4,\qquad b=2,\qquad c=1,\qquad d=3. \] Therefore, \[ \begin{aligned} |A| &=(4)(3)-(2)(1)\\[2mm] &=12-2\\[2mm] &=10. \end{aligned} \]

\[ \boxed{|A|=10} \]

The determinant is a fundamental quantity that helps determine whether a matrix is invertible. Since \[ |A|=10\neq0, \] the matrix is non-singular and therefore has an inverse. Hence, the correct answer is Option (B).

DIFFERENTIAL CALCULUS • EASY-MODERATE

Question 14

If \[ f(x)=e^{2x}\sin x, \] then \[ \frac{d}{dx}\left(e^{2x}\sin x\right) \] is equal to

(A) \(e^{2x}(2\sin x+\cos x)\)
(B) \(e^{2x}(2\cos x+\sin x)\)
(C) \(e^{2x}(\sin x-\cos x)\)
(D) \(e^{2x}(2\sin x-\cos x)\)


Correct Answer

(A) \(e^{2x}(2\sin x+\cos x)\)


Detailed Explanation

Since the function is a product of two differentiable functions, we apply the Product Rule.

\[ \boxed{ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx} } \]

Let \[ u=e^{2x},\qquad v=\sin x. \] Then, \[ \frac{du}{dx}=2e^{2x}, \qquad \frac{dv}{dx}=\cos x. \] Applying the Product Rule, \[ \begin{aligned} \frac{d}{dx}(e^{2x}\sin x) &=e^{2x}\cos x+2e^{2x}\sin x\\[2mm] &=e^{2x}(2\sin x+\cos x). \end{aligned} \]

\[ \boxed{ \frac{d}{dx}(e^{2x}\sin x) = e^{2x}(2\sin x+\cos x) } \]

Hence, the correct answer is Option (A).

REAL ANALYSIS • EASY-MODERATE

Question 15

Which one of the following sequences converges to \(0\) as \(n\to\infty\)?

(A) \(a_n=n\)
(B) \(a_n=(-1)^n\)
(C) \(a_n=\dfrac{1}{n}\)
(D) \(a_n=\sqrt{n}\)


Correct Answer

(C) \(\dfrac{1}{n}\)


Detailed Explanation

A sequence \(\{a_n\}\) converges to a limit \(L\) if

\[ \boxed{ \lim_{n\to\infty}a_n=L } \]

Now examine each option.

  • \(n\to\infty\), so it does not converge.
  • \((-1)^n\) oscillates between \(-1\) and \(1\), so the limit does not exist.
  • \(\dfrac1n\to0\) as \(n\to\infty\).
  • \(\sqrt n\to\infty\), so it diverges.

Hence,

\[ \boxed{ \lim_{n\to\infty}\frac1n=0 } \]

Therefore, the only sequence that converges to zero is Option (C).

LINEAR ALGEBRA • MODERATE

Question 16

If \[ A= \begin{bmatrix} 1 & 2\\ 2 & 4 \end{bmatrix}, \] then the rank of the matrix \(A\) is

(A) \(0\)
(B) \(1\)
(C) \(2\)
(D) \(3\)


Correct Answer

(B) \(1\)


Detailed Explanation

The rank of a matrix is the maximum number of linearly independent rows or columns.

\[ \boxed{\operatorname{Rank}(A)=\text{Maximum number of linearly independent rows (or columns)}} \]

Observe that \[ R_2=2R_1. \] Thus, the second row is a scalar multiple of the first row, so the two rows are linearly dependent. Also, \[ |A| = \begin{vmatrix} 1&2\\ 2&4 \end{vmatrix} = 4-4 = 0, \] which confirms that the matrix is singular. Since there is only one linearly independent row (or column),

\[ \boxed{\operatorname{Rank}(A)=1} \]

Hence, the correct answer is Option (B).

DIFFERENTIAL CALCULUS • MODERATE

Question 17

If \[ f(x)=x^3-6x^2+9x+1, \] then the value of \(x\) at which the function attains its local minimum is

(A) \(1\)
(B) \(2\)
(C) \(3\)
(D) \(4\)


Correct Answer

(C) \(3\)


Detailed Explanation

To locate the local extrema, first determine the critical points by solving

\[ \boxed{f'(x)=0} \]

Differentiate the function: \[ \begin{aligned} f'(x) &=3x^2-12x+9\\[2mm] &=3(x-1)(x-3). \end{aligned} \] Hence, the critical points are \[ x=1,\qquad x=3. \] Now compute the second derivative: \[ f''(x)=6x-12. \] At \(x=1\), \[ f''(1)=-6<0, \] so \(x=1\) is a local maximum. At \(x=3\), \[ f''(3)=6>0, \] which indicates a local minimum.

\[ \boxed{f''(3)>0\;\Rightarrow\;\text{Local Minimum at }x=3} \]

Therefore, the function attains its local minimum at \(x=3\). Hence, the correct answer is Option (C).

INTEGRAL CALCULUS • MODERATE

Question 18

Evaluate \[ \int x e^x\,dx. \]

(A) \(xe^x+C\)
(B) \(e^x(x-1)+C\)
(C) \(e^x(x+1)+C\)
(D) \(\dfrac{x^2e^x}{2}+C\)


Correct Answer

(B) \(e^x(x-1)+C\)


Detailed Explanation

The given integral is evaluated using the method of Integration by Parts.

\[ \boxed{ \int u\,dv = uv-\int v\,du } \]

Choose \[ u=x, \qquad dv=e^x\,dx. \] Then, \[ du=dx, \qquad v=e^x. \] Applying Integration by Parts, \[ \begin{aligned} \int xe^x\,dx &=xe^x-\int e^x\,dx\\[2mm] &=xe^x-e^x+C\\[2mm] &=e^x(x-1)+C. \end{aligned} \]

\[ \boxed{ \int xe^x\,dx=e^x(x-1)+C } \]

Hence, the correct answer is Option (B).

COMPLEX ANALYSIS • MODERATE

Question 19

If \[ z=\frac{1+i}{1-i}, \] then the value of \(z\) is

(A) \(i\)
(B) \(-i\)
(C) \(1\)
(D) \(-1\)


Correct Answer

(A) \(i\)


Detailed Explanation

To simplify a complex fraction, multiply both numerator and denominator by the conjugate of the denominator.

\[ \boxed{ (a-bi)\text{ is the conjugate of }(a+bi) } \]

Thus, \[ \begin{aligned} \frac{1+i}{1-i} &= \frac{(1+i)(1+i)} {(1-i)(1+i)}\\[2mm] &= \frac{1+2i+i^2} {1-i^2}\\[2mm] &= \frac{2i}{2}\\[2mm] &=i. \end{aligned} \]

\[ \boxed{ \frac{1+i}{1-i}=i } \]

Therefore, the value of the given complex number is \(i\). Hence, the correct answer is Option (A).

VECTOR CALCULUS • MODERATE

Question 20

If \[ \vec{F}(x,y,z)=yz\,\hat{i}+xz\,\hat{j}+xy\,\hat{k}, \] then the divergence of the vector field \(\vec{F}\) is

(A) \(x+y+z\)
(B) \(0\)
(C) \(xyz\)
(D) \(1\)


Correct Answer

(B) \(0\)


Detailed Explanation

The divergence of a vector field is defined by

\[ \boxed{ \nabla\cdot\vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} } \]

For \[ \vec{F}(x,y,z)=yz\,\hat{i}+xz\,\hat{j}+xy\,\hat{k}, \] we have \[ F_x=yz,\qquad F_y=xz,\qquad F_z=xy. \] Therefore, \[ \begin{aligned} \nabla\cdot\vec{F} &= \frac{\partial (yz)}{\partial x} + \frac{\partial (xz)}{\partial y} + \frac{\partial (xy)}{\partial z}\\[2mm] &=0+0+0\\[2mm] &=0. \end{aligned} \]

\[ \boxed{ \nabla\cdot\vec{F}=0 } \]

Hence, the given vector field is solenoidal, since its divergence is zero. Therefore, the correct answer is Option (B).

NUMERICAL ANALYSIS • MODERATE

Question 21

Which one of the following methods always requires two initial approximations that bracket the root of the equation?

(A) Newton-Raphson Method
(B) Bisection Method
(C) Fixed Point Iteration Method
(D) Euler's Method


Correct Answer

(B) Bisection Method


Detailed Explanation

The Bisection Method is a bracketing method. It begins with two points \(a\) and \(b\) such that the function changes sign over the interval.

\[ \boxed{ f(a)\,f(b)<0 } \]

The condition \[ f(a)\,f(b)<0 \] guarantees that at least one real root lies in the interval \((a,b)\), provided the function is continuous. The interval is repeatedly divided into two equal parts until the desired accuracy is achieved. Unlike the Bisection Method, the Newton-Raphson Method starts with only one initial approximation, while Euler's Method is used to solve differential equations.

\[ \boxed{ f(a)\,f(b)<0 \Longrightarrow \text{Root exists in }(a,b) } \]

Hence, the correct answer is Option (B).

PROBABILITY & STATISTICS • MODERATE

Question 22

Two fair coins are tossed simultaneously. What is the probability of obtaining exactly one head?

(A) \(\frac14\)
(B) \(\frac12\)
(C) \(\frac34\)
(D) \(1\)


Correct Answer

(B) \(\frac12\)


Detailed Explanation

The sample space for tossing two fair coins is \[ S=\{HH,HT,TH,TT\}. \] The event of obtaining exactly one head is \[ E=\{HT,TH\}. \] Thus, the number of favorable outcomes is \(2\), while the total number of outcomes is \(4\).

\[ \boxed{ P(E)=\frac{n(E)}{n(S)} } \]

Therefore, \[ P(E)=\frac24=\frac12. \]

\[ \boxed{ P(\text{Exactly One Head})=\frac12 } \]

Hence, the probability of obtaining exactly one head is \(\frac12\). Therefore, the correct answer is Option (B).

DISCRETE MATHEMATICS • MODERATE

Question 23

How many different ways can the letters of the word \[ \text{MATH} \] be arranged?

(A) \(12\)
(B) \(16\)
(C) \(24\)
(D) \(48\)


Correct Answer

(C) \(24\)


Detailed Explanation

The word \[ \text{MATH} \] contains four distinct letters. The number of arrangements of \(n\) distinct objects is given by the factorial formula.

\[ \boxed{ {}^{n}P_n=n! } \]

Since there are four distinct letters, \[ 4!=4\times3\times2\times1=24. \] Hence, there are exactly \(24\) different arrangements.

\[ \boxed{ 4!=24 } \]

Therefore, the correct answer is Option (C).

