JEE Main 2014 — Mathematics Calculus

Given f(x)={\begin{matrix}\frac{\sqrt{2+cosx}-1}{{(\pi -x)}^{2}}, & x\neq \pi \\ k, & x=\pi \end{matrix} is continuous at $x=\pi$x=π

Hence, $f(\pi )=\underset{x\rightarrow \pi }{\mathrm{lim}}f(x)$f(π)=x→πlim​f(x)

$$\Rightarrow k=\underset{x\rightarrow \pi }{\mathrm{lim}}\frac{\sqrt{2+\mathrm{cos}x}-1}{{(\pi -x)}^{2}}$$⇒k=x→πlim​(π−x)22+cosx​−1​

Consider $L.H.L.=\underset{x\rightarrow {\pi }^{-}}{\mathrm{lim}}\frac{\sqrt{2+\mathrm{cos}x}-1}{{(\pi -x)}^{2}}$L.H.L.=x→π−lim​(π−x)22+cosx​−1​

$$\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{2+\mathrm{cos}(\pi -h)}-1}{{(\pi -(\pi -h))}^{2}}$$h→0lim​(π−(π−h))22+cos(π−h)​−1​

Also, $\mathrm{cos}(\pi -h)=-\mathrm{cos}h$cos(π−h)=−cosh

$$\Rightarrow k=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{2-\mathrm{cos}h}-1}{{h}^{2}}$$⇒k=h→0lim​h22−cosh​−1​

$$\Rightarrow k=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{\sqrt{2-\mathrm{cos}h}-1}{{h}^{2}})\times (\frac{\sqrt{2-\mathrm{cos}h}+1}{\sqrt{2-\mathrm{cos}h}+1})$$⇒k=h→0lim​(h22−cosh​−1​)×(2−cosh​+12−cosh​+1​)

$$\Rightarrow k=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(2-\mathrm{cos}h)-1}{{h}^{2}{\sqrt{2-\mathrm{cos}h}+1}}$$⇒k=h→0lim​h22−cosh​+1(2−cosh)−1​

$$\Rightarrow k=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{1-\mathrm{cos}h}{{h}^{2}})\underset{h\rightarrow 0}{\mathrm{lim}}\frac{1}{{\sqrt{2-\mathrm{cos}h}+1}}$$⇒k=h→0lim​(h21−cosh​)h→0lim​2−cosh​+11​

Now, using $\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x,\Rightarrow 1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}x,$cos2x=1−2sin2x,⇒1−cos2x=2sin2x,

$$\Rightarrow k=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{2{sin}^{2}(\frac{h}{2})}{4{(\frac{h}{2})}^{2}}\times \frac{1}{(\sqrt{2-1}+1)}$$⇒k=h→0lim​4(2h​)22sin2(2h​)​×(2−1​+1)1​

$$\Rightarrow k=\frac{1}{2}\times \frac{1}{2}$$⇒k=21​×21​

$$\Rightarrow k=\frac{1}{4}.$$⇒k=41​.