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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
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
    $\frac{1}{e^{2}-3 e+13}$
  • ✓
    $\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}$
✓

Answer

Correct option: D.
$\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}$
d
$f(x)=\left\{\begin{array}{ll}a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4\end{array}\right.$

For continuity at $\mathrm{x}=1$

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$

$\Rightarrow \quad a e+b e^{-1}=c \Rightarrow \quad b=c e-a e^{2}$

For continuity at $x=3$

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$

$\Rightarrow 9 c=9 a+6 c$

$\Rightarrow \quad c=3 a$

$f^{\prime}(0)+f^{\prime}(2)=e$

$\left(a e^{x}-b e^{x}\right)_{x}=0+(2 c x)_{x}=2=e$

$\Rightarrow \quad a-b+4 c=e$

From $(1),(2) \;and\;(3)$

$a-3 a e+a e^{2}+12 a=e$

$\Rightarrow \mathrm{a}\left(\mathrm{e}^{2}+13-3 \mathrm{e}\right)=\mathrm{e}$

$\Rightarrow \mathrm{a}=\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}$

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