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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If $f(x)$ is invertible and twice differentiable function satisfying $f '(x) = \int\limits_0^{f(x)} {{f^{ - 1}}} (t)dt,\,\forall x\, \in \,R$ and $f '(0) =1$ then $f '(1)$ can be 
  • A
    $e$
  • B
    $e^2$
  • C
    $\frac{1}{e}$
  • ✓
    $\sqrt e $
✓

Answer

Correct option: D.
$\sqrt e $
d
${f^\prime }({\rm{x}}) = \int\limits_0^{f({\rm{x}})} {{f^{ - 1}}} ({\rm{t}}){\rm{dt}}\quad {f^\prime }(0) = 1$

$f^{\prime \prime}(x)=f^{-1}(f(x)) \cdot f^{\prime}(x)$

$f^{\prime \prime}(x)=x f^{\prime}(x)$

$\ln \left|f^{\prime}(\mathrm{x})\right|=\frac{\mathrm{x}^{2}}{2}+\mathrm{c}$

$c=0$

$\ln \left|f^{\prime}(\mathrm{x})\right|=\frac{\mathrm{x}^{2}}{2}$ Hence $\left|f^{\prime}(1)\right|=\mathrm{e}^{\frac{1}{2}}=\sqrt{\mathrm{e}}$

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