JEE Main 2023 — Limits Question with Solution

Given that

$\Rightarrow \underset{x\to 0}{\lim }\frac{e^{ax}-\cos \left(bx\right)-\frac{cx}{2}{e}^{-cx}}{1-\cos 2x}=17$

We know that $e^{ax}=1+ax+\frac{(ax{)}^2}{2!}+\dots$ and $\cos \left(bx\right)=1-\frac{(bx{)}^2}{2!}+\dots$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(1+ax+\frac{{\left(ax\right)}^2}{2!}+\dots \right)-\left(1-\frac{{\left(bx\right)}^2}{2!}+\dots \right)-\frac{cx}{2}\left(1-\left(cx\right)+\frac{{\left(cx\right)}^2}{2!}-\dots \right)}{\left(\frac{1-\cos 2x}{{\left(2x\right)}^2}\right)\times 4x^2}=17$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(a-\frac{c}{2}\right)x+\left(\frac{a^2+b^2+c^2}{2}\right)x^2+\dots }{\frac{1}{2}\times 4x^2}=17$

Now for limit to exist, $a-\frac{c}{2}=0$$\Rightarrow c=2a$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\left(\frac{a^2+b^2+c^2}{2}\right)x^2+\dots }{\frac{1}{2}\times 4x^2}=17$

$\Rightarrow \frac{a^2+b^2+c^2}{4}=17$

$\Rightarrow a^2+b^2+4a^2=68$

$\Rightarrow 5a^2+b^2=68$

Hence this is the correct option.