If $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \ a x, & 1 \lt x \leq 2\end{array} ight.$. If $\lim _{x ightarrow 1} f(x)$ exists then the sum of the cubes of the possible values of $a$ is

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If $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \ a x, & 1 \lt x \leq 2\end{array}ight.$. If $\lim _{x ightarrow 1} f(x)$ exists then the sum of the cubes of the possible values of $a$ is

Aicharya · Accepted Answer

$\because \lim _{x \rightarrow 1} f(x)$ exists $\Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$ $\Rightarrow 1+\frac{2}{a}=a \Rightarrow a^2-a-2=0 \Rightarrow a=-1,2$ Sum of the cubes $=(-1)^3+(2)^3=-1+8=7$.

If $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \\ a x, & 1 \lt x \leq…

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