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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
In order that the function $f(x) = {(x + 1)^{\cot \,x}}$ is continuous at $x = 0 , f(0)$ must be defined as
  • A
    $f(0) = \frac{1}{e}$
  • B
    $f(0) = 0$
  • ✓
    $f(0) = e$
  • D
    None of these
✓

Answer

Correct option: C.
$f(0) = e$
c
(c) For continuity at $x = 0$, we must have $f(0) = \mathop {{\rm{lim}}}\limits_{x \to 0} f(x)$

$\mathop {\lim }\limits_{x \to \alpha } \,\frac{{1 - \cos \,(a{x^2} + bx + c)}}{{{{(x - \alpha )}^2}}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} {\left\{ {{{(1 + x)}^{\frac{1}{x}}}} \right\}^{x\cot x}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} {\left\{ {{{(1 + x)}^{\frac{1}{x}}}} \right\}^{\mathop {{\rm{lim}}}\limits_{x \to 0} \,\left( {\frac{x}{{\tan x}}} \right)}}$ $ = {e^1} = e$.

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