Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $$a \in R$$ and the equation $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$ (where [$$x$$] denotes the greater integer $$ \le x$$) has no integral solution, then all possible values of a lie in the interval :

A

$$\left( { - 2, - 1} \right)$$

B

$$\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)$$

C

$$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$

D

$$\left( {1,2} \right)$$

Given, $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$

As we know, $$\left[ x \right] + \left\{ x \right\} = x$$

where $$\left[ x \right]$$ is integral part and $$\left\{ x \right\}$$ is fractional part.

$$\therefore$$$$\left\{ x \right\} = x - \left[ x \right]$$

Now put $$\left\{ x \right\}$$ inplace of $$x - \left[ x \right]$$ in the equation.

The new equation is $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$$

= $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$

As $$\left\{ x \right\}$$ is fractional part so it is lies between 0 to 1($$0 \le \left\{ x \right\} < 1$$).

By considering positive sign, we get

$$0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$ \Rightarrow $$$$0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6$$

$$ \Rightarrow $$$$2 \ge + \sqrt {4 + 12{a^2}} > - 4$$

$$\because$$$$ + \sqrt {4 + 12{a^2}} $$ is always positive which is greater than any negative no. So can ignore the inequality $$ + \sqrt {4 + 12{a^2}} > - 4$$

Consider this inequality,

$$2 \ge + \sqrt {4 + 12{a^2}} $$

$$ \Rightarrow $$ $$4 \ge 4 + 12{a^2}$$

$$ \Rightarrow $$ $$12{a^2} \le 0$$

$$ \Rightarrow $$ $${a^2} \le 0$$

$$ \Rightarrow $$ $${a^2} = 0$$

$$ \Rightarrow $$ $${a} = 0$$

If $$a$$ = 0 then $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$$ so $$\left\{ x \right\}$$ becomes 0 but question says $$\left\{ x \right\}$$ $$ \ne $$ 0.

So $$a$$ can't be 0.

Now by considering negative sign, we get

$$0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$ \Rightarrow $$$$0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6$$

$$ \Rightarrow $$$$2 \ge - \sqrt {4 + 12{a^2}} > - 4$$

As 2 is always greater than $${ - \sqrt {4 + 12{a^2}} }$$. Ignore this inequality.

Now consider this inequality,

$$ - \sqrt {4 + 12{a^2}} > - 4$$

$$ \Rightarrow $$ $$\sqrt {4 + 12{a^2}} < 4$$

$$ \Rightarrow $$ $$4 + 12{a^2} < 16$$

$$ \Rightarrow $$ $$12{a^2} < 12$$

$$ \Rightarrow $$ $${a^2} < 1$$

$$ \Rightarrow $$ $$\left( {{a^2} - 1} \right) < 0$$

$$ \Rightarrow $$ $$\left( {a + 1} \right)\left( {a - 1} \right) < 0$$

$$ \Rightarrow $$ $$ - 1 < a < 1$$

But earlier we found that $$a$$ $$ \ne $$ 0.

So, the range of $$a$$ is = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$

As we know, $$\left[ x \right] + \left\{ x \right\} = x$$

where $$\left[ x \right]$$ is integral part and $$\left\{ x \right\}$$ is fractional part.

$$\therefore$$$$\left\{ x \right\} = x - \left[ x \right]$$

Now put $$\left\{ x \right\}$$ inplace of $$x - \left[ x \right]$$ in the equation.

The new equation is $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$$

[**Note :** Question says this equation has no integral solution, it means $$\left\{ x \right\} \ne $$ 0. So, $$x$$ is not a integer.]

= $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$

As $$\left\{ x \right\}$$ is fractional part so it is lies between 0 to 1($$0 \le \left\{ x \right\} < 1$$).

By considering positive sign, we get

$$0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$ \Rightarrow $$$$0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6$$

$$ \Rightarrow $$$$2 \ge + \sqrt {4 + 12{a^2}} > - 4$$

$$\because$$$$ + \sqrt {4 + 12{a^2}} $$ is always positive which is greater than any negative no. So can ignore the inequality $$ + \sqrt {4 + 12{a^2}} > - 4$$

Consider this inequality,

$$2 \ge + \sqrt {4 + 12{a^2}} $$

$$ \Rightarrow $$ $$4 \ge 4 + 12{a^2}$$

$$ \Rightarrow $$ $$12{a^2} \le 0$$

$$ \Rightarrow $$ $${a^2} \le 0$$

$$ \Rightarrow $$ $${a^2} = 0$$

$$ \Rightarrow $$ $${a} = 0$$

If $$a$$ = 0 then $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$$ so $$\left\{ x \right\}$$ becomes 0 but question says $$\left\{ x \right\}$$ $$ \ne $$ 0.

So $$a$$ can't be 0.

Now by considering negative sign, we get

$$0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$ \Rightarrow $$$$0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6$$

$$ \Rightarrow $$$$2 \ge - \sqrt {4 + 12{a^2}} > - 4$$

As 2 is always greater than $${ - \sqrt {4 + 12{a^2}} }$$. Ignore this inequality.

Now consider this inequality,

$$ - \sqrt {4 + 12{a^2}} > - 4$$

$$ \Rightarrow $$ $$\sqrt {4 + 12{a^2}} < 4$$

$$ \Rightarrow $$ $$4 + 12{a^2} < 16$$

$$ \Rightarrow $$ $$12{a^2} < 12$$

$$ \Rightarrow $$ $${a^2} < 1$$

$$ \Rightarrow $$ $$\left( {{a^2} - 1} \right) < 0$$

$$ \Rightarrow $$ $$\left( {a + 1} \right)\left( {a - 1} \right) < 0$$

$$ \Rightarrow $$ $$ - 1 < a < 1$$

But earlier we found that $$a$$ $$ \ne $$ 0.

So, the range of $$a$$ is = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$

2

MCQ (Single Correct Answer)

If the equations $${x^2} + 2x + 3 = 0$$ and $$a{x^2} + bx + c = 0,$$ $$a,\,b,\,c\, \in \,R,$$ have a common root, then $$a\,:b\,:c\,$$ is

A

$$1:2:3$$

B

$$3:2:1$$

C

$$1:3:2$$

D

$$3:1:2$$

Given equations are

$$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$$

Roots of equation $$(i)$$ are imaginary roots.

According to the question $$(ii)$$ will also have both roots same as $$(i).$$

Thus $${a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)$$

$$ \Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda $$

Hence, required ratio is $$1:2:3$$

$$\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)$$

Roots of equation $$(i)$$ are imaginary roots.

According to the question $$(ii)$$ will also have both roots same as $$(i).$$

Thus $${a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)$$

$$ \Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda $$

Hence, required ratio is $$1:2:3$$

3

MCQ (Single Correct Answer)

The number of values of $$k$$, for which the system of equations : $$$\matrix{
{\left( {k + 1} \right)x + 8y = 4k} \cr
{kx + \left( {k + 3} \right)y = 3k - 1} \cr
} $$$

has no solution, is

has no solution, is

A

infinite

B

1

C

2

D

3

From the given system, we have

$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$

( as System has no solution)

$$ \Rightarrow {k^2} + 4k + 3 = 8k$$

$$ \Rightarrow k = 1,3$$

If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false

And if $$k = 3$$

Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$

Hence for only one value of $$k.$$ System has no solution.

$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$

( as System has no solution)

$$ \Rightarrow {k^2} + 4k + 3 = 8k$$

$$ \Rightarrow k = 1,3$$

If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false

And if $$k = 3$$

Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$

Hence for only one value of $$k.$$ System has no solution.

4

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.

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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