1

### JEE Main 2019 (Online) 10th January Evening Slot

The number of values of $\theta$ $\in$ (0, $\pi$) for which the system of linear equations

x + 3y + 7z = 0

$-$ x + 4y + 7z = 0

(sin3$\theta$)x + (cos2$\theta$)y + 2z = 0.

has a non-trival solution, is -
A
two
B
one
C
four
D
three

## Explanation

$\left| {\matrix{ 1 & 3 & 7 \cr { - 1} & 4 & 7 \cr {\sin 3\theta } & {\cos 2\theta } & 2 \cr } } \right| = 0$

(8 $-$ 7 cos 2$\theta$) $-$ 3($-$2 $-$ 7 sin 3$\theta$)

+7 ($-$ cos 2$\theta$ $-$ 4 sin 3$\theta$) = 0

14 $-$ 7 cos 2$\theta$ + 21 sin 3$\theta$ $-$ 7 cos 2$\theta$

$-$ 28 sin 3$\theta$ = 0

14 $-$ 7 sin 3$\theta$ $-$ 14 cos 2$\theta$ = 0

14 $-$ 7 (3 sin $\theta$ $-$ 4 sin3$\theta$ ) $-$ 14 (1 $-$ 2 sin2 $\theta$) = 0

$-$ 21 sin $\theta$ + 28 sin3 $\theta$ + 28 sin2 $\theta$ = 0

7 sin $\theta$ [$-$ 3 + 4 sin2 $\theta$ + 4 sin $\theta$] = 0 sin$\theta$,

4 sin2 $\theta$ + 6 sin $\theta$ $-$ 2 sin $\theta$ $-$ 3 = 0

2 sin $\theta$(2 sin $\theta$ + 3) $-$ 1 (2 sin $\theta$ + 3) = 0

sin $\theta$ = ${{ - 3} \over 2}$; sin$\theta$ = ${1 \over 2}$

Hence, 2 solutions in (0, $\pi$)
2

### JEE Main 2019 (Online) 11th January Morning Slot

If the system of linear equations
2x + 2y + 3z = a
3x – y + 5z = b
x – 3y + 2z = c
where a, b, c are non zero real numbers, has more one solution, then :
A
b – c – a = 0
B
a + b + c = 0
C
b – c + a = 0
D
b + c – a = 0

## Explanation

P1 : 2x + 2y + 3z = a

P2 : 3x $-$ y + 5z = b

P3 : x $-$ 3y + 2z = c

We find

P1 + P3 = P2 $\Rightarrow$ a + c = b
3

### JEE Main 2019 (Online) 11th January Morning Slot

Let A = $\left( {\matrix{ 0 & {2q} & r \cr p & q & { - r} \cr p & { - q} & r \cr } } \right).$   If  AAT = I3,   then   $\left| p \right|$ is
A
${1 \over {\sqrt 2 }}$
B
${1 \over {\sqrt 5 }}$
C
${1 \over {\sqrt 6 }}$
D
${1 \over {\sqrt 3 }}$

## Explanation

A is orthogonal matrix

$\Rightarrow$  02 + p2 + p2 = 1

$\Rightarrow$  $\left| p \right| = {1 \over {\sqrt 2 }}$
4

### JEE Main 2019 (Online) 11th January Evening Slot

If  $\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

= (a + b + c) (x + a + b + c)2, x $\ne$ 0,

then x is equal to :
A
–2(a + b + c)
B
2(a + b + c)
C
abc
D
–(a + b + c)

## Explanation

$\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

R1 $\to$ R1 + R2 + R3

$= \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

$= \left( {a + b + c} \right)\left| {\matrix{ 1 & 0 & 0 \cr {2b} & { - \left( {a + b + c} \right)} & 0 \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

$=$ (a + b + c) (a + b + c)2

$\Rightarrow$  x $=$ $-$ 2(a + b + c)

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