Question

Previous year Question of jee mains
Chapter:- application and derivatives​

Answer 1

Answer:

9

Step-by-step explanation:

Given equation

sinx4+1−sinx1=a    ....(1)

f(x)=sinx4+1−sinx1

f′(x)=cosx((1−sinx)21−sin2x4)

=cosxsin2x(1−sinx)2(2−sinx)(3sinx−2)

For maxima or minima, f′(x)=0

⇒sinx=2(not possible) , sinx=32 and cosx=0⇒x=2π

⇒x=sin−132

f(sin−132)=6+3=9

f(0+)→

⇒a=9   (by (1))

Answer 2

Answer:

$\frac{4}{ \sin x} + \frac{1}{ 1 - \sin x}$

$→ \frac{4(1 - \sin x) + \sin x }{ \sin x (1 - \sin x) } = \alpha$

$→ \frac{4 -3 \sin x }{ \sin x - { \sin }^{2} x } = \alpha$

$→4 -3 \sin x = \alpha \sin x - \alpha { \sin }^{2} x$

$→ \alpha { \sin }^{2} x - ( \alpha + 3) \sin x + 4 = 0$

$\sin x = \frac{ \alpha + 3± \sqrt{ {( \alpha + 3)}^{2} - 4.4 \alpha } }{2 \alpha } \\ = \frac{ \alpha + 3± \sqrt{ { \alpha }^{2} + 9 + 6 \alpha - 16 \alpha } }{2 \alpha } \\ = \frac{ \alpha + 3± \sqrt{ { \alpha }^{2} - 10 \alpha + 9 } }{2 \alpha } \\ = \frac{ \alpha + 3± \sqrt{( \alpha -9)( \alpha - 1)} }{2 \alpha }$

Now, for sinx to be real,,

$\sqrt{( \alpha -9)( \alpha - 1)} = 0$

$→( \alpha -9)( \alpha - 1) = 0$

$→\alpha = 9, 1$

putting value of α in the value of sinx,,

$\sin x = \frac{9 + 3}{2 \times 9} = \frac{2}{3} €(0 ,1)$

and

$\sin x = \frac{1 + 3}{2 \times 1} = \frac{4}{2} = 2 \: which \: is \: not \: possible.$

So,

$\alpha = 9 \: is \: the \: minimum \: value \: for (0,π/2)$