JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If a function \(f(x)\) defined by \(f(x)=\left\{\begin{array}{ll}a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4\end{array}\right.\) be continuous for some \(a, b, c \in R\) and \(f ^{\prime}(0)+ f ^{\prime}(2)= e ,\) then the value of of \(a\) is
- A \(\frac{e}{e^{2}-3 e-13}\)
- B \(\frac{e}{e^{2}+3 e+13}\)
- C \(\frac{1}{e^{2}-3 e+13}\)
- D \(\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\left\{\begin{array}{ll}a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4\end{array}\right.\) For continuity at \(\mathrm{x}=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)\)…
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