ALGEBRA • MODERATE

Question 24

Let \(G\) be a cyclic group of order \(12\). The number of generators of \(G\) is

(A) \(2\)
(B) \(4\)
(C) \(6\)
(D) \(12\)


Correct Answer

(B) \(4\)


Detailed Explanation

For a cyclic group of order \(n\), the number of generators is given by Euler's Totient Function.

\[ \boxed{\text{Number of Generators}=\phi(n)} \]

Here, \[ n=12. \] Using Euler's Totient Function, \[ \begin{aligned} \phi(12) &=12\left(1-\frac12\right)\left(1-\frac13\right)\\[2mm] &=12\times\frac12\times\frac23\\[2mm] &=4. \end{aligned} \] Thus, there are exactly four generators of the cyclic group of order \(12\).

\[ \boxed{\phi(12)=4} \]

Hence, the correct answer is Option (B).

LINEAR ALGEBRA • MODERATE

Question 25

If \[ A= \begin{bmatrix} 2 & 0\\ 0 & 5 \end{bmatrix}, \] then the trace of the matrix \(A\) is

(A) \(7\)
(B) \(10\)
(C) \(3\)
(D) \(0\)


Correct Answer

(A) \(7\)


Detailed Explanation

The trace of a square matrix is defined as the sum of its principal diagonal elements.

\[ \boxed{\operatorname{tr}(A)=\sum_{i=1}^{n}a_{ii}} \]

For the given matrix, \[ A= \begin{bmatrix} 2 & 0\\ 0 & 5 \end{bmatrix}, \] the diagonal elements are \[ 2\quad\text{and}\quad5. \] Therefore, \[ \operatorname{tr}(A)=2+5=7. \] It is worth noting that the trace of a matrix is also equal to the sum of its eigenvalues.

\[ \boxed{\operatorname{tr}(A)=7} \]

Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • MODERATE

Question 26

If \[ y=\ln(\sin x), \] then \(\dfrac{dy}{dx}\) is

(A) \(\tan x\)
(B) \(\cot x\)
(C) \(\sec x\)
(D) \(\csc x\)


Correct Answer

(B) \(\cot x\)


Detailed Explanation

The given function is a composite function. Therefore, we apply the Chain Rule.

\[ \boxed{ \frac{d}{dx}\bigl(\ln u\bigr)=\frac{1}{u}\cdot\frac{du}{dx} } \]

Let \[ u=\sin x. \] Then, \[ \frac{du}{dx}=\cos x. \] Hence, \[ \begin{aligned} \frac{dy}{dx} &=\frac{1}{\sin x}\cdot\cos x\\[2mm] &=\frac{\cos x}{\sin x}\\[2mm] &=\cot x. \end{aligned} \]

\[ \boxed{ \frac{d}{dx}\bigl(\ln(\sin x)\bigr)=\cot x } \]

Hence, the correct answer is Option (B).

REAL ANALYSIS • MODERATE

Question 27

The infinite geometric series \[ 1+\frac12+\frac14+\frac18+\cdots \] has the sum

(A) \(1\)
(B) \(2\)
(C) \(3\)
(D) Diverges


Correct Answer

(B) \(2\)


Detailed Explanation

This is an infinite geometric series with \[ a=1,\qquad r=\frac12. \] Since \[ |r|<1, \] the series converges.

\[ \boxed{ S=\frac{a}{1-r}, \qquad |r|<1 } \]

Substituting the given values, \[ \begin{aligned} S &=\frac{1}{1-\frac12}\\[2mm] &=\frac{1}{\frac12}\\[2mm] &=2. \end{aligned} \] Therefore, the infinite geometric series converges to the value \(2\).

\[ \boxed{ 1+\frac12+\frac14+\frac18+\cdots=2 } \]

Hence, the correct answer is Option (B).

LINEAR ALGEBRA • MODERATE

Question 28

If \[ A= \begin{bmatrix} 3 & 0\\ 0 & 7 \end{bmatrix}, \] then the characteristic polynomial of the matrix \(A\) is

(A) \(\lambda^2-10\lambda+21\)
(B) \(\lambda^2+10\lambda+21\)
(C) \(\lambda^2-21\)
(D) \(\lambda^2-10\lambda-21\)


Correct Answer

(A) \(\lambda^2-10\lambda+21\)


Detailed Explanation

The characteristic polynomial of a square matrix is obtained by solving

\[ \boxed{ \det(A-\lambda I)=0 } \]

For the given matrix, \[ A-\lambda I= \begin{bmatrix} 3-\lambda & 0\\ 0 & 7-\lambda \end{bmatrix}. \] Hence, \[ \begin{aligned} \det(A-\lambda I) &=(3-\lambda)(7-\lambda)\\[2mm] &=\lambda^2-10\lambda+21. \end{aligned} \] Thus, the characteristic polynomial is

\[ \boxed{ \lambda^2-10\lambda+21 } \]

The roots of this polynomial are the eigenvalues of the matrix, namely \[ \lambda=3,\;7. \] Hence, the correct answer is Option (A).

DIFFERENTIAL EQUATIONS • MODERATE

Question 29

The integrating factor (I.F.) of the differential equation \[ \frac{dy}{dx}+3y=e^x \] is

(A) \(e^{x}\)
(B) \(e^{2x}\)
(C) \(e^{3x}\)
(D) \(3e^x\)


Correct Answer

(C) \(e^{3x}\)


Detailed Explanation

A first-order linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x) \] is solved using an integrating factor.

\[ \boxed{ \text{I.F.}=e^{\int P(x)\,dx} } \]

For the given equation, \[ P(x)=3. \] Therefore, \[ \begin{aligned} \text{I.F.} &=e^{\int3\,dx}\\[2mm] &=e^{3x}. \end{aligned} \] Multiplying the entire differential equation by this integrating factor converts the left-hand side into an exact derivative.

\[ \boxed{ \text{I.F.}=e^{3x} } \]

Hence, the correct answer is Option (C).

REAL ANALYSIS • MODERATE

Question 30

The sequence \[ a_n=\frac{3n+1}{5n-2} \] converges to

(A) \(0\)
(B) \(\frac35\)
(C) \(\frac53\)
(D) Diverges


Correct Answer

(B) \(\frac35\)


Detailed Explanation

To evaluate the limit of a rational sequence, divide both the numerator and denominator by the highest power of \(n\).

\[ \boxed{ \lim_{n\to\infty}\frac{a_n}{b_n} = \frac{\text{Leading Coefficient of }a_n} {\text{Leading Coefficient of }b_n} } \]

Therefore, \[ \begin{aligned} \lim_{n\to\infty}\frac{3n+1}{5n-2} &= \lim_{n\to\infty} \frac{3+\frac1n} {5-\frac2n}\\[2mm] &=\frac35. \end{aligned} \] As \(n\to\infty\), \[ \frac1n\to0, \qquad \frac2n\to0. \] Hence, the sequence converges to

\[ \boxed{ \lim_{n\to\infty}\frac{3n+1}{5n-2}=\frac35 } \]

Therefore, the correct answer is Option (B).

COMPLEX ANALYSIS • MODERATE

Question 31

If \[ z=2-3i, \] then the complex conjugate of \(z\) is

(A) \(2+3i\)
(B) \(-2+3i\)
(C) \(-2-3i\)
(D) \(3+2i\)


Correct Answer

(A) \(2+3i\)


Detailed Explanation

The complex conjugate of a complex number is obtained by changing the sign of its imaginary part.

\[ \boxed{ \overline{a+bi}=a-bi } \]

The given complex number is \[ z=2-3i. \] Changing the sign of the imaginary part, \[ \overline{z}=2+3i. \] Also, \[ z\,\overline{z}=|z|^2. \] Indeed, \[ (2-3i)(2+3i)=4+9=13. \] This verifies the property of complex conjugates.

\[ \boxed{ \overline{2-3i}=2+3i } \]

Hence, the correct answer is Option (A).

VECTOR CALCULUS • MODERATE

Question 32

If \[ \phi(x,y,z)=x^2+y^2+z^2, \] then the gradient of the scalar function \(\phi\) is

(A) \(2x\hat{i}+2y\hat{j}+2z\hat{k}\)
(B) \(x\hat{i}+y\hat{j}+z\hat{k}\)
(C) \(2\hat{i}+2\hat{j}+2\hat{k}\)
(D) \(0\)


Correct Answer

(A) \(2x\hat{i}+2y\hat{j}+2z\hat{k}\)


Detailed Explanation

The gradient of a scalar function gives the direction of the maximum rate of increase of the function.

\[ \boxed{ \nabla\phi= \frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} + \frac{\partial\phi}{\partial z}\hat{k} } \]

For \[ \phi=x^2+y^2+z^2, \] we have \[ \frac{\partial\phi}{\partial x}=2x,\qquad \frac{\partial\phi}{\partial y}=2y,\qquad \frac{\partial\phi}{\partial z}=2z. \] Hence, \[ \nabla\phi = 2x\hat{i}+2y\hat{j}+2z\hat{k}. \]

\[ \boxed{ \nabla(x^2+y^2+z^2) = 2x\hat{i}+2y\hat{j}+2z\hat{k} } \]

Hence, the correct answer is Option (A).

NUMERICAL ANALYSIS • MODERATE

Question 33

The Newton-Raphson iteration formula for solving \[ f(x)=0 \] is

(A) \[ x_{n+1}=x_n+\frac{f(x_n)}{f'(x_n)} \]
(B) \[ x_{n+1}=x_n-\frac{f'(x_n)}{f(x_n)} \]
(C) \[ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \]
(D) \[ x_{n+1}=\frac{x_n+f(x_n)}2 \]


Correct Answer

(C) \[ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \]


Detailed Explanation

The Newton-Raphson Method is based on the tangent line approximation of a nonlinear function.

\[ \boxed{ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} } \]

Starting from an initial approximation \(x_0\), successive approximations are computed until convergence. For the method to work efficiently,

  • \(f(x)\) should be differentiable.
  • \(f'(x)\neq0\) near the root.
  • The initial approximation should be chosen close to the actual root.

Under suitable conditions, the Newton-Raphson Method exhibits quadratic convergence, making it one of the fastest numerical methods for solving nonlinear equations.

\[ \boxed{ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} } \]

Hence, the correct answer is Option (C).

PROBABILITY & STATISTICS • MODERATE

Question 34

If the random variable \(X\) follows a binomial distribution with parameters \[ n=8,\qquad p=\frac12, \] then the mean of the distribution is

(A) \(2\)
(B) \(4\)
(C) \(6\)
(D) \(8\)


Correct Answer

(B) \(4\)


Detailed Explanation

For a binomial random variable, \[ X\sim B(n,p), \] the mean is given by

\[ \boxed{\mu=np} \]

Here, \[ n=8,\qquad p=\frac12. \] Therefore, \[ \begin{aligned} \mu &=np\\[2mm] &=8\times\frac12\\[2mm] &=4. \end{aligned} \]

\[ \boxed{\mu=4} \]

Hence, the mean of the given binomial distribution is \(4\). Therefore, the correct answer is Option (B).

DISCRETE MATHEMATICS • MODERATE

Question 35

How many diagonals does a polygon with \(10\) sides have?

