Joint Entrance Examination

Graduate Aptitude Test in Engineering

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General Aptitude

1

If

$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$$

then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to :

$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right| = 0} \right\},$$

then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to :

A

$$4 + 2\sqrt 3 $$

B

$$ - 2 + \sqrt 3 $$

C

$$ - 2 - \sqrt 3 $$

D

$$-\,\,4 - 2\sqrt 3 $$

Given,

$$\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$$ = 0

$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos^{2}x) $$-$$ sinx(sin^{2}x) = 0

$$ \Rightarrow $$$$\,\,\,$$ cos^{3}x $$-$$ sin^{3} x = 0

$$ \Rightarrow $$$$\,\,\,$$ tan^{3}x = 1

$$ \Rightarrow $$$$\,\,\,$$ tanx = 1

$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$

= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$

= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$

= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$

= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$

= $$-$$ 2 $$-$$ $$\sqrt 3 $$

$$\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$$ = 0

$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos

$$ \Rightarrow $$$$\,\,\,$$ cos

$$ \Rightarrow $$$$\,\,\,$$ tan

$$ \Rightarrow $$$$\,\,\,$$ tanx = 1

$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$

= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$

= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$

= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$

= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$

= $$-$$ 2 $$-$$ $$\sqrt 3 $$

2

For two 3 × 3 matrices A and B, let A + B = 2B^{T} and 3A + 2B = I_{3}, where B^{T} is
the transpose of B and I_{3} is 3 × 3 identity matrix. Then :

A

5A + 10B = 2I_{3}

B

10A + 5B = 3I_{3}

C

B + 2A = I_{3}

D

3A + 6B = 2I_{3}

Given, A + B = 2B^{T} .......(1)

$$ \Rightarrow $$ (A + B)^{T} = (2B^{T})^{T}

$$ \Rightarrow $$ A^{T} + B^{T} = 2B

$$ \Rightarrow $$ B = $${{{A^T} + {B^T}} \over 2}$$

Now put this in equation (1)

So, A + $${{{A^T} + {B^T}} \over 2}$$ = 2B^{T}

$$ \Rightarrow $$2A + A^{T} = 3B^{T}

$$ \Rightarrow $$ A = $${{3{B^T} - {A^T}} \over 2}$$

Also, 3A + 2B = I_{3} .......(2)

$$ \Rightarrow $$ $$3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$$ = I_{3}

$$ \Rightarrow $$ 11B^{T} - A^{T} = 2I_{3}

$$ \Rightarrow $$ (11B^{T} - A^{T})^{T} = (2I_{3})^{T}

$$ \Rightarrow $$ 11B - A = 2I_{3} ........(3)

Multiply (3) by 3 and then adding (2) and (3) we get,

35B = 7I_{3}

$$ \Rightarrow $$ B = $${{{I_3}} \over 5}$$

From (3), 11$${{{I_3}} \over 5}$$ - A = 2I_{3}

$$ \Rightarrow $$ A = $${{{I_3}} \over 5}$$

$$ \therefore $$ 5A = 5B = I_{3}

$$ \Rightarrow $$ 10A + 5B = 3I_{3}

$$ \Rightarrow $$ (A + B)

$$ \Rightarrow $$ A

$$ \Rightarrow $$ B = $${{{A^T} + {B^T}} \over 2}$$

Now put this in equation (1)

So, A + $${{{A^T} + {B^T}} \over 2}$$ = 2B

$$ \Rightarrow $$2A + A

$$ \Rightarrow $$ A = $${{3{B^T} - {A^T}} \over 2}$$

Also, 3A + 2B = I

$$ \Rightarrow $$ $$3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$$ = I

$$ \Rightarrow $$ 11B

$$ \Rightarrow $$ (11B

$$ \Rightarrow $$ 11B - A = 2I

Multiply (3) by 3 and then adding (2) and (3) we get,

35B = 7I

$$ \Rightarrow $$ B = $${{{I_3}} \over 5}$$

From (3), 11$${{{I_3}} \over 5}$$ - A = 2I

$$ \Rightarrow $$ A = $${{{I_3}} \over 5}$$

$$ \therefore $$ 5A = 5B = I

$$ \Rightarrow $$ 10A + 5B = 3I

3

If $$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}$$

then the ordered pair (A, B) is equal to

then the ordered pair (A, B) is equal to

A

(4, 5)

