JEE Main 2023 — Mathematics Calculus

Given that

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{{e}^{ax}-\mathrm{cos}(bx)-\frac{cx}{2}{e}^{-cx}}{1-\mathrm{cos}2x}=17$$⇒x→0lim​1−cos2xeax−cos(bx)−2cx​e−cx​=17

We know that ${e}^{ax}=1+ax+\frac{(ax{)}^{2}}{2!}+\ldots$eax=1+ax+2!(ax)2​+… and $\mathrm{cos}(bx)=1-\frac{(bx{)}^{2}}{2!}+\ldots$cos(bx)=1−2!(bx)2​+…

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{(1+ax+\frac{{(ax)}^{2}}{2!}+\ldots )-(1-\frac{{(bx)}^{2}}{2!}+\ldots )-\frac{cx}{2}(1-(cx)+\frac{{(cx)}^{2}}{2!}-\ldots )}{(\frac{1-\mathrm{cos}2x}{{(2x)}^{2}})\times 4{x}^{2}}=17$$⇒x→0lim​((2x)21−cos2x​)×4x2(1+ax+2!(ax)2​+…)−(1−2!(bx)2​+…)−2cx​(1−(cx)+2!(cx)2​−…)​=17

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{(a-\frac{c}{2})x+(\frac{{a}^{2}+{b}^{2}+{c}^{2}}{2}){x}^{2}+\ldots }{\frac{1}{2}\times 4{x}^{2}}=17$$⇒x→0lim​21​×4x2(a−2c​)x+(2a2+b2+c2​)x2+…​=17

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

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{(\frac{{a}^{2}+{b}^{2}+{c}^{2}}{2}){x}^{2}+\ldots }{\frac{1}{2}\times 4{x}^{2}}=17$$⇒x→0lim​21​×4x2(2a2+b2+c2​)x2+…​=17

$$\Rightarrow \frac{{a}^{2}+{b}^{2}+{c}^{2}}{4}=17$$⇒4a2+b2+c2​=17

$$\Rightarrow {a}^{2}+{b}^{2}+4{a}^{2}=68$$⇒a2+b2+4a2=68

$$\Rightarrow 5{a}^{2}+{b}^{2}=68$$⇒5a2+b2=68

Hence this is the correct option.