JEE Main 2021 — Mathematics Calculus

$f(x)$f(x) is continuous at $x=0$x=0

$$\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}f(x)=f(0)=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}f(x)...(1)$$x→0+lim​f(x)=f(0)=x→0−lim​f(x)...(1)

$$f(0)=b...(2)$$f(0)=b...(2)

$$\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}(\frac{\mathrm{sin}(a+1)x}{2x}+\frac{\mathrm{sin}2x}{2x})$$x→0−lim​f(x)=x→0−lim​(2xsin(a+1)x​+2xsin2x​)

$$=\frac{a+1}{2}+1...(3)$$=2a+1​+1...(3)

$$\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{\sqrt{x+b{x}^{3}}-\sqrt{x}}{b{x}^{5/2}}$$x→0+lim​f(x)=x→0+lim​bx5/2x+bx3​−x​​

$$=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{(x+b{x}^{3}-x)}{b{x}^{5/2}(\sqrt{x+b{x}^{3}}+\sqrt{x})}$$=x→0+lim​bx5/2(x+bx3​+x​)(x+bx3−x)​

$$=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{\sqrt{x}}{\sqrt{x}(\sqrt{1+b{x}^{2}}+1)}=\frac{1}{2}...(4)$$=x→0+lim​x​(1+bx2​+1)x​​=21​...(4)

Use (2),(3)&(4) in $(1)$(1)

$$\frac{1}{2}=b=\frac{a+1}{2}+1$$21​=b=2a+1​+1

$$\Rightarrow b=\frac{1}{2},a=-2$$⇒b=21​,a=−2

$$a+b=\frac{-3}{2}$$a+b=2−3​