(A) \(30\)
(B) \(35\)
(C) \(40\)
(D) \(45\)


Correct Answer

(B) \(35\)


Detailed Explanation

The total number of diagonals in an \(n\)-sided polygon is given by

\[ \boxed{ \text{Number of Diagonals} = \frac{n(n-3)}{2} } \]

For a decagon, \[ n=10. \] Hence, \[ \begin{aligned} \frac{10(10-3)}2 &=\frac{10\times7}{2}\\[2mm] &=35. \end{aligned} \] This can also be verified by noting that each vertex is connected to \(10-3=7\) vertices by diagonals, and every diagonal is counted twice.

\[ \boxed{ \frac{10(10-3)}2=35 } \]

Hence, the polygon has 35 diagonals. Therefore, the correct answer is Option (B).

LINEAR ALGEBRA • MODERATE+

Question 36

If \[ A= \begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix}, \] then the eigenvalues of the matrix \(A\) are

(A) \(1,\;3\)
(B) \(2,\;2\)
(C) \(3,\;3\)
(D) \(2,\;4\)


Correct Answer

(A) \(1,\;3\)


Detailed Explanation

The eigenvalues of a square matrix are obtained from the characteristic equation.

\[ \boxed{ \det(A-\lambda I)=0 } \]

Since the given matrix is upper triangular, \[ A= \begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix}, \] its eigenvalues are simply the diagonal entries. Alternatively, \[ \begin{aligned} \det(A-\lambda I) &= \begin{vmatrix} 1-\lambda & 2\\ 0 & 3-\lambda \end{vmatrix}\\[2mm] &=(1-\lambda)(3-\lambda)=0. \end{aligned} \] Therefore, \[ \lambda=1,\;3. \]

\[ \boxed{ \lambda_1=1,\qquad \lambda_2=3 } \]

Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • MODERATE+

Question 37

If \[ f(x)=x^2-4x+7, \] then the minimum value of the function is

(A) \(1\)
(B) \(2\)
(C) \(3\)
(D) \(4\)


Correct Answer

(C) \(3\)


Detailed Explanation

To determine the minimum value, first locate the critical point.

\[ \boxed{ f'(x)=0 } \]

Differentiate the function: \[ f'(x)=2x-4. \] Setting \[ 2x-4=0, \] gives \[ x=2. \] Now compute the second derivative: \[ f''(x)=2>0, \] which confirms that \(x=2\) is a point of local (and global) minimum. Finally, \[ \begin{aligned} f(2) &=2^2-4(2)+7\\[2mm] &=4-8+7\\[2mm] &=3. \end{aligned} \]

\[ \boxed{ \min f(x)=3 } \]

Hence, the minimum value of the function is \(3\). Therefore, the correct answer is Option (C).

INTEGRAL CALCULUS • MODERATE+

Question 38

Evaluate \[ \int_{0}^{\pi}\sin x\,dx. \]

(A) \(0\)
(B) \(1\)
(C) \(2\)
(D) \(\pi\)


Correct Answer

(C) \(2\)


Detailed Explanation

The Fundamental Theorem of Calculus states that

\[ \boxed{ \int_a^b f(x)\,dx=F(b)-F(a) } \]

Since \[ \int\sin x\,dx=-\cos x, \] we obtain \[ \begin{aligned} \int_{0}^{\pi}\sin x\,dx &=\left[-\cos x\right]_{0}^{\pi}\\[2mm] &=(-\cos\pi)-(-\cos0)\\[2mm] &=1+1\\[2mm] &=2. \end{aligned} \]

\[ \boxed{ \int_{0}^{\pi}\sin x\,dx=2 } \]

Hence, the correct answer is Option (C).

COMPLEX ANALYSIS • MODERATE+

Question 39

If \[ f(z)=z^2, \] then \[ \frac{df}{dz} \] is equal to

(A) \(z\)
(B) \(2z\)
(C) \(z^2\)
(D) \(2z^2\)


Correct Answer

(B) \(2z\)


Detailed Explanation

The function \[ f(z)=z^2 \] is a polynomial in the complex variable \(z\). Every polynomial is analytic over the entire complex plane. Differentiation follows exactly the same rules as ordinary calculus.

\[ \boxed{ \frac{d}{dz}(z^n)=nz^{\,n-1} } \]

Applying the power rule, \[ \begin{aligned} \frac{d}{dz}(z^2) &=2z^{2-1}\\[2mm] &=2z. \end{aligned} \] Since \(f(z)\) is analytic everywhere, the derivative exists for every complex number \(z\).

\[ \boxed{ \frac{d}{dz}(z^2)=2z } \]

Hence, the correct answer is Option (B).

DIFFERENTIAL CALCULUS • MODERATE+

Question 40

The function \[ f(x)=x^2-4x+3 \] satisfies the hypotheses of Rolle's Theorem on the interval \([1,3]\). The value of \(c\in(1,3)\) satisfying Rolle's Theorem is

(A) \(1\)
(B) \(2\)
(C) \(3\)
(D) \(4\)


Correct Answer

(B) \(2\)


Detailed Explanation

Rolle's Theorem states that if a function is continuous on a closed interval, differentiable on the corresponding open interval, and satisfies \[ f(a)=f(b), \] then there exists at least one point \(c\in(a,b)\) such that

\[ \boxed{ f'(c)=0 } \]

First verify the endpoint values: \[ \begin{aligned} f(1)&=1-4+3=0,\\ f(3)&=9-12+3=0. \end{aligned} \] Thus, \[ f(1)=f(3), \] so Rolle's Theorem is applicable. Now differentiate: \[ f'(x)=2x-4. \] Setting the derivative equal to zero, \[ 2x-4=0, \] gives \[ x=2. \] Since \(2\in(1,3)\), the required value is

\[ \boxed{ c=2 } \]

Hence, the correct answer is Option (B).

LINEAR ALGEBRA • MODERATE+

Question 41

Let \[ A= \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}. \] The sum of the eigenvalues of the matrix \(A\) is

(A) \(2\)
(B) \(3\)
(C) \(4\)
(D) \(5\)


Correct Answer

(C) \(4\)


Detailed Explanation

A fundamental result in Linear Algebra states that the sum of the eigenvalues of a square matrix is equal to its trace.

\[ \boxed{ \lambda_1+\lambda_2+\cdots+\lambda_n = \operatorname{tr}(A) } \]

For the given matrix, \[ A= \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}, \] the trace is \[ \operatorname{tr}(A)=2+2=4. \] Indeed, the eigenvalues of this matrix are \[ 1 \quad\text{and}\quad 3, \] whose sum is also \[ 1+3=4. \] Thus, both methods give the same answer.

\[ \boxed{ \operatorname{tr}(A)=4 } \]

Hence, the correct answer is Option (C).

LINEAR ALGEBRA • MODERATE+

Question 42

If \[ A= \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix}, \] then the characteristic equation of the matrix is

(A) \((\lambda-2)^2=0\)
(B) \((\lambda+2)^2=0\)
(C) \(\lambda^2-5\lambda+4=0\)
(D) \(\lambda^2-2=0\)


Correct Answer

(A) \((\lambda-2)^2=0\)


Detailed Explanation

The characteristic equation is obtained from

\[ \boxed{ \det(A-\lambda I)=0 } \]

For the given matrix, \[ A-\lambda I= \begin{bmatrix} 2-\lambda & 1\\ 0 & 2-\lambda \end{bmatrix}. \] Hence, \[ \begin{aligned} \det(A-\lambda I) &=(2-\lambda)^2\\[2mm] &=(\lambda-2)^2. \end{aligned} \] Therefore, \[ (\lambda-2)^2=0. \] The matrix has a repeated eigenvalue \[ \lambda=2 \] with algebraic multiplicity \(2\).

\[ \boxed{ (\lambda-2)^2=0 } \]

Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • MODERATE+

Question 43

If \[ y=x\ln x, \] then \[ \frac{dy}{dx} \] is equal to

(A) \(\ln x\)
(B) \(1+\ln x\)
(C) \(\dfrac1x+\ln x\)
(D) \(x+\ln x\)


Correct Answer

(B) \(1+\ln x\)


Detailed Explanation

Since the given function is the product of two differentiable functions, we apply the Product Rule.

\[ \boxed{ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx} } \]

Let \[ u=x, \qquad v=\ln x. \] Then, \[ \frac{du}{dx}=1, \qquad \frac{dv}{dx}=\frac1x. \] Applying the Product Rule, \[ \begin{aligned} \frac{dy}{dx} &=x\left(\frac1x\right)+(\ln x)(1)\\[2mm] &=1+\ln x. \end{aligned} \]

\[ \boxed{ \frac{d}{dx}(x\ln x)=1+\ln x } \]

Hence, the correct answer is Option (B).

REAL ANALYSIS • MODERATE+

Question 44

The series \[ \sum_{n=1}^{\infty}\frac{1}{n^2} \] is

(A) Divergent
(B) Conditionally convergent
(C) Absolutely convergent
(D) Oscillatory


Correct Answer

(C) Absolutely convergent


Detailed Explanation

The given series is a \(p\)-series of the form \[ \sum_{n=1}^{\infty}\frac{1}{n^p}. \] A \(p\)-series converges if and only if

\[ \boxed{ \sum_{n=1}^{\infty}\frac{1}{n^p} \text{ converges if }p>1 } \]

Here, \[ p=2>1. \] Hence, \[ \sum_{n=1}^{\infty}\frac1{n^2} \] is convergent. Since every term is positive, \[ \sum_{n=1}^{\infty}\left|\frac1{n^2}\right| = \sum_{n=1}^{\infty}\frac1{n^2} \] also converges. Therefore, the series is absolutely convergent.

\[ \boxed{ p=2>1 \Longrightarrow \sum_{n=1}^{\infty}\frac1{n^2} \text{ converges} } \]

Hence, the correct answer is Option (C).

COMPLEX ANALYSIS • MODERATE+

Question 45

If \[ f(z)=u(x,y)+iv(x,y) \] is analytic, then the Cauchy-Riemann equations are

(A) \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \]
(B) \[ \frac{\partial u}{\partial x} = - \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} \]
(C) \[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} \]
(D) \[ \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} \]


Correct Answer

(A)


Detailed Explanation

A complex function \[ f(z)=u(x,y)+iv(x,y) \] is analytic if the first-order partial derivatives exist, are continuous, and satisfy the Cauchy-Riemann equations.

\[ \boxed{ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} } \]

These equations are necessary conditions for differentiability of a complex function. For many entrance examinations, checking the Cauchy-Riemann equations is the first step in determining whether a function is analytic. Hence, whenever these equations hold (along with continuity of the partial derivatives), the function is analytic at that point.

\[ \boxed{ u_x=v_y, \qquad u_y=-v_x } \]

Therefore, the correct answer is Option (A).

