Q50 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics
Let $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$ then
$\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=$
- A$0$
- Correct answerB$-2 \sin 1 \cos (\cos 1)$
- C$\infty$
- D$-2 \sin 1 \cos 1$
Solution
Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$
$\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}$
When limit it applied, it gives $\frac{0}{0}$ form, so we apply L'Hospital rule.
$\Rightarrow \lim _{x \rightarrow 1} \frac{\cos \left(\cos x^2\right)\left(-\sin x^2\right)(2 x)-0}{1-0}$
Apply the limit,
$\Rightarrow-2(1) \sin (1) \cos (\cos 1)$
$=-2 \sin (1) \cos (\cos 1)$
Hence, the correct answer is option: 'B'.