Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The real number $$k$$ for which the equation, $$2{x^3} + 3x + k = 0$$ has two distinct real roots in $$\left[ {0,\,1} \right]$$

A

lies between 1 and 2

B

lies between 2 and 3

C

lies between $$ - 1$$ and 0

D

does not exist.

$$f\left( x \right) = 2{x^3} + 3x + k$$

$$f'\left( x \right) = 6{x^2} + 3 > 0$$

$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)

$$ \Rightarrow f\left( x \right)$$ is strictly increasing function

$$ \Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.

$$f'\left( x \right) = 6{x^2} + 3 > 0$$

$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)

$$ \Rightarrow f\left( x \right)$$ is strictly increasing function

$$ \Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.

2

MCQ (Single Correct Answer)

The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has:

A

infinite number of real roots

B

no real roots

C

exactly one real root

D

exactly four real roots

Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$

Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,

we get $${t^2} - 4t - 1 = 0$$

$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $$ and

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$

and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected

Hence given equation has no solution.

$$\therefore$$ The equation has no real roots.

Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,

we get $${t^2} - 4t - 1 = 0$$

$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $$ and

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$

and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected

Hence given equation has no solution.

$$\therefore$$ The equation has no real roots.

3

MCQ (Single Correct Answer)

If $$\alpha $$ and $$\beta $$ are the roots of the equation $${x^2} - x + 1 = 0,$$ then $${\alpha ^{2009}} + {\beta ^{2009}} = $$

A

$$\, - 1$$

B

$$\, 1$$

C

$$\, 2$$

D

$$\, - 2$$

$${x^2} - x + 1 = 0$$

$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$

$$x = {{1 \pm \sqrt 3 i} \over 2}$$

$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$

$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$

$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$

$$ = - {\omega ^2} - \omega = 1$$

$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$

$$x = {{1 \pm \sqrt 3 i} \over 2}$$

$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$

$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$

$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$

$$ = - {\omega ^2} - \omega = 1$$

4

MCQ (Single Correct Answer)

If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is :

A

less than $$4ab$$

B

greater than $$-4ab$$

C

less than $$-4ab$$

D

greater than $$4ab$$

Given that roots of the equation

$$b{x^2} + cx + a = 0$$ are imaginary

$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$

$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$

As $$x$$ is real, $$D \ge 0$$

$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$

$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$

$$ \Rightarrow {c^2} + y \ge 0$$

$$ \Rightarrow y \ge - {c^2}$$

But from eqn. $$(i),$$ $${c^2} < 4ab$$

or $$ - {c^2} > - 4ab$$

$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$

$$y > - 4ab$$

$$b{x^2} + cx + a = 0$$ are imaginary

$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$

$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$

As $$x$$ is real, $$D \ge 0$$

$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$

$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$

$$ \Rightarrow {c^2} + y \ge 0$$

$$ \Rightarrow y \ge - {c^2}$$

But from eqn. $$(i),$$ $${c^2} < 4ab$$

or $$ - {c^2} > - 4ab$$

$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$

$$y > - 4ab$$

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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

Trigonometric Functions & Equations

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