VECTOR CALCULUS • MODERATE+

Question 46

If \[ \vec{F}(x,y,z)=x^2\hat{i}+y^2\hat{j}+z^2\hat{k}, \] then the divergence of \(\vec{F}\) is

(A) \(2x+2y+2z\)
(B) \(x+y+z\)
(C) \(2(x+y+z)\)
(D) \(0\)


Correct Answer

(C) \(2(x+y+z)\)


Detailed Explanation

The divergence of a vector field is defined by

\[ \boxed{ \nabla\cdot\vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} } \]

For the given vector field, \[ F_x=x^2,\qquad F_y=y^2,\qquad F_z=z^2. \] Therefore, \[ \begin{aligned} \nabla\cdot\vec{F} &= \frac{\partial(x^2)}{\partial x} + \frac{\partial(y^2)}{\partial y} + \frac{\partial(z^2)}{\partial z}\\[2mm] &=2x+2y+2z\\[2mm] &=2(x+y+z). \end{aligned} \]

\[ \boxed{ \nabla\cdot\vec{F} = 2(x+y+z) } \]

Hence, the correct answer is Option (C).

DIFFERENTIAL EQUATIONS • MODERATE+

Question 47

The differential equation \[ \frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0 \] has the auxiliary equation

(A) \(m^2+5m+6=0\)
(B) \(m^2-5m+6=0\)
(C) \(m^2-6=0\)
(D) \(m^2+6=0\)


Correct Answer

(B) \(m^2-5m+6=0\)


Detailed Explanation

For a linear homogeneous differential equation with constant coefficients, \[ a\frac{d^2y}{dx^2} +b\frac{dy}{dx} +cy=0, \] the corresponding auxiliary (characteristic) equation is obtained by replacing \[ \frac{d}{dx} \quad\text{with}\quad m. \]

\[ \boxed{ am^2+bm+c=0 } \]

Comparing with \[ \frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0, \] we have \[ a=1,\qquad b=-5,\qquad c=6. \] Hence, \[ m^2-5m+6=0. \] Factoring, \[ (m-2)(m-3)=0. \] Thus, the complementary solution is \[ y=C_1e^{2x}+C_2e^{3x}. \]

\[ \boxed{ m^2-5m+6=0 } \]

Hence, the correct answer is Option (B).

LINEAR ALGEBRA • MODERATE+

Question 48

If \[ A= \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}, \] then the inverse of the matrix \(A\) is

(A) \[ \begin{bmatrix} -2 & 1\\ \frac32 & -\frac12 \end{bmatrix} \]
(B) \[ \begin{bmatrix} 2 & -1\\ -3 & 1 \end{bmatrix} \]
(C) \[ \begin{bmatrix} 4 & -2\\ -3 & 1 \end{bmatrix} \]
(D) \[ \begin{bmatrix} 1 & -2\\ -3 & 4 \end{bmatrix} \]


Correct Answer

(A)


Detailed Explanation

A square matrix is invertible if its determinant is non-zero. The inverse of a \(2\times2\) matrix is given by

\[ \boxed{ A^{-1} = \frac1{ad-bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} } \]

First compute the determinant: \[ |A| = (1)(4)-(2)(3) = -2. \] Therefore, \[ \begin{aligned} A^{-1} &= \frac1{-2} \begin{bmatrix} 4 & -2\\ -3 & 1 \end{bmatrix}\\[2mm] &= \begin{bmatrix} -2 & 1\\ \frac32 & -\frac12 \end{bmatrix}. \end{aligned} \]

\[ \boxed{ A^{-1} = \begin{bmatrix} -2 & 1\\ \frac32 & -\frac12 \end{bmatrix} } \]

Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • MODERATE+

Question 49

If \[ y=\sin^{-1}x, \] then \[ \frac{dy}{dx} \] is equal to

(A) \[ \frac1{\sqrt{1-x^2}} \]
(B) \[ \frac1{1-x^2} \]
(C) \[ \sqrt{1-x^2} \]
(D) \[ -\frac1{\sqrt{1-x^2}} \]


Correct Answer

(A) \[ \frac1{\sqrt{1-x^2}} \]


Detailed Explanation

The derivative of an inverse trigonometric function is a standard result in Differential Calculus.

\[ \boxed{ \frac{d}{dx}\left(\sin^{-1}x\right) = \frac1{\sqrt{1-x^2}} } \]

This formula is valid for \[ -1

\[ \boxed{ \frac{d}{dx}\left(\sin^{-1}x\right) = \frac1{\sqrt{1-x^2}} } \]

Hence, the correct answer is Option (A).

LINEAR ALGEBRA • MODERATE+

Question 50

If \[ A= \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}, \] then the determinant of the matrix \(A\) is

(A) \(2\)
(B) \(3\)
(C) \(4\)
(D) \(5\)


Correct Answer

(B) \(3\)


Detailed Explanation

The determinant of a \(2\times2\) matrix is computed using

\[ \boxed{ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc } \]

For the given matrix, \[ A= \begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}, \] we obtain \[ \begin{aligned} |A| &=(2)(2)-(1)(1)\\[2mm] &=4-1\\[2mm] &=3. \end{aligned} \] Since \[ |A|\neq0, \] the matrix is invertible.

\[ \boxed{ |A|=3 } \]

Hence, the correct answer is Option (B).

DIFFERENTIAL CALCULUS • MODERATE+

Question 51

If \[ f(x)=\frac{x^2+1}{x}, \] then the derivative \(f'(x)\) is

(A) \[ 1-\frac1{x^2} \]
(B) \[ 1+\frac1{x^2} \]
(C) \[ 2x+\frac1{x} \]
(D) \[ 2-\frac1{x} \]


Correct Answer

(A) \[ 1-\frac1{x^2} \]


Detailed Explanation

First simplify the function: \[ \begin{aligned} f(x) &=\frac{x^2+1}{x}\\[2mm] &=x+\frac1x. \end{aligned} \] Now differentiate each term separately.

\[ \boxed{ \frac{d}{dx}\left(x+\frac1x\right) = 1-\frac1{x^2} } \]

Indeed, \[ \frac{d}{dx}(x)=1, \] and \[ \frac{d}{dx}\left(x^{-1}\right) =-x^{-2} =-\frac1{x^2}. \] Therefore, \[ \begin{aligned} f'(x) &=1-\frac1{x^2}. \end{aligned} \] This derivative exists for all \[ x\neq0. \]

\[ \boxed{ f'(x) = 1-\frac1{x^2} } \]

Hence, the correct answer is Option (A).

LINEAR ALGEBRA • MODERATE+

Question 52

Let \[ A= \begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}. \] The eigenvalues of the matrix \(A\) are

(A) \(1,\;4,\;7\)
(B) \(12,\;7,\;4\)
(C) \(1,\;5,\;7\)
(D) \(4,\;7,\;28\)


Correct Answer

(A) \(1,\;4,\;7\)


Detailed Explanation

A diagonal matrix has a very simple eigenvalue structure.

\[ \boxed{ \text{Eigenvalues of a diagonal matrix are its diagonal entries.} } \]

The given matrix is \[ A= \begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}. \] Hence, its eigenvalues are directly obtained from the principal diagonal: \[ \lambda_1=1,\qquad \lambda_2=4,\qquad \lambda_3=7. \] Also, \[ \operatorname{tr}(A)=1+4+7=12, \] which is equal to the sum of the eigenvalues.

\[ \boxed{ \lambda=1,\;4,\;7 } \]

Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • MODERATE+

Question 53

If \[ y=e^{3x}\cos x, \] then \[ \frac{dy}{dx} \] is equal to

(A) \[ e^{3x}(3\cos x-\sin x) \]
(B) \[ e^{3x}(3\sin x+\cos x) \]
(C) \[ e^{3x}(3\cos x+\sin x) \]
(D) \[ e^{3x}(\cos x-\sin x) \]


Correct Answer

(A) \[ e^{3x}(3\cos x-\sin x) \]


Detailed Explanation

The function is the product of two differentiable functions. Therefore, we use the Product Rule.

\[ \boxed{ \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} } \]

Let \[ u=e^{3x}, \qquad v=\cos x. \] Then, \[ \frac{du}{dx}=3e^{3x}, \qquad \frac{dv}{dx}=-\sin x. \] Applying the Product Rule, \[ \begin{aligned} \frac{dy}{dx} &=e^{3x}(-\sin x)+3e^{3x}\cos x\\[2mm] &=e^{3x}(3\cos x-\sin x). \end{aligned} \]

\[ \boxed{ \frac{d}{dx}\left(e^{3x}\cos x\right) = e^{3x}(3\cos x-\sin x) } \]

Hence, the correct answer is Option (A).

INTEGRAL CALCULUS • MODERATE+

Question 54

Evaluate \[ \int \frac{1}{x}\,dx. \]

(A) \(\ln x+C\)
(B) \(\ln|x|+C\)
(C) \(\dfrac1x+C\)
(D) \(x\ln x+C\)


Correct Answer

(B) \(\ln|x|+C\)


Detailed Explanation

This is one of the fundamental standard integrals in calculus.

\[ \boxed{ \int\frac1x\,dx=\ln|x|+C } \]

The absolute value is essential because \[ \frac{d}{dx}\left(\ln|x|\right)=\frac1x, \] for every \[ x\neq0. \] If we write only \(\ln x\), the formula is valid only for \(x>0\). Hence, the correct indefinite integral is \[ \ln|x|+C. \]

\[ \boxed{ \int\frac1x\,dx=\ln|x|+C } \]

Hence, the correct answer is Option (B).

COMPLEX ANALYSIS • MODERATE+

Question 55

The modulus of the complex number \[ z=3-4i \] is

(A) \(3\)
(B) \(4\)
(C) \(5\)
(D) \(7\)


Correct Answer

(C) \(5\)


Detailed Explanation

For any complex number \[ z=a+bi, \] its modulus is defined by

\[ \boxed{ |z|=\sqrt{a^2+b^2} } \]

Here, \[ a=3, \qquad b=-4. \] Therefore, \[ \begin{aligned} |z| &=\sqrt{3^2+(-4)^2}\\[2mm] &=\sqrt{9+16}\\[2mm] &=\sqrt{25}\\[2mm] &=5. \end{aligned} \] Also, observe that \[ z\overline z=|z|^2=25. \] This is a standard property of complex numbers.

\[ \boxed{ |3-4i|=5 } \]

Hence, the correct answer is Option (C).

LINEAR ALGEBRA • MODERATE+

Question 56

Let \[ A= \begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}. \] The characteristic polynomial of the matrix \(A\) is

(A) \(\lambda^2-2\lambda-3\)
(B) \(\lambda^2+2\lambda-3\)
(C) \(\lambda^2-3\lambda+2\)
(D) \(\lambda^2-4\lambda+3\)


Correct Answer

(A) \(\lambda^2-2\lambda-3\)


Detailed Explanation

The characteristic polynomial of a square matrix is obtained from

\[ \boxed{ \det(A-\lambda I)=0 } \]

For \[ A= \begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}, \] we have \[ A-\lambda I= \begin{bmatrix} 1-\lambda & 2\\ 2 & 1-\lambda \end{bmatrix}. \] Therefore, \[ \begin{aligned} \det(A-\lambda I) &=(1-\lambda)^2-4\\[2mm] &=\lambda^2-2\lambda-3. \end{aligned} \] Hence, the characteristic equation is \[ \lambda^2-2\lambda-3=0. \] Factoring, \[ (\lambda-3)(\lambda+1)=0. \] Thus, the eigenvalues are \[ 3,\;-1. \]

\[ \boxed{ \lambda^2-2\lambda-3 } \]

Hence, the correct answer is Option (A).

