Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

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Reinforced Cement Concrete

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General Aptitude

1

Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$$-$$ 1 and it leaves remainder 6 when divided by x + 1; then :

A

p(2) = 11

B

p(2) = 19

C

p($$-$$ 2) = 19

D

p($$-$$ 2) = 11

Let, P(x) = ax^{2} + bx + c

As, P(0) = 1,

$$\therefore\,\,\,$$ a(0)^{2} + b(0) + c = 1

$$ \Rightarrow $$$$\,\,\,$$ c = 1

$$\therefore\,\,\,$$ P(x) = ax^{2} + bx + 1

If P(x) is divided by x $$-$$ 1, remainder = 4

$$ \Rightarrow $$$$\,\,\,$$ P$$\left( 1 \right) = 4$$

$$\therefore\,\,\,$$ a + b + 1 = 4 . . . . . (1)

If P(x) is divided by x + 1, remainder = 6

$$ \Rightarrow $$$$\,\,\,$$ P($$-$$ 1) = 6

$$\therefore\,\,\,$$ a $$-$$ b + 1 = 6 . . . .(2)

By solving (1) and (2) we get,

a = 4, and b = $$-$$1

$$\therefore\,\,\,$$ P(x) = 4x^{2} $$-$$ x + 1

P(2) = 4(2)^{2} $$-$$ 2 + 1 = 15

P($$-$$ 2) = 4 ($$-$$2)^{2} $$-$$ ($$-$$ 2) + 1 = 19

As, P(0) = 1,

$$\therefore\,\,\,$$ a(0)

$$ \Rightarrow $$$$\,\,\,$$ c = 1

$$\therefore\,\,\,$$ P(x) = ax

If P(x) is divided by x $$-$$ 1, remainder = 4

$$ \Rightarrow $$$$\,\,\,$$ P$$\left( 1 \right) = 4$$

$$\therefore\,\,\,$$ a + b + 1 = 4 . . . . . (1)

If P(x) is divided by x + 1, remainder = 6

$$ \Rightarrow $$$$\,\,\,$$ P($$-$$ 1) = 6

$$\therefore\,\,\,$$ a $$-$$ b + 1 = 6 . . . .(2)

By solving (1) and (2) we get,

a = 4, and b = $$-$$1

$$\therefore\,\,\,$$ P(x) = 4x

P(2) = 4(2)

P($$-$$ 2) = 4 ($$-$$2)

2

The sum of all the real values of x satisfying the equation

2^{(x$$-$$1)(x2 + 5x $$-$$ 50)} = 1 is :

2

A

16

B

14

C

$$-$$4

D

$$-$$ 5

We know, 2^{x} = 1 only when x = 0.

Similarly, 2^{(x$$-$$1)(x2 + 5x $$-$$ 50)} = 1 when

(x$$-$$1)(x^{2} + 5x $$-$$ 50) = 0

$$ \Rightarrow $$ (x - 1)(x + 10)(x - 5) = 0

$$ \therefore $$ x = 1, -10, 5

Sum of real values of x = 1 + (-10) + 5 = -4

Similarly, 2

(x$$-$$1)(x

$$ \Rightarrow $$ (x - 1)(x + 10)(x - 5) = 0

$$ \therefore $$ x = 1, -10, 5

Sum of real values of x = 1 + (-10) + 5 = -4

3

Let S = { $$x$$ $$ \in $$ R : $$x$$ $$ \ge $$ 0 and

$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$}. Then S

$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$}. Then S

A

contains exactly four elements

B

is an empty set

C

contains exactly one element

D

contains exactly two elements

Given,

$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

__Case 1__ :

When $$\sqrt x - 3 \ge 0,$$ then equation becomes

$$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

$$ \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0$$

$$ \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0$$

$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 0,4$$

but as $$\sqrt x - 3 \ge 0$$ or $$\sqrt x \ge 3$$ then $$\sqrt x \ne 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 4$$ value is possible.

__Case 2__ :

When $$\sqrt x - 3 < 0$$ or $$\sqrt x < 3.$$ There equation becomes

$$ - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

$$ \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0$$

$$ \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0$$

$$ \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0$$

$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0$$

$$ \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 2,6$$

as $$\sqrt x < 3$$ so $$\sqrt x \ne 6$$

$$\therefore\,\,\,$$ $$\sqrt x = 2$$ is possible.

So, total possible value of $$\sqrt x = 2,4$$

or for x possible values are 4, 16.

$$\therefore\,\,\,$$ Set S contains exactly two elements 4 and 16.

$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

When $$\sqrt x - 3 \ge 0,$$ then equation becomes

$$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

$$ \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0$$

$$ \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0$$

$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 0,4$$

but as $$\sqrt x - 3 \ge 0$$ or $$\sqrt x \ge 3$$ then $$\sqrt x \ne 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 4$$ value is possible.

When $$\sqrt x - 3 < 0$$ or $$\sqrt x < 3.$$ There equation becomes

$$ - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

$$ \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0$$

$$ \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0$$

$$ \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0$$

$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0$$

$$ \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 2,6$$

as $$\sqrt x < 3$$ so $$\sqrt x \ne 6$$

$$\therefore\,\,\,$$ $$\sqrt x = 2$$ is possible.

