JEE Advanced 2021 — Paper 2 — Question 19

$\begin{aligned} & \text { (A) } \int_{0}^{\pi / 3} g(x) d x=\int_{0}^{\pi / 3}(f(x)-3 \cos 3 x) d x=\int_{0}^{\pi / 3} f(x) d x-\int_{0}^{\pi / 3} 3 \cos 3 x d x=0 \\& \Rightarrow g(x)=0 \end{aligned}$ has at least one solution in $[0, \pi / 3]$

(B) Let $h(x)=f(x)-3 \sin 3 x+6 / \pi$ $\int_{0}^{\pi / 3} h(x)=\int_{0}^{\pi / 3}\left(f(x)-3 \sin 3 x+\frac{6}{\pi}\right) d x=0$

$\Rightarrow \mathrm{h}(\mathrm{x})=0$ has atleast one solution in $[0, \pi / 3]$

(C) $\lim _{x \rightarrow 0} \frac{\int_{0}^{x} f(t) d t}{x}=\lim _{x \rightarrow 0} \frac{f(x)}{1}=f(0)=1$

$$\Rightarrow \lim _{x \rightarrow 0} \frac{x \int_{0}^{x} f(t) d t}{x^{2}\left(\frac{1-e^{x^{2}}}{x^{2}}\right)}=\lim _{x \rightarrow 0} \frac{-\int_{0}^{x} f(t)}{x} \cdot \frac{x^{2}}{\left(e^{x^{2}}-1\right)}=-1$$

(D) $\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{\int_{0}^{x} f(t)}{x}=1$