DIFFERENTIAL CALCULUS • MODERATE+

Question 57

If \[ f(x)=\tan x, \] then the second derivative \[ \frac{d^2y}{dx^2} \] is

(A) \[ 2\sec^2x\tan x \]
(B) \[ \sec^2x \]
(C) \[ 2\tan^2x \]
(D) \[ \sec x\tan x \]


Correct Answer

(A) \[ 2\sec^2x\tan x \]


Detailed Explanation

First compute the first derivative.

\[ \boxed{ \frac{d}{dx}(\tan x)=\sec^2x } \]

Now differentiate again. Using the Chain Rule, \[ \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx}(\sec^2x)\\[2mm] &=2\sec x\cdot(\sec x\tan x)\\[2mm] &=2\sec^2x\tan x. \end{aligned} \] Thus, the required second derivative is \[ 2\sec^2x\tan x. \]

\[ \boxed{ \frac{d^2}{dx^2}(\tan x) = 2\sec^2x\tan x } \]

Hence, the correct answer is Option (A).

LINEAR ALGEBRA • MODERATE+

Question 58

Let \[ A= \begin{bmatrix} 4 & 1\\ 2 & 3 \end{bmatrix}. \] The trace of the matrix \(A\) is

(A) \(5\)
(B) \(6\)
(C) \(7\)
(D) \(10\)


Correct Answer

(C) \(7\)


Detailed Explanation

The trace of a square matrix is defined as the sum of its principal diagonal entries.

\[ \boxed{ \operatorname{tr}(A)=\sum_{i=1}^{n}a_{ii} } \]

For the given matrix, \[ A= \begin{bmatrix} 4 & 1\\ 2 & 3 \end{bmatrix}, \] the principal diagonal elements are \[ 4 \quad\text{and}\quad 3. \] Hence, \[ \operatorname{tr}(A)=4+3=7. \] Also, the trace is equal to the sum of the eigenvalues of the matrix.

\[ \boxed{ \operatorname{tr}(A)=7 } \]

Hence, the correct answer is Option (C).

INTEGRAL CALCULUS • MODERATE+

Question 59

Evaluate \[ \int e^{2x}\,dx. \]

(A) \[ e^{2x}+C \]
(B) \[ 2e^{2x}+C \]
(C) \[ \frac12e^{2x}+C \]
(D) \[ \frac1{2x}e^{2x}+C \]


Correct Answer

(C) \[ \frac12e^{2x}+C \]


Detailed Explanation

To integrate an exponential function of the form \[ e^{ax}, \] we use the standard integration formula.

\[ \boxed{ \int e^{ax}\,dx = \frac1a e^{ax}+C, \qquad a\neq0 } \]

Here, \[ a=2. \] Therefore, \[ \begin{aligned} \int e^{2x}\,dx &= \frac12e^{2x}+C. \end{aligned} \] To verify, \[ \frac{d}{dx}\left(\frac12e^{2x}\right) = \frac12\cdot2e^{2x} = e^{2x}. \] Thus, the result is correct.

\[ \boxed{ \int e^{2x}\,dx = \frac12e^{2x}+C } \]

Hence, the correct answer is Option (C).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 60

If \[ A= \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}, \] then the rank of the matrix \(A\) is

(A) \(0\)
(B) \(1\)
(C) \(2\)
(D) \(3\)


Correct Answer

(B) \(1\)


Detailed Explanation

The rank of a matrix is the maximum number of linearly independent rows (or columns).

\[ \boxed{ \operatorname{Rank}(A) = \text{Maximum number of linearly independent rows} } \]

For \[ A= \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}, \] the second row is identical to the first row. Thus, \[ R_2=R_1, \] which means only one row is linearly independent. Also, \[ |A| = 1\times1-1\times1 = 0, \] confirming that the matrix is singular. Hence,

\[ \boxed{ \operatorname{Rank}(A)=1 } \]

Therefore, the correct answer is Option (B).

DIFFERENTIAL CALCULUS • ENTRANCE LEVEL

Question 61

If \[ y=\frac{\sin x}{x}, \qquad x\neq0, \] then \[ \frac{dy}{dx} \] is equal to

(A) \[ \frac{x\cos x-\sin x}{x^2} \]
(B) \[ \frac{x\cos x+\sin x}{x^2} \]
(C) \[ \frac{\cos x}{x} \]
(D) \[ \frac{\sin x}{x^2} \]


Correct Answer

(A) \[ \frac{x\cos x-\sin x}{x^2} \]


Detailed Explanation

The function is a quotient of two differentiable functions. Hence, we apply the Quotient Rule.

\[ \boxed{ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,u'-u\,v'}{v^2} } \]

Let \[ u=\sin x, \qquad v=x. \] Then, \[ u'=\cos x, \qquad v'=1. \] Applying the Quotient Rule, \[ \begin{aligned} \frac{dy}{dx} &= \frac{x(\cos x)-(\sin x)(1)}{x^2}\\[2mm] &= \frac{x\cos x-\sin x}{x^2}. \end{aligned} \] This derivative appears frequently in limits, Fourier analysis, and special function problems.

\[ \boxed{ \frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{x\cos x-\sin x}{x^2} } \]

Hence, the correct answer is Option (A).

LAPLACE TRANSFORM • ENTRANCE LEVEL

Question 62

The Laplace Transform of \[ \sin(at) \] is

(A) \[ \frac{s}{s^2+a^2} \]
(B) \[ \frac{a}{s^2+a^2} \]
(C) \[ \frac{1}{s-a} \]
(D) \[ \frac{s}{s-a} \]


Correct Answer

(B) \[ \frac{a}{s^2+a^2} \]


Detailed Explanation

The Laplace Transform of a function \(f(t)\) is defined by

\[ \boxed{ \mathcal{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t)\,dt } \]

One of the standard Laplace Transform formulas is \[ \mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2}. \] Similarly, \[ \mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}. \] Students frequently confuse these two formulas in entrance examinations, so they should be memorized carefully.

\[ \boxed{ \mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2} } \]

Hence, the correct answer is Option (B).

FOURIER SERIES • ENTRANCE LEVEL

Question 63

The Fourier Series of a periodic function is generally expressed in terms of

(A) Exponential functions only
(B) Polynomial functions only
(C) Sine and Cosine functions
(D) Logarithmic functions


Correct Answer

(C) Sine and Cosine functions


Detailed Explanation

A Fourier Series represents a periodic function as an infinite sum of sine and cosine functions having different frequencies.

\[ \boxed{ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n\cos nx + b_n\sin nx \right) } \]

Here, \[ a_n \] and \[ b_n \] are called the Fourier coefficients. The Fourier Series is widely used in heat conduction, signal processing, vibration analysis, electrical engineering, and many branches of applied mathematics. Thus, a periodic function is expanded using both sine and cosine functions.

\[ \boxed{ \text{Fourier Series} = \text{Sine Series} + \text{Cosine Series} } \]

Hence, the correct answer is Option (C).

BETA & GAMMA FUNCTIONS • ENTRANCE LEVEL

Question 64

The Gamma Function satisfies the identity

(A) \[ \Gamma(n)=n! \]
(B) \[ \Gamma(n+1)=n\,\Gamma(n) \]
(C) \[ \Gamma(n)=n\,\Gamma(n+1) \]
(D) \[ \Gamma(n)=\frac1{n!} \]


Correct Answer

(B) \[ \Gamma(n+1)=n\,\Gamma(n) \]


Detailed Explanation

The Gamma Function is defined by

\[ \boxed{ \Gamma(x) = \int_0^\infty t^{x-1}e^{-t}\,dt, \qquad x>0 } \]

One of its most important recurrence relations is \[ \Gamma(n+1)=n\,\Gamma(n). \] Using this relation repeatedly, \[ \Gamma(n+1)=n!. \] For example, \[ \Gamma(5)=4!=24. \] This recurrence formula is extensively used in probability theory, special functions, and definite integrals.

\[ \boxed{ \Gamma(n+1)=n\,\Gamma(n) } \]

Hence, the correct answer is Option (B).

ABSTRACT ALGEBRA • ENTRANCE LEVEL

Question 65

Which one of the following groups is always an abelian group?

(A) Symmetric Group \(S_3\)
(B) Dihedral Group \(D_4\)
(C) \((\mathbb{Z},+)\)
(D) Symmetric Group \(S_4\)


Correct Answer

(C) \((\mathbb{Z},+)\)


Detailed Explanation

A group \(G\) is called abelian if its binary operation satisfies the commutative law.

\[ \boxed{ a*b=b*a, \qquad \forall\,a,b\in G } \]

The group \[ (\mathbb{Z},+) \] is abelian because addition of integers is commutative: \[ a+b=b+a. \] However, - \(S_3\) is non-abelian. - \(S_4\) is non-abelian. - \(D_4\) (the symmetry group of a square) is also non-abelian. Therefore, among the given options, only \[ (\mathbb{Z},+) \] is always an abelian group.

\[ \boxed{ (\mathbb{Z},+) \text{ is an abelian group} } \]

Hence, the correct answer is Option (C).

ABSTRACT ALGEBRA • ENTRANCE LEVEL

Question 66

The order of the symmetric group \(S_n\) is

(A) \(n\)
(B) \(n^2\)
(C) \(2^n\)
(D) \(n!\)


Correct Answer

(D) \(n!\)


Detailed Explanation

The symmetric group \(S_n\) consists of all possible permutations of \(n\) distinct objects. The number of such permutations is

\[ \boxed{ |S_n|=n! } \]

For example, \[ |S_3|=3!=6, \] and \[ |S_4|=4!=24. \] Since the order of a finite group equals the total number of its elements, the order of the symmetric group is \[ n!. \]

\[ \boxed{ |S_n|=n! } \]

Hence, the correct answer is Option (D).

TOPOLOGY • ENTRANCE LEVEL

Question 67

Which one of the following sets is an open interval in the real line?

(A) \([2,5]\)
(B) \((2,5)\)
(C) \([2,5)\)
(D) \((2,5]\)


Correct Answer

(B) \((2,5)\)


Detailed Explanation

In the usual topology on the real line, a set is called open if every point of the set has a neighborhood that lies completely inside the set.

\[ \boxed{ (a,b) = \{x\in\mathbb{R}:a

The interval \[ (2,5) \] does not include either endpoint. Hence, every point inside the interval has a small open neighborhood contained entirely within the interval. The remaining options contain one or both endpoints, making them either closed or half-open intervals. Therefore, \[ (2,5) \] is the only open interval among the given choices.

\[ \boxed{ (2,5) \text{ is an open interval} } \]

Hence, the correct answer is Option (B).

METRIC SPACES • ENTRANCE LEVEL

Question 68

Which one of the following is NOT a property of a metric \(d(x,y)\)?

