If fx= log⁡sec^2⁡xcot^2⁡x for x≠0 K for x=0 is continuous at x=0 then K is

Correct answer: (B) 1. Since f(x) is continuous at x=0 ⇒ f0=limx→0⁡log⁡sec^2⁡xcot⁡2x ⇒ k= limx→0⁡cot^2⁡x⋅log⁡(1+tan^2⁡x) =limx→0⁡(log⁡(1+tan^2⁡x))/(tan^2⁡x) ⇒ k=1

Q85 / 150 · MHT CET 2017 · Mathematics

If $f (x) = \{\begin{matrix} \log {(\sec^{2} x)}^{\cot^{2} x} & f o r & x \neq 0 \\ K & f o r & x = 0 \end{matrix}$ is continuous at $x = 0$ then $K$ is

A
$e^{- 1}$
Correct answer
B
$1$
C
$e$
D
$0$

Solution

Since

f (x)

is continuous at

x = 0

\Rightarrow

f (0) = \lim_{x \to 0} {\log (\sec^{2} x)}^{\cot 2 x}

\Rightarrow

k = \lim_{x \to 0} \cot^{2} x \cdot l o g (1 + \tan^{2} x)

= \lim_{x \to 0} \frac{l o g (1 + \tan^{2} x)}{\tan^{2} x}

\Rightarrow

k = 1

Hence, the correct answer is option: 'B'.

If fx=log⁡sec2⁡xcot2⁡xforx≠0Kforx=0 is continuous at x=0 then K is

Solution

Relevant Questions

If $f (x) = \{\begin{matrix} \log {(\sec^{2} x)}^{\cot^{2} x} & f o r & x \neq 0 \\ K & f o r & x = 0 \end{matrix}$ is continuous at $x = 0$ then $K$ is