B

(-4, -5)

C

(-4, 3)

D

(-4, 5)

$$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right|$$

Applying c_{1} $$ \to $$ c_{1} + c_{2} + c_{3}

$$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$

Taking common (5x $$-$$ 4) from c_{1}

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$$

Apply R_{2} $$ \to $$R_{2} $$-$$ R_{1} and R_{3} $$ \to $$R_{3} $$-$$ R_{1}

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$$

$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$

So, (A + Bx) (x $$-$$ A)^{2} = (5x $$-$$ 4) (x + 4)^{2}

By comparing both sides we get, A = $$-$$ 4 and B = 5

Applying c

$$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$

Taking common (5x $$-$$ 4) from c

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$$

Apply R

$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$$

$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$

So, (A + Bx) (x $$-$$ A)

By comparing both sides we get, A = $$-$$ 4 and B = 5

4

If the system of linear equations

x + ky + 3z = 0

3x + ky - 2z = 0

2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then $${{xz} \over {{y^2}}}$$ is equal to

x + ky + 3z = 0

3x + ky - 2z = 0

2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then $${{xz} \over {{y^2}}}$$ is equal to

A

30

B

-10

C

10

D

-30

System of equations has non-zero solution when determinant of coefficient = 0.

So, in this questions,

$$\left| {\matrix{ 1 & K & 3 \cr 3 & K & { - 2} \cr 2 & 4 & { - 3} \cr } } \right| = 0$$

$$ \Rightarrow \,\,\,\,$$ ($$-$$ 3K + 8) $$-$$ K ($$-$$9 + 4) + 3(12 $$-$$ 2K) = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 3K + 8 + 9K $$-$$ 4K + 36 $$-$$ 6K = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 4K + 44 = 0

$$ \Rightarrow \,\,\,\,$$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $$-$$ 2z = 0 . . . (2)

2x + 4y $$-$$ 3z = 0 . . . (3)

By adding equation (1) and (3) we get,

3x + 15y = 0

$$ \Rightarrow \,\,\,\,$$ x = $$-$$ 5y

Putting x = $$-$$ 5y in equation (1) we get

$$-$$ 5y + 11y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ 6y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ z = $$-$$ 2y

$$\therefore\,\,\,\,$$ $${{xz} \over {{y^2}}}$$

$$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$$

$$ = {{10{y^2}} \over {{y^2}}}$$

$$ = 10$$

So, in this questions,

$$\left| {\matrix{ 1 & K & 3 \cr 3 & K & { - 2} \cr 2 & 4 & { - 3} \cr } } \right| = 0$$

$$ \Rightarrow \,\,\,\,$$ ($$-$$ 3K + 8) $$-$$ K ($$-$$9 + 4) + 3(12 $$-$$ 2K) = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 3K + 8 + 9K $$-$$ 4K + 36 $$-$$ 6K = 0

$$ \Rightarrow \,\,\,\,$$ $$-$$ 4K + 44 = 0

$$ \Rightarrow \,\,\,\,$$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $$-$$ 2z = 0 . . . (2)

2x + 4y $$-$$ 3z = 0 . . . (3)

By adding equation (1) and (3) we get,

3x + 15y = 0

$$ \Rightarrow \,\,\,\,$$ x = $$-$$ 5y

Putting x = $$-$$ 5y in equation (1) we get

$$-$$ 5y + 11y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ 6y + 3z = 0

$$ \Rightarrow \,\,\,\,$$ z = $$-$$ 2y

$$\therefore\,\,\,\,$$ $${{xz} \over {{y^2}}}$$

$$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$$

$$ = {{10{y^2}} \over {{y^2}}}$$

$$ = 10$$

Number in Brackets after Paper Name Indicates No of Questions

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

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Complex Numbers *keyboard_arrow_right*

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