(A) \[ d(x,y)\ge0 \]
(B) \[ d(x,y)=d(y,x) \]
(C) \[ d(x,z)\le d(x,y)+d(y,z) \]
(D) \[ d(x,y)<0 \]


Correct Answer

(D) \[ d(x,y)<0 \]


Detailed Explanation

A metric is a function that measures the distance between two points. For every metric space \((X,d)\), the following axioms must hold.

\[ \boxed{ \begin{aligned} &d(x,y)\ge0,\\ &d(x,y)=0\iff x=y,\\ &d(x,y)=d(y,x),\\ &d(x,z)\le d(x,y)+d(y,z). \end{aligned} } \]

These are known as:

  • Non-negativity
  • Identity of indiscernibles
  • Symmetry
  • Triangle inequality
Since a distance can never be negative, \[ d(x,y)<0 \] can never be satisfied.

\[ \boxed{ d(x,y)\ge0 \text{ for all }x,y\in X } \]

Hence, the correct answer is Option (D).

PROBABILITY THEORY • ENTRANCE LEVEL

Question 69

If \(X\) follows a Poisson distribution with parameter \[ \lambda=5, \] then the variance of \(X\) is

(A) \(1\)
(B) \(5\)
(C) \(10\)
(D) \(25\)


Correct Answer

(B) \(5\)


Detailed Explanation

A Poisson random variable with parameter \(\lambda\) has two remarkable properties.

\[ \boxed{ \text{Mean}=\lambda, \qquad \text{Variance}=\lambda } \]

Given \[ \lambda=5, \] we obtain \[ \operatorname{Var}(X)=5. \] Similarly, \[ E(X)=5. \] This equality of mean and variance is one of the identifying characteristics of the Poisson distribution and is frequently tested in entrance examinations.

\[ \boxed{ \operatorname{Var}(X)=\lambda=5 } \]

Hence, the correct answer is Option (B).

GRAPH THEORY • ENTRANCE LEVEL

Question 70

According to the Handshaking Lemma, the sum of the degrees of all vertices in an undirected graph having \(m\) edges is

(A) \(m\)
(B) \(2m\)
(C) \(m^2\)
(D) \(2m+1\)


Correct Answer

(B) \(2m\)


Detailed Explanation

The Handshaking Lemma is one of the fundamental theorems of Graph Theory. It states that every edge contributes exactly two incidences—one at each endpoint.

\[ \boxed{ \sum_{v\in V}\deg(v)=2|E| } \]

If a graph contains \[ m \] edges, then \[ |E|=m. \] Therefore, \[ \sum\deg(v)=2m. \] An important consequence of this theorem is that the number of vertices having odd degree is always even.

\[ \boxed{ \sum\deg(v)=2m } \]

Hence, the correct answer is Option (B).

BOOLEAN ALGEBRA • ENTRANCE LEVEL

Question 71

In Boolean Algebra, the complement of \(1\) is

(A) \(1\)
(B) \(0\)
(C) \(-1\)
(D) Undefined


Correct Answer

(B) \(0\)


Detailed Explanation

In Boolean Algebra, every element has a unique complement. The complement of an element \(A\) is denoted by \(A'\) (or \(\overline{A}\)).

\[ \boxed{ 1'=0, \qquad 0'=1 } \]

This follows directly from the complement laws: \[ A+A'=1, \] and \[ A\cdot A'=0. \] Substituting \[ A=1, \] gives \[ 1'=0. \] Boolean complements play a fundamental role in digital logic design, switching circuits, and computer science.

\[ \boxed{ 1'=0 } \]

Hence, the correct answer is Option (B).

VECTOR CALCULUS • ENTRANCE LEVEL

Question 72

If \[ \vec{F}(x,y,z)=yz\,\hat{i}+xz\,\hat{j}+xy\,\hat{k}, \] then the divergence of \(\vec{F}\) is

(A) \(x+y+z\)
(B) \(xy+yz+xz\)
(C) \(0\)
(D) \(3xyz\)


Correct Answer

(C) \(0\)


Detailed Explanation

The divergence of a vector field \[ \vec{F}=P\hat{i}+Q\hat{j}+R\hat{k} \] is defined as

\[ \boxed{ \nabla\cdot\vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} } \]

Here, \[ P=yz,\qquad Q=xz,\qquad R=xy. \] Therefore, \[ \frac{\partial P}{\partial x}=0,\qquad \frac{\partial Q}{\partial y}=0,\qquad \frac{\partial R}{\partial z}=0. \] Hence, \[ \nabla\cdot\vec{F}=0+0+0=0. \] Thus, the vector field is solenoidal.

\[ \boxed{ \nabla\cdot\vec{F}=0 } \]

Hence, the correct answer is Option (C).

VECTOR CALCULUS • ENTRANCE LEVEL

Question 73

If \[ \vec{F}(x,y,z)=(-y)\,\hat{i}+x\,\hat{j}, \] then the curl of the vector field is

(A) \(0\)
(B) \(2\hat{k}\)
(C) \(-2\hat{k}\)
(D) \(\hat{i}+\hat{j}\)


Correct Answer

(B) \(2\hat{k}\)


Detailed Explanation

For a vector field \[ \vec{F}=P\hat{i}+Q\hat{j}+R\hat{k}, \] the curl is defined as

\[ \boxed{ \nabla\times\vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ P & Q & R \end{vmatrix} } \]

Here, \[ P=-y,\qquad Q=x,\qquad R=0. \] Only the \(k\)-component is non-zero. \[ \begin{aligned} (\nabla\times\vec{F})_k &= \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\\[2mm] &= 1-(-1)\\[2mm] &=2. \end{aligned} \] Therefore, \[ \nabla\times\vec{F} = 2\hat{k}. \]

\[ \boxed{ \nabla\times\vec{F} = 2\hat{k} } \]

Hence, the correct answer is Option (B).

VECTOR CALCULUS • ENTRANCE LEVEL

Question 74

If \[ \phi(x,y,z)=x^2+y^2-z^2, \] then the Laplacian of \(\phi\), \[ \nabla^2\phi, \] is equal to

(A) \(0\)
(B) \(2\)
(C) \(4\)
(D) \(2\)


Correct Answer

(B) \(2\)


Detailed Explanation

The Laplacian operator of a scalar function is defined as

\[ \boxed{ \nabla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} } \]

For \[ \phi=x^2+y^2-z^2, \] we obtain \[ \frac{\partial^2\phi}{\partial x^2}=2, \] \[ \frac{\partial^2\phi}{\partial y^2}=2, \] and \[ \frac{\partial^2\phi}{\partial z^2}=-2. \] Hence, \[ \begin{aligned} \nabla^2\phi &=2+2-2\\[2mm] &=2. \end{aligned} \]

\[ \boxed{ \nabla^2(x^2+y^2-z^2)=2 } \]

Hence, the correct answer is Option (B).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 75

If \[ A= \begin{bmatrix} 3 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & 5 \end{bmatrix}, \] then the determinant of \(A\) is

(A) \(-30\)
(B) \(30\)
(C) \(8\)
(D) \(6\)


Correct Answer

(A) \(-30\)


Detailed Explanation

The determinant of a diagonal matrix equals the product of its diagonal entries.

\[ \boxed{ |A| = a_{11}a_{22}\cdots a_{nn} } \]

For the given matrix, \[ A= \begin{bmatrix} 3 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & 5 \end{bmatrix}, \] the determinant is \[ \begin{aligned} |A| &=3\times(-2)\times5\\[2mm] &=-30. \end{aligned} \] Since \[ |A|\neq0, \] the matrix is non-singular and hence invertible.

\[ \boxed{ |A|=-30 } \]

Hence, the correct answer is Option (A).

REAL ANALYSIS • ENTRANCE LEVEL

Question 76

The sequence \[ a_n=\frac{(-1)^n}{n} \] is

(A) Divergent
(B) Convergent with limit \(0\)
(C) Convergent with limit \(1\)
(D) Unbounded


Correct Answer

(B) Convergent with limit \(0\)


Detailed Explanation

The given sequence alternates in sign because of the factor \[ (-1)^n. \] However, its magnitude decreases as \[ \frac1n\longrightarrow0. \]

\[ \boxed{ \lim_{n\to\infty}\frac{(-1)^n}{n}=0 } \]

Indeed, \[ -\frac1n \le \frac{(-1)^n}{n} \le \frac1n. \] Since \[ \lim_{n\to\infty}\frac1n = \lim_{n\to\infty}\left(-\frac1n\right) =0, \] the Squeeze Theorem gives \[ \lim_{n\to\infty}\frac{(-1)^n}{n}=0. \] Thus, the sequence converges to zero.

\[ \boxed{ \lim_{n\to\infty}\frac{(-1)^n}{n}=0 } \]

Hence, the correct answer is Option (B).

DIFFERENTIAL EQUATIONS • ENTRANCE LEVEL

Question 77

The order of the differential equation \[ \frac{d^3y}{dx^3} + 2\frac{dy}{dx} +y=0 \] is

(A) \(1\)
(B) \(2\)
(C) \(3\)
(D) \(4\)


Correct Answer

(C) \(3\)


Detailed Explanation

The order of a differential equation is determined by the highest order derivative appearing in the equation.

\[ \boxed{ \text{Order} = \text{Highest order derivative present} } \]

In the given equation, \[ \frac{d^3y}{dx^3} + 2\frac{dy}{dx} +y=0, \] the highest derivative is \[ \frac{d^3y}{dx^3}. \] Therefore, \[ \text{Order}=3. \] Note that the coefficients of the derivatives do not affect the order.

\[ \boxed{ \text{Order}=3 } \]

Hence, the correct answer is Option (C).

GREEN'S THEOREM • ENTRANCE LEVEL

Question 78

Green's Theorem establishes the relationship between

(A) A line integral and a double integral over a plane region
(B) A surface integral and a volume integral
(C) A line integral and a triple integral
(D) A double integral and a triple integral


Correct Answer

(A) A line integral and a double integral over a plane region


Detailed Explanation

Green's Theorem converts a line integral around a simple closed curve into a double integral over the enclosed plane region.

\[ \boxed{ \oint_C(P\,dx+Q\,dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA } \]

Here,

  • \(C\) is a positively oriented simple closed curve.
  • \(R\) is the region enclosed by \(C\).
Green's Theorem is widely used for evaluating line integrals and proving important results in Vector Calculus.

\[ \boxed{ \text{Line Integral} \Longleftrightarrow \text{Double Integral} } \]

Hence, the correct answer is Option (A).

STOKES' THEOREM • ENTRANCE LEVEL

Question 79

Stokes' Theorem relates

(A) A line integral and a surface integral
(B) A double integral and a triple integral
(C) A line integral and a volume integral
(D) A surface integral and a volume integral


Correct Answer

(A) A line integral and a surface integral


Detailed Explanation

Stokes' Theorem converts the circulation of a vector field around a closed curve into the flux of its curl across the surface bounded by that curve.

\[ \boxed{ \oint_C\vec F\cdot d\vec r = \iint_S (\nabla\times\vec F)\cdot\hat n\,dS } \]

Where,

  • \(C\) is the positively oriented boundary curve.
  • \(S\) is any smooth surface bounded by \(C\).
  • \(\hat n\) is the unit normal vector to the surface.
Stokes' Theorem is one of the fundamental integral theorems of Vector Calculus and is frequently asked in postgraduate entrance examinations.

\[ \boxed{ \text{Line Integral} \Longleftrightarrow \text{Surface Integral} } \]

Hence, the correct answer is Option (A).

