Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Let $$f:R \to R$$ be a continuous function defined by
$$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$$

**Statement - 1 :** $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$.

**Statement - 2 :** $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$

A

Statement - 1 is true, Statement -2 is true; Statement - 2 is **not** a correct explanation for Statement - 1.

B

Statement - 1 is true, Statement - 2 is false.

C

Statement - 1 is false, Statement - 2 is true.

D

Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1.

$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$

$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$

$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$

$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$

maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$

$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$

Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$

$$f\left( c \right) = {1 \over 3}$$

$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$

$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$

$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$

maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$

$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$

Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$

$$f\left( c \right) = {1 \over 3}$$

2

MCQ (Single Correct Answer)

The equation of the tangent to the curve $$y = x + {4 \over {{x^2}}}$$, that

is parallel to the $$x$$-axis, is

is parallel to the $$x$$-axis, is

A

$$y=1$$

B

$$y=2$$

C

$$y=3$$

D

$$y=0$$

Since tangent is parallel to $$x$$-axis,

$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$

Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$

$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$

Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$

3

MCQ (Single Correct Answer)

Let $$f:R \to R$$ be defined by
$$$f\left( x \right) = \left\{ {\matrix{
{k - 2x,\,\,if} & {x \le - 1} \cr
{2x + 3,\,\,if} & {x > - 1} \cr
} } \right.$$$

If $$f$$has a local minimum at $$x=-1$$, then a possible value of $$k$$ is

A

$$0$$

B

$$ - {1 \over 2}$$

C

$$-1$$

D

$$1$$

$$f\left( x \right) = \left\{ {\matrix{
{k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr
{2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr
} } \right.$$

This is true where $$k=-1$$

This is true where $$k=-1$$

4

MCQ (Single Correct Answer)

Given $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ such that $$x=0$$ is the only

real root of $$P'\,\left( x \right) = 0.$$ If $$P\left( { - 1} \right) < P\left( 1 \right),$$ then in the interval $$\left[ { - 1,1} \right]:$$

real root of $$P'\,\left( x \right) = 0.$$ If $$P\left( { - 1} \right) < P\left( 1 \right),$$ then in the interval $$\left[ { - 1,1} \right]:$$

A

$$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$

B

$$P(-1)$$ is the minimum but $$P(1)$$ is not the maximum of $$P$$

C

Neither $$P(-1)$$ is the minimum nor $$P(1)$$ is the maximum of $$P$$

D

$$P(-1)$$ is the minimum and $$P(1)$$ is the maximum of $$P$$

We have $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$

$$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$

But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$

$$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$

As given that $$P\left( { - 1} \right) < P\left( a \right)$$

$$ \Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$$

Now $$P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$$

As $$P'\left( x \right) = 0,$$ there is only one solution $$x = 0,$$

therefore $$4{x^2} + 3ax + 2b = 0$$ should not have any real roots i.e. $$D < 0$$

$$ \Rightarrow 9{a^2} - 32b < 0$$

$$ \Rightarrow b > {{9{a^2}} \over {32}} > 0$$

Hence $$a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$$

$$\forall x > 0$$

$$\therefore$$ $$P(x)$$ is an increasing function on $$\left( {0,1} \right)$$

$$\therefore$$ $$P\left( 0 \right) < P\left( a \right)$$

Similarly we can prove $$P\left( x \right)$$ is decreasing on $$\left( { - 1,0} \right)$$

$$\therefore$$ $$P\left( { - 1} \right) > P\left( 0 \right)$$

So we can conclude that

Max $$P\left( x \right) = P\left( 1 \right)$$ and Min $$P\left( x \right) = P\left( 0 \right)$$

$$ \Rightarrow P\left( { - 1} \right)$$ is not minimum but $$P\left( 1 \right)$$ is the maximum of $$P.$$

$$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$

But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$

$$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$

As given that $$P\left( { - 1} \right) < P\left( a \right)$$

$$ \Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$$

Now $$P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$$

As $$P'\left( x \right) = 0,$$ there is only one solution $$x = 0,$$

therefore $$4{x^2} + 3ax + 2b = 0$$ should not have any real roots i.e. $$D < 0$$

$$ \Rightarrow 9{a^2} - 32b < 0$$

$$ \Rightarrow b > {{9{a^2}} \over {32}} > 0$$

Hence $$a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$$

$$\forall x > 0$$

$$\therefore$$ $$P(x)$$ is an increasing function on $$\left( {0,1} \right)$$

$$\therefore$$ $$P\left( 0 \right) < P\left( a \right)$$

Similarly we can prove $$P\left( x \right)$$ is decreasing on $$\left( { - 1,0} \right)$$

$$\therefore$$ $$P\left( { - 1} \right) > P\left( 0 \right)$$

So we can conclude that

Max $$P\left( x \right) = P\left( 1 \right)$$ and Min $$P\left( x \right) = P\left( 0 \right)$$

$$ \Rightarrow P\left( { - 1} \right)$$ is not minimum but $$P\left( 1 \right)$$ is the maximum of $$P.$$

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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