JEE Main 2021 — Mathematics Calculus

We have,

f(x)={\begin{matrix}a\mathrm{sin}\frac{\pi }{2}(x-1),\mathrm{for}x\leq 0 \\ \frac{\mathrm{tan}2x-\mathrm{sin}2x}{b{x}^{3}},\mathrm{for}x>0\end{matrix}

If function is continuous at $x=0$x=0, then

$$\mathrm{LHL}=\mathrm{RHL}=f(0)$$LHL=RHL=f(0)

Now,

$$\mathrm{LHL}=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}[a\mathrm{sin}\frac{\pi }{2}(x-1)]$$LHL=x→0−lim​[asin2π​(x−1)]

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}[a\mathrm{sin}\frac{\pi }{2}(0-h-1)]$$⇒LHL=h→0lim​[asin2π​(0−h−1)]

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}[a\mathrm{sin}\frac{\pi }{2}(-1)]$$⇒LHL=h→0lim​[asin2π​(−1)]

$$\Rightarrow \mathrm{LHL}=-a$$⇒LHL=−a

Now,

$$\mathrm{RHL}=\underset{x\rightarrow {0}^{+}}{lim}(\frac{\mathrm{tan}2x-\mathrm{sin}2x}{b{x}^{3}})$$RHL=x→0+lim​(bx3tan2x−sin2x​)

$$\because \mathrm{sin}x=x-\frac{{x}^{3}}{3!}+\frac{{x}^{5}}{5!}+\ldots \ldots$$∵sinx=x−3!x3​+5!x5​+……

$$\because \mathrm{tan}x=x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\ldots .$$∵tanx=x+3x3​+152x5​+….

$$\mathrm{RHL}=\underset{x\rightarrow {0}^{+}}{lim}[\frac{\frac{(2x{)}^{3}}{3}+\frac{(2x{)}^{3}}{6}}{b{x}^{3}}]=\frac{\frac{8}{3}+\frac{8}{6}}{b}=\frac{4}{b}$$RHL=x→0+lim​[bx33(2x)3​+6(2x)3​​]=b38​+68​​=b4​

And,

$$f(0)=a\mathrm{sin}[\frac{\pi }{2}(0-1)]=a\mathrm{sin}(-\frac{\pi }{2})=-a$$f(0)=asin[2π​(0−1)]=asin(−2π​)=−a

Then,

$$\mathrm{RHL}=f(0)$$RHL=f(0)

$$\Rightarrow \frac{4}{b}=-a\Rightarrow -ab=4\Rightarrow 10-ab=14$$⇒b4​=−a⇒−ab=4⇒10−ab=14