LAPLACE TRANSFORM • ENTRANCE LEVEL

Question 80

The Laplace Transform of the constant function \[ 1 \] is

(A) \[ \frac1s \]
(B) \[ \frac1{s^2} \]
(C) \[ s \]
(D) \[ 1 \]


Correct Answer

(A) \[ \frac1s \]


Detailed Explanation

By definition,

\[ \boxed{ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt } \]

For \[ f(t)=1, \] we have \[ \begin{aligned} \mathcal{L}\{1\} &= \int_0^\infty e^{-st}\,dt\\[2mm] &= \left[ -\frac{e^{-st}}{s} \right]_0^\infty\\[2mm] &= \frac1s, \qquad s>0. \end{aligned} \] This is one of the most frequently used standard Laplace transforms.

\[ \boxed{ \mathcal{L}\{1\} = \frac1s } \]

Hence, the correct answer is Option (A).

ABSTRACT ALGEBRA • ENTRANCE LEVEL

Question 81

The identity element of the group \[ (\mathbb{Z},+) \] is

(A) \(0\)
(B) \(1\)
(C) \(-1\)
(D) Does not exist


Correct Answer

(A) \(0\)


Detailed Explanation

In any group, the identity element is an element that leaves every group element unchanged under the group operation.

\[ \boxed{ a+e=e+a=a, \qquad \forall a\in\mathbb{Z} } \]

For the additive group \[ (\mathbb{Z},+), \] the integer \[ 0 \] satisfies \[ a+0=0+a=a \] for every integer \(a\). Hence, \[ 0 \] is the identity element. The inverse of any integer \(a\) in this group is \[ -a. \]

\[ \boxed{ e=0 } \]

Hence, the correct answer is Option (A).

ABSTRACT ALGEBRA • ENTRANCE LEVEL

Question 82

Which one of the following sets forms a group under ordinary multiplication?

(A) \(\mathbb{N}\)
(B) \(\mathbb{Z}\)
(C) \(\mathbb{Q}\setminus\{0\}\)
(D) \(\mathbb{R}\)


Correct Answer

(C) \(\mathbb{Q}\setminus\{0\}\)


Detailed Explanation

A set forms a group under multiplication if it satisfies the group axioms:

  • Closure
  • Associativity
  • Identity element
  • Inverse for every element

\[ \boxed{ (\mathbb{Q}\setminus\{0\},\times) \text{ is a group} } \]

The multiplicative identity is \[ 1, \] and every non-zero rational number \[ a \] has the inverse \[ \frac1a. \] The remaining options fail because:

  • \(\mathbb N\): inverses are not always natural numbers.
  • \(\mathbb Z\): inverses such as \(1/2\) are not integers.
  • \(\mathbb R\): contains \(0\), which has no multiplicative inverse.
Therefore, \[ (\mathbb{Q}\setminus\{0\},\times) \] is the correct choice.

\[ \boxed{ (\mathbb{Q}\setminus\{0\},\times) } \]

Hence, the correct answer is Option (C).

METRIC SPACES • ENTRANCE LEVEL

Question 83

In a metric space \((X,d)\), the distance between a point and itself is

(A) \(1\)
(B) \(-1\)
(C) \(0\)
(D) Depends on the metric


Correct Answer

(C) \(0\)


Detailed Explanation

One of the axioms of a metric is the identity property.

\[ \boxed{ d(x,y)=0 \iff x=y } \]

Therefore, when the two points are identical, \[ x=y, \] we obtain \[ d(x,x)=0. \] This property holds for every metric space, regardless of the specific distance function being used. It is one of the four defining axioms of a metric.

\[ \boxed{ d(x,x)=0 } \]

Hence, the correct answer is Option (C).

ABSTRACT ALGEBRA • ENTRANCE LEVEL

Question 84

The order of the cyclic group \[ \mathbb{Z}_{12} \] is

(A) \(6\)
(B) \(10\)
(C) \(12\)
(D) Infinite


Correct Answer

(C) \(12\)


Detailed Explanation

The cyclic group \[ \mathbb{Z}_n \] consists of the residue classes \[ \{0,1,2,\ldots,n-1\} \] under addition modulo \(n\).

\[ \boxed{ |\mathbb{Z}_n|=n } \]

Since \[ n=12, \] the group contains exactly \[ 12 \] distinct elements. Therefore, \[ |\mathbb{Z}_{12}|=12. \] Every cyclic group of finite order has exactly as many elements as its modulus.

\[ \boxed{ |\mathbb{Z}_{12}|=12 } \]

Hence, the correct answer is Option (C).

ABSTRACT ALGEBRA • ENTRANCE LEVEL

Question 85

A subgroup \(H\) of a group \(G\) must always contain

(A) Every element of \(G\)
(B) Only generators of \(G\)
(C) The identity element of \(G\)
(D) Exactly two elements


Correct Answer

(C) The identity element of \(G\)


Detailed Explanation

A non-empty subset \(H\) of a group \(G\) is called a subgroup if it satisfies the subgroup criteria. Every subgroup must contain the identity element of the parent group.

\[ \boxed{ e\in H } \]

Indeed, if \[ a\in H, \] then \[ a^{-1}\in H. \] Hence, \[ aa^{-1}=e\in H. \] Therefore, every subgroup necessarily contains the identity element. This is one of the most fundamental results in Group Theory and appears frequently in postgraduate entrance examinations.

\[ \boxed{ \text{Every subgroup contains the identity element} } \]

Hence, the correct answer is Option (C).

REAL ANALYSIS • ENTRANCE LEVEL

Question 86

Evaluate \[ \lim_{x\to0}\frac{\sin x}{x}. \]

(A) \(0\)
(B) \(1\)
(C) \(\infty\)
(D) Does not exist


Correct Answer

(B) \(1\)


Detailed Explanation

This is one of the most fundamental limits in Calculus and Real Analysis.

\[ \boxed{ \lim_{x\to0}\frac{\sin x}{x}=1 } \]

The result can be established using the Squeeze Theorem: \[ \cos x \le \frac{\sin x}{x} \le 1, \qquad 0

\[ \boxed{ \lim_{x\to0}\frac{\sin x}{x}=1 } \]

Hence, the correct answer is Option (B).

PROBABILITY THEORY • ENTRANCE LEVEL

Question 87

If two events \(A\) and \(B\) are independent, then \[ P(A\cap B) \] is equal to

(A) \[ P(A)+P(B) \]
(B) \[ P(A)-P(B) \]
(C) \[ P(A)\,P(B) \]
(D) \[ \frac{P(A)}{P(B)} \]


Correct Answer

(C) \[ P(A)\,P(B) \]


Detailed Explanation

Two events are called independent if the occurrence of one event does not affect the probability of the other.

\[ \boxed{ P(A\cap B) = P(A)\,P(B) } \]

For dependent events, \[ P(A\cap B) = P(A)\,P(B|A). \] However, for independent events, \[ P(B|A)=P(B). \] Therefore, \[ P(A\cap B) = P(A)\,P(B). \] This is one of the most frequently tested formulas in probability theory.

\[ \boxed{ P(A\cap B) = P(A)\,P(B) } \]

Hence, the correct answer is Option (C).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 88

If \[ A= \begin{bmatrix} 2 & 0\\ 0 & 5 \end{bmatrix}, \] then the eigenvalues of \(A\) are

(A) \(2,\;5\)
(B) \(7,\;10\)
(C) \(2,\;-5\)
(D) \(10,\;25\)


Correct Answer

(A) \(2,\;5\)


Detailed Explanation

The eigenvalues of a diagonal matrix are simply its diagonal entries.

\[ \boxed{ \text{If } A=\operatorname{diag}(a_1,a_2,\ldots,a_n), \text{ then } \lambda_i=a_i. } \]

Since \[ A= \begin{bmatrix} 2 & 0\\ 0 & 5 \end{bmatrix}, \] its diagonal entries are \[ 2 \quad\text{and}\quad 5. \] Hence, the eigenvalues are \[ 2,\;5. \] Also, \[ \operatorname{tr}(A)=2+5=7, \] which equals the sum of the eigenvalues, and \[ |A|=2\times5=10, \] which equals their product.

\[ \boxed{ \lambda_1=2,\qquad \lambda_2=5 } \]

Hence, the correct answer is Option (A).

COMPLEX ANALYSIS • ENTRANCE LEVEL

Question 89

If \[ z=1+i, \] then the complex conjugate of \(z\) is

(A) \(1+i\)
(B) \(-1+i\)
(C) \(1-i\)
(D) \(-1-i\)


Correct Answer

(C) \(1-i\)


Detailed Explanation

For any complex number \[ z=a+bi, \] its complex conjugate is obtained by changing the sign of the imaginary part.

\[ \boxed{ \overline{a+bi}=a-bi } \]

Here, \[ z=1+i. \] Therefore, \[ \overline{z}=1-i. \] Also, \[ z\overline{z} = (1+i)(1-i) = 1-i^2 = 2, \] which is equal to \[ |z|^2. \] This property is frequently used in simplifying complex expressions.

\[ \boxed{ \overline{1+i}=1-i } \]

Hence, the correct answer is Option (C).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 90

Let \[ A= \begin{bmatrix} 2 & 1\\ 0 & 3 \end{bmatrix}. \] The determinant of the matrix \(A\) is

(A) \(5\)
(B) \(6\)
(C) \(3\)
(D) \(1\)


Correct Answer

(B) \(6\)


Detailed Explanation

The determinant of a \(2\times2\) matrix is calculated using

\[ \boxed{ \begin{vmatrix} a & b\\ c & d \end{vmatrix} =ad-bc } \]

For \[ A= \begin{bmatrix} 2 & 1\\ 0 & 3 \end{bmatrix}, \] we have \[ \begin{aligned} |A| &=(2)(3)-(1)(0)\\[2mm] &=6. \end{aligned} \] Since \[ |A|\neq0, \] the matrix is invertible.

\[ \boxed{ |A|=6 } \]

Hence, the correct answer is Option (B).

DIFFERENTIAL EQUATIONS • ENTRANCE LEVEL

Question 91

The differential equation \[ \frac{dy}{dx}=3y \] has the general solution

(A) \[ y=Ce^{3x} \]
(B) \[ y=C+3x \]
(C) \[ y=3Ce^x \]
(D) \[ y=Cx^3 \]


Correct Answer

(A) \[ y=Ce^{3x} \]


Detailed Explanation

The given equation is a first-order separable differential equation. Separate the variables: \[ \frac{dy}{y}=3\,dx. \] Integrating both sides, \[ \int\frac1y\,dy = \int3\,dx. \]

\[ \boxed{ \ln|y| = 3x+C } \]

Exponentiating both sides, \[ |y| = e^{3x+C} = e^C e^{3x}. \] Let \[ e^C=C_1, \] where \(C_1\) is an arbitrary non-zero constant. Thus, the general solution is \[ y=Ce^{3x}. \] This is the standard solution of the differential equation \[ \frac{dy}{dx}=ky, \] namely, \[ y=Ce^{kx}. \]

\[ \boxed{ y=Ce^{3x} } \]

Hence, the correct answer is Option (A).