So, total possible value of $$\sqrt x = 2,4$$

or for x possible values are 4, 16.

$$\therefore\,\,\,$$ Set S contains exactly two elements 4 and 16.

4

If $$\lambda $$ $$ \in $$ **R** is such that the sum of the cubes of the roots of the equation,

x^{2} + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :

x

A

$$4\sqrt 2 $$

B

$$2\sqrt 5 $$

C

$$2\sqrt 7 $$

D

20

Let $$\alpha $$, $$\beta $$ are the roots of the equation,

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 2 and $$\alpha $$$$\beta $$ = 10 $$-$$ $$\lambda $$

$${\alpha ^3} + {\beta ^3}$$ = ($$\alpha $$ + $$\beta $$)^{3} $$-$$ 3$$\alpha $$$$\beta $$ ($$\alpha $$ + $$\beta $$)

= ($$\lambda $$ $$-$$ 2)^{3} $$-$$ 3(10 $$-$$ $$\lambda $$)($$\lambda $$ $$-$$ 2)

= $$\lambda ^3$$ $$-$$ 3$$\lambda ^2$$ $$-$$ 24$$\lambda $$ + 52

Let $$f(\lambda $$) = $$\lambda ^3$$ $$-$$ 3$$\lambda ^2$$ $$-$$ 24$$\lambda $$ + 52

$$ \therefore $$ $${{df(\lambda )} \over {d\lambda }}$$ = 3$$\lambda ^2$$ $$-$$ 6$$\lambda $$ $$-$$ 24

$$ \therefore $$ at maximum of minimum $${{df(\lambda )} \over {d\lambda }}$$ = 0

$$ \therefore $$ $$\lambda ^2$$ $$-$$ 2$$\lambda $$ $$-$$ 8 = 0

$$ \Rightarrow $$ ($$\lambda $$ + 2) ($$\lambda $$ $$-$$ 4) = 0

$$ \Rightarrow $$ $$\lambda $$ = $$-$$2, 4

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 2$$\lambda $$ $$-$$ 2

When $$\lambda $$ = $$-$$2

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = $$-$$ 6 < 0

$$ \therefore $$ at $$\lambda $$ = $$-$$2, f($$\lambda $$) has maximum value.

When $$\lambda $$ = 4

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 6 > 0

$$ \therefore $$ at $$\lambda $$ = 4, f($$\lambda $$) has minimum value.

$$ \therefore $$ When $$\lambda $$ = 4 equation is,

x^{2} $$-$$ 2x + 6 = 0

$$ \therefore $$ ($$\alpha $$ $$-$$ $$\beta $$)^{2} = ($$\alpha $$ + $$\beta $$)^{2} $$-$$ 4$$\alpha \beta$$

$$ \Rightarrow $$ x^{2} $$-$$ 4 $$ \times $$ 6

= $$-$$ 20

$$ \Rightarrow $$ ($$\alpha $$ $$-$$ $$\beta $$) = $$2\sqrt 5 i$$

$$ \Rightarrow $$ $$\left| {\alpha - \beta } \right|$$ = $$2\sqrt 5$$ (ans)

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 2 and $$\alpha $$$$\beta $$ = 10 $$-$$ $$\lambda $$

$${\alpha ^3} + {\beta ^3}$$ = ($$\alpha $$ + $$\beta $$)

= ($$\lambda $$ $$-$$ 2)

= $$\lambda ^3$$ $$-$$ 3$$\lambda ^2$$ $$-$$ 24$$\lambda $$ + 52

Let $$f(\lambda $$) = $$\lambda ^3$$ $$-$$ 3$$\lambda ^2$$ $$-$$ 24$$\lambda $$ + 52

$$ \therefore $$ $${{df(\lambda )} \over {d\lambda }}$$ = 3$$\lambda ^2$$ $$-$$ 6$$\lambda $$ $$-$$ 24

$$ \therefore $$ at maximum of minimum $${{df(\lambda )} \over {d\lambda }}$$ = 0

$$ \therefore $$ $$\lambda ^2$$ $$-$$ 2$$\lambda $$ $$-$$ 8 = 0

$$ \Rightarrow $$ ($$\lambda $$ + 2) ($$\lambda $$ $$-$$ 4) = 0

$$ \Rightarrow $$ $$\lambda $$ = $$-$$2, 4

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 2$$\lambda $$ $$-$$ 2

When $$\lambda $$ = $$-$$2

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = $$-$$ 6 < 0

$$ \therefore $$ at $$\lambda $$ = $$-$$2, f($$\lambda $$) has maximum value.

When $$\lambda $$ = 4

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 6 > 0

$$ \therefore $$ at $$\lambda $$ = 4, f($$\lambda $$) has minimum value.

$$ \therefore $$ When $$\lambda $$ = 4 equation is,

x

$$ \therefore $$ ($$\alpha $$ $$-$$ $$\beta $$)

$$ \Rightarrow $$ x

= $$-$$ 20

$$ \Rightarrow $$ ($$\alpha $$ $$-$$ $$\beta $$) = $$2\sqrt 5 i$$

$$ \Rightarrow $$ $$\left| {\alpha - \beta } \right|$$ = $$2\sqrt 5$$ (ans)

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