1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

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 :
A
$$4 + 2\sqrt 3 $$
B
$$ - 2 + \sqrt 3 $$
C
$$ - 2 - \sqrt 3 $$
D
$$-\,\,4 - 2\sqrt 3 $$

Explanation

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 $$-$$ cos2x) $$-$$ sinx(sin2x) = 0

$$ \Rightarrow $$$$\,\,\,$$ cos3x $$-$$ sin3 x = 0

$$ \Rightarrow $$$$\,\,\,$$ tan3x = 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 $$-$$ $$\sqrt 3 $$
2
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

For two 3 × 3 matrices A and B, let A + B = 2BT and 3A + 2B = I3, where BT is the transpose of B and I3 is 3 × 3 identity matrix. Then :
A
5A + 10B = 2I3
B
10A + 5B = 3I3
C
B + 2A = I3
D
3A + 6B = 2I3

Explanation

Given, A + B = 2BT .......(1)

$$ \Rightarrow $$ (A + B)T = (2BT)T

$$ \Rightarrow $$ AT + BT = 2B

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

Now put this in equation (1)

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

$$ \Rightarrow $$2A + AT = 3BT

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

Also, 3A + 2B = I3 .......(2)

$$ \Rightarrow $$ $$3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$$ = I3

$$ \Rightarrow $$ 11BT - AT = 2I3

$$ \Rightarrow $$ (11BT - AT)T = (2I3)T

$$ \Rightarrow $$ 11B - A = 2I3 ........(3)

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

35B = 7I3

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

From (3), 11$${{{I_3}} \over 5}$$ - A = 2I3

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

$$ \therefore $$ 5A = 5B = I3

$$ \Rightarrow $$ 10A + 5B = 3I3
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

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
A
(4, 5)
B
(-4, -5)
C
(-4, 3)
D
(-4, 5)

Explanation

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

Applying c1 $$ \to $$ c1 + c2 + c3

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

Taking common (5x $$-$$ 4) from c1

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

Apply R2 $$ \to $$R2 $$-$$ R1 and R3 $$ \to $$R3 $$-$$ R1

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

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

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

By comparing both sides we get, A = $$-$$ 4 and B = 5
4
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

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
A
30
B
-10
C
10
D
-30

Explanation

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

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

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

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

$$ = 10$$

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