REAL ANALYSIS • ENTRANCE LEVEL

Question 92

Evaluate \[ \lim_{n\to\infty}\frac{2n+1}{5n-3}. \]

(A) \[ \frac25 \]
(B) \[ \frac52 \]
(C) \[ 0 \]
(D) \[ 1 \]


Correct Answer

(A) \[ \frac25 \]


Detailed Explanation

Both the numerator and denominator are first-degree polynomials in \(n\). Divide the numerator and denominator by \(n\).

\[ \boxed{ \lim_{n\to\infty} \frac{2n+1}{5n-3} = \lim_{n\to\infty} \frac{2+\frac1n}{5-\frac3n} } \]

Since \[ \frac1n\to0 \quad\text{and}\quad \frac3n\to0, \] we obtain \[ \begin{aligned} \lim_{n\to\infty} \frac{2n+1}{5n-3} &= \frac25. \end{aligned} \] A useful rule is that when the degrees of the numerator and denominator are equal, the limit equals the ratio of their leading coefficients.

\[ \boxed{ \lim_{n\to\infty} \frac{2n+1}{5n-3} = \frac25 } \]

Hence, the correct answer is Option (A).

PROBABILITY THEORY • ENTRANCE LEVEL

Question 93

A fair coin is tossed once. The probability of getting a head is

(A) \[ 0 \]
(B) \[ \frac14 \]
(C) \[ \frac12 \]
(D) \[ 1 \]


Correct Answer

(C) \[ \frac12 \]


Detailed Explanation

A fair coin has two equally likely outcomes.

\[ \boxed{ S=\{H,T\} } \]

The total number of possible outcomes is \[ 2. \] The favorable outcome for obtaining a head is \[ \{H\}, \] which contains one outcome. Therefore, \[ \begin{aligned} P(\text{Head}) &= \frac{\text{Number of favorable outcomes}} {\text{Total number of outcomes}}\\[2mm] &= \frac12. \end{aligned} \] This is one of the basic applications of classical probability.

\[ \boxed{ P(\text{Head})=\frac12 } \]

Hence, the correct answer is Option (C).

COMPLEX ANALYSIS • ENTRANCE LEVEL

Question 94

If \[ z=i, \] then the value of \[ i^{\,4} \] is

(A) \(i\)
(B) \(-1\)
(C) \(1\)
(D) \(-i\)


Correct Answer

(C) \(1\)


Detailed Explanation

The imaginary unit satisfies the fundamental relation

\[ \boxed{ i^2=-1 } \]

Therefore, \[ \begin{aligned} i^4 &=(i^2)^2\\[2mm] &=(-1)^2\\[2mm] &=1. \end{aligned} \] The powers of \(i\) repeat after every four powers: \[ i,\; -1,\; -i,\; 1,\; i,\ldots \] Hence, \[ i^4=1. \]

\[ \boxed{ i^4=1 } \]

Hence, the correct answer is Option (C).

DIFFERENTIAL CALCULUS • ENTRANCE LEVEL

Question 95

If \[ f(x)=\ln x, \] then \[ f'(x) \] is equal to

(A) \[ x \]
(B) \[ \frac1x \]
(C) \[ \ln x \]
(D) \[ \frac1{x^2} \]


Correct Answer

(B) \[ \frac1x \]


Detailed Explanation

The natural logarithmic function has one of the most important standard derivatives in calculus.

\[ \boxed{ \frac{d}{dx}(\ln x)=\frac1x, \qquad x>0 } \]

The derivative exists only for \[ x>0, \] because the natural logarithm is defined only for positive real numbers. This formula is frequently used in logarithmic differentiation, integration, and solving differential equations.

\[ \boxed{ \frac{d}{dx}(\ln x)=\frac1x } \]

Hence, the correct answer is Option (B).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 96

If \[ A= \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}, \] then the trace of the matrix \(A\) is

(A) \(4\)
(B) \(5\)
(C) \(6\)
(D) \(10\)


Correct Answer

(B) \(5\)


Detailed Explanation

The trace of a square matrix is the sum of its principal diagonal elements.

\[ \boxed{ \operatorname{tr}(A) = \sum_{i=1}^{n}a_{ii} } \]

For \[ A= \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}, \] the principal diagonal entries are \[ 1 \quad\text{and}\quad 4. \] Hence, \[ \operatorname{tr}(A)=1+4=5. \] The trace is also equal to the sum of the eigenvalues of the matrix.

\[ \boxed{ \operatorname{tr}(A)=5 } \]

Hence, the correct answer is Option (B).

INTEGRAL CALCULUS • ENTRANCE LEVEL

Question 97

Evaluate \[ \int_0^1 x^2\,dx. \]

(A) \[ \frac13 \]
(B) \[ \frac12 \]
(C) \[ 1 \]
(D) \[ \frac14 \]


Correct Answer

(A) \[ \frac13 \]


Detailed Explanation

Use the Power Rule for integration.

\[ \boxed{ \int x^n\,dx = \frac{x^{n+1}}{n+1}+C, \qquad n\neq-1 } \]

Therefore, \[ \begin{aligned} \int_0^1x^2\,dx &= \left[ \frac{x^3}{3} \right]_0^1\\[2mm] &= \frac13-0\\[2mm] &= \frac13. \end{aligned} \] This is a standard definite integral frequently used in calculus and applications.

\[ \boxed{ \int_0^1x^2\,dx=\frac13 } \]

Hence, the correct answer is Option (A).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 98

If \[ A= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \] then \(A^5\) is equal to

(A) \[ \begin{bmatrix} 5 & 0\\ 0 & 5 \end{bmatrix} \]
(B) \[ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \]
(C) \[ \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \]
(D) \[ \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} \]


Correct Answer

(B) \[ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \]


Detailed Explanation

The given matrix is the identity matrix.

\[ \boxed{ I^n=I, \qquad n\ge1 } \]

Since \[ A=I, \] we have \[ A^2=I,\qquad A^3=I,\qquad \ldots,\qquad A^5=I. \] The identity matrix behaves like the number \(1\) under matrix multiplication.

\[ \boxed{ A^5=I } \]

Hence, the correct answer is Option (B).

DIFFERENTIAL CALCULUS • ENTRANCE LEVEL

Question 99

If \[ y=x^5, \] then \[ \frac{dy}{dx} \] is equal to

(A) \[ 5x^4 \]
(B) \[ x^4 \]
(C) \[ 5x \]
(D) \[ x^5 \]


Correct Answer

(A) \[ 5x^4 \]


Detailed Explanation

Use the Power Rule of differentiation.

\[ \boxed{ \frac{d}{dx}(x^n)=nx^{\,n-1} } \]

Applying the formula with \[ n=5, \] we get \[ \begin{aligned} \frac{d}{dx}(x^5) &=5x^{5-1}\\[2mm] &=5x^4. \end{aligned} \] This is one of the most frequently used differentiation formulas in calculus.

\[ \boxed{ \frac{d}{dx}(x^5)=5x^4 } \]

Hence, the correct answer is Option (A).

LINEAR ALGEBRA • ENTRANCE LEVEL

Question 100

Let \[ A= \begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}. \] The characteristic equation of the matrix \(A\) is

(A) \[ (\lambda-2)^2=0 \]
(B) \[ (\lambda+2)^2=0 \]
(C) \[ \lambda^2-2=0 \]
(D) \[ \lambda^2+4=0 \]


Correct Answer

(A) \[ (\lambda-2)^2=0 \]


Detailed Explanation

The characteristic equation of a square matrix is obtained by solving

\[ \boxed{ \det(A-\lambda I)=0 } \]

For \[ A= \begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix}, \] we have \[ A-\lambda I= \begin{bmatrix} 2-\lambda & 0\\ 0 & 2-\lambda \end{bmatrix}. \] Therefore, \[ \begin{aligned} \det(A-\lambda I) &=(2-\lambda)(2-\lambda)\\[2mm] &=(2-\lambda)^2. \end{aligned} \] Equating the determinant to zero, \[ (2-\lambda)^2=0, \] which is equivalent to \[ (\lambda-2)^2=0. \] Thus, the matrix has a repeated eigenvalue \[ \lambda=2. \]

\[ \boxed{ (\lambda-2)^2=0 } \]

Hence, the correct answer is Option (A).


🎉 Set–1 Completed Successfully

This completes Set–1 (Questions 1–100) for the M.Sc./M.A. Mathematics Entrance Examination. The paper covers a broad range of B.Sc. Mathematics topics commonly asked in entrance examinations of universities across India, including:

  • Linear Algebra
  • Real Analysis
  • Differential Calculus
  • Integral Calculus
  • Differential Equations
  • Complex Analysis
  • Vector Calculus
  • Laplace Transform
  • Fourier Series
  • Abstract Algebra
  • Metric Spaces
  • Topology
  • Probability Theory
  • Graph Theory
  • Boolean Algebra
  • Beta & Gamma Functions
  • Green's Theorem
  • Stokes' Theorem

The next paper (Set–2: Questions 101–200) should ideally contain a significantly higher proportion of advanced, university-level entrance questions, including more conceptual and numerical problems from: Cayley–Hamilton Theorem, Rank–Nullity Theorem, Jordan Canonical Form, Eigenvalue Applications, Riemann Integration, Sequences & Series, Uniform Convergence, Group Homomorphism, Ring Theory, Field Theory, Metric Spaces, Topological Spaces, Multiple Integrals, Gamma/Beta Applications, PDEs, Fourier Transform, Probability Distributions, and advanced PYQ-style problems. This will ensure that the complete **200-question bank** closely matches the difficulty level of actual **M.Sc./M.A. Mathematics Entrance Examinations** conducted by universities such as DDU, BHU, CUET PG, AMU, JNU, DU, HPU, Allahabad University, Lucknow University, and other state universities.

नोट: यह हमारी MA/MSc Mathematics Entrance Exam सीरीज का पहला भाग (Part 1) है। हम जल्द ही Part 2 लेकर आएंगे जिसमें Abstract Algebra (अमूर्त बीजगणित) और Differential Equations (अवकल समीकरण) के महत्वपूर्ण प्रश्न शामिल होंगे। तब तक इन प्रश्नों का अच्छे से रिवीजन करते रहें!

हमसे जुड़ें (Join Our Community)

यूनिवर्सिटी की हर एक छोटी-बड़ी अपडेट, परीक्षा शेड्यूल, स्टडी मटेरियल और अगली क्विज़ सीरीज़ तुरंत पाने के लिए हमारे सोशल मीडिया हैंडल्स को ज़रूर फॉलो करें:

Post a Comment

Thanks for your comments.

Previous Post Next Post