If $f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \ 3 x-3, & 1 \lt x \leq 2 \ b x^3+1, & x\gt2\end{array} ight.$ is differentiable $\forall x \in R$, then $a-b=$

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If $f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \ 3 x-3, & 1 \lt x \leq 2 \ b x^3+1, & x\gt2\end{array}ight.$ is differentiable $\forall x \in R$, then $a-b=$

Aicharya · Accepted Answer

$f(x)$ is differentiable $\forall x \in R \Rightarrow f(x)$ is continuous $\forall x \in R$ $\begin{aligned} & 	herefore \lim _{x ightarrow 2^{+}} f(x)=\lim _{x ightarrow 2^{-}} f(x) \Rightarrow \lim _{x ightarrow 2}\left(6 x^3+1ight)=\lim _{x ightarrow 2}(3 x-3) \ & \Rightarrow 8 b+1=3 \Rightarrow b=\frac{1}{4} 	ext { Also, } \lim _{x ightarrow 1^{-}} f(x)=\lim _{x ightarrow 1^{+}} f(x) \ & \Rightarrow \lim _{x ightarrow 1}\left(a x^2+b x-\frac{13}{8}ight)=\lim _{x ightarrow 1}(3 x-3) \ & \Rightarrow a+b-\frac{13}{8}=0 \Rightarrow a+\frac{1}{4}-\frac{13}{8}=0 \Rightarrow a=\frac{11}{8} \ & 	herefore a-b=\frac{11}{8}-\frac{1}{4}=\frac{9}{8} \end{aligned}$

If $f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 \lt x \leq 2 \\ b x^3+1, &…

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