The value of k which the function fx= 45^(tan⁡4xtan⁡5x), 0<x<(π)/2 k+2/5, x=(π)/2 is continuous at x=(π)/2, is

Correct answer: (D) 3/5. Given function is continuous at x=(π)/2 ⇒f(π)/2=f(π^(+))/2=f(π^(-))/2 ⇒k+2/5=45^(tan⁡4xtan⁡5x) ⇒k+2/5=45^(tan2πcot5π2) ⇒k+2/5=45^0 ⇒k+2/5=1 ⇒k=3/5

Q34 / 90 · JEE Main 2017 (09 Apr Online) · Mathematics

The value of $k$ which the function $f (x) = \{\begin{matrix} {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}, & 0 < x < \frac{π}{2} \\ k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ is

A
$\frac{2}{5}$
B
$- \frac{2}{5}$
C
$\frac{17}{20}$
Correct answer
D
$\frac{3}{5}$

Solution

Given function is continuous at $x = \frac{π}{2}$

$\Rightarrow f (\frac{π}{2}) = f (\frac{π^{+}}{2}) = f (\frac{π^{-}}{2})$

$\Rightarrow k + \frac{2}{5} = {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}$

$\Rightarrow k + \frac{2}{5} = {(\frac{4}{5})}^{\tan (2 π) \cot (\frac{5 π}{2})}$

$\Rightarrow k + \frac{2}{5} = {(\frac{4}{5})}^{0}$

$\Rightarrow k + \frac{2}{5} = 1$

$\Rightarrow k = \frac{3}{5}$

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

The value of k which the function fx= 45tan⁡4xtan⁡5x,0<x<π2k+25,x=π2 is continuous at x=π2, is

Solution

Relevant Questions

The value of $k$ which the function $f (x) = \{\begin{matrix} {(\frac{4}{5})}^{\frac{\tan 4 x}{\tan 5 x}}, & 0 < x < \frac{π}{2} \\ k + \frac{2}{5}, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ is