Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If$$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$ for $$x \ne 0,y \ne 0,$$ then $$D$$ is

A

divisible by $$x$$ but not $$y$$

B

divisible by $$y$$ but not $$x$$

C

divisible by neither $$x$$ nor $$y$$

D

divisible by both $$x$$ and $$y$$

Given, $$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$

Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$

and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$

$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$

Hence, $$D$$ is divisible by both $$x$$ and $$y$$

Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$

and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$

$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$

Hence, $$D$$ is divisible by both $$x$$ and $$y$$

2

MCQ (Single Correct Answer)

Let $$A = \left| {\matrix{
5 & {5\alpha } & \alpha \cr
0 & \alpha & {5\alpha } \cr
0 & 0 & 5 \cr
} } \right|.$$ If $$\,\,\left| {{A^2}} \right| = 25,$$ then $$\,\left| \alpha \right|$$ equals

A

$$1/5$$

B

$$5$$

C

$${5^2}$$

D

$$1$$

$$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$$

$$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$$

$$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$$

3

MCQ (Single Correct Answer)

Let $$A = \left( {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right)$$ and $$B = \left( {\matrix{
a & 0 \cr
0 & b \cr
} } \right),a,b \in N.$$ Then

A

there cannot exist any $$B$$ such that $$AB=BA$$

B

there exist more then one but finite number of $$B'$$s such that $$AB=BA$$

C

there exists exactly one $$B$$ such that $$AB=BA$$

D

there exist infinitely many $$B'$$s such that $$AB=BA$$

$$A = \left[ {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right]\,\,\,\,B = \left[ {\matrix{
a & 0 \cr
0 & b \cr
} } \right]$$

$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$

$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$

Hence, $$AB=BA$$ only when $$a=b$$

$$\therefore$$ There can be infinitely many $$B's$$

for which $$AB=BA$$

$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$

$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$

Hence, $$AB=BA$$ only when $$a=b$$

$$\therefore$$ There can be infinitely many $$B's$$

for which $$AB=BA$$

4

MCQ (Single Correct Answer)

If $$A$$ and $$B$$ are square matrices of size $$n\, \times \,n$$ such that

$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right),$$ then which of the following will be always true?

$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right),$$ then which of the following will be always true?

A

$$A=B$$

B

$$AB=BA$$

C

either of $$A$$ or $$B$$ is a zero matrix

D

either of $$A$$ or $$B$$ is identity matrix

$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$$

$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$

$$ \Rightarrow AB = BA$$

$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$

$$ \Rightarrow AB = BA$$

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

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Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations