Jee Main 2024 31 01 2024 Shift2 - Question9

Answer: (4)

Solution:

Take $e^{\sin x}=t(t>0)$

$\Rightarrow t-\frac{2}{t}=2$

$\Rightarrow \frac{t^{2}-2}{t}=2$

$\Rightarrow t^{2}-2 t-2=0$

$\Rightarrow t^{2}-2 t+1=3$

$\Rightarrow(t-1)^{2}=3$

$\Rightarrow t=1 \pm \sqrt{3}$

$\Rightarrow t=1 \pm 1.73$

$\Rightarrow t=2.73$ or -0.73 (rejected as $t>0$ )

$\Rightarrow e^{\sin x}=2.73$

$\Rightarrow \log _e e^{\sin x}=\log _e 2.73$

$\Rightarrow \sin x=\log _e 2.73>1$

So no solution.