1

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

Let  d $\in$ R, and

$A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr } } \right],$

$\theta \in \left[ {0,2\pi } \right]$ If the minimum value of det(A) is 8, then a value of d is -
A
$-$ 7
B
$2\left( {\sqrt 2 + 2} \right)$
C
$-$ 5
D
$2\left( {\sqrt 2 + 1} \right)$

## Explanation

$\det A = \left| {\matrix{ { - 2} & {4 + d} & {\sin \theta - 2} \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr } } \right|$

(R1 $\to$ R1 + R3 $-$ 2R2)

$= \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$

= (2 + sin $\theta$) ( 2 + 2d $-$ sin$\theta$) $-$ d(2sin$\theta$ $-$ d)

= 4 + 4d $-$ 2sin$\theta$ + 2sin$\theta$ + 2dsin$\theta$ $-$ sin2$\theta$ $-$ 2dsin$\theta$ + d2

d2 + 4d + 4 $-$ sin2$\theta$

= (d + 2)2 $-$ sin2$\theta$

For a given d, minimum value of

det(A) = (d + 2)2 $-$ 1 = 8

$\Rightarrow$  d = 1 or $-$ 5
2

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

Let A = $\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$ where b > 0.

Then the minimum value of ${{\det \left( A \right)} \over b}$ is -
A
$\sqrt 3$
B
$-$ $2\sqrt 3$
C
$- \sqrt 3$
D
$2\sqrt 3$

## Explanation

A = $\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$ (b > 0)

$\left| A \right|$ = 2(2b2 + 2 $-$ b2) $-$ b(2b $-$ b) + 1(b2 $-$ b2 $-$ 1)

$\left| A \right|$ = 2(b2 + 2) $-$ b2 $-$ 1

$\left| A \right|$ = b2 + 3

${{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3$

$b + {3 \over b} \ge 2\sqrt 3$
3

### 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, $\pi$)
4

### 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