JEE Main 2019 — Mathematics Calculus

\because f(x)={\begin{matrix}\frac{\mathrm{sin}⁡(p+1)x +sinx}{\begin{matrix}x\end{matrix}} \\ q \\ \frac{\sqrt{{x}^{2}+x}-\sqrt{x}}{{x}^{3/2}}\end{matrix} \begin{matrix}:x<0 \\ :x=0 \\ :x>0\end{matrix}

$\because f$∵f is continuous at $x=0$x=0

$$L.H.L=$$L.H.L=

$$\underset{h\rightarrow 0}{lim} \frac{\mathrm{sin}(p+1)(-h)+sin(-h)}{-h}=\underset{h\rightarrow 0}{lim}\frac{\mathrm{sin}(p+1)h+\mathrm{sin}h}{h}$$h→0lim​−hsin(p+1)(−h)+sin(−h)​=h→0lim​hsin(p+1)h+sinh​

$$=\underset{h\rightarrow 0}{lim}\frac{\mathrm{sin}(p+1)h}{h(p+1)}(p+1)+\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}h}{h}=p+1+1$$=h→0lim​h(p+1)sin(p+1)h​(p+1)+h→0lim​hsinh​=p+1+1

$$=p+2$$=p+2

$$R.H.L= \underset{h\rightarrow 0}{lim} \frac{\sqrt{{h}^{2}+h-} \sqrt{h}}{{h}^{\frac{3}{2}}}\times \frac{\sqrt{{h}^{2}+h}+\sqrt{h}}{\sqrt{{h}^{2}+h}+\sqrt{h}}=\underset{h\rightarrow 0}{lim}\frac{{h}^{2}+h-h}{{h}^{\frac{3}{2}} {h}^{\frac{1}{2}}(\sqrt{h+1}+1)}$$R.H.L=h→0lim​h23​h2+h−​h​​×h2+h​+h​h2+h​+h​​=h→0lim​h23​h21​(h+1​+1)h2+h−h​

$$\underset{h\rightarrow 0}{lim}\frac{1}{\sqrt{1+h}+1}=\frac{1}{2}$$h→0lim​1+h​+11​=21​

$$\Rightarrow L.H.L.=R.H.L.=f(0)\Rightarrow p+2=\frac{1}{2}=q$$⇒L.H.L.=R.H.L.=f(0)⇒p+2=21​=q

$$=\begin{matrix}p=-\frac{3}{2} & q=\frac{1}{2}\end{matrix}$$=p=−23​​q=21​​