The value of k for which the function f(x) = begincases ( (4/5) )( tan 4x/ tan 5x) &, 0

Question

The value of k for which the function f(x) = begincases ( (4/5) )( tan 4x/ tan 5x) &, 0 < x < (π/2) k + (2/5) &, x = (π/2) endcases is continuous at x = (π/2), is : (A) (17/20) (B) (2/5) (C) (3/5) (D)  - (2/5)

Tardigrade Admin · Accepted Answer

k + 2 /5 = (4 /5) =k+(2/4)=1 K=(3/5)

Q. The value of $k$ for which the function $f (x) = {(\frac{4}{5})^{\frac{t a n 4 x}{t a n 5 x}} k + \frac{2}{5