1

### 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

### 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

### 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

### 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

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

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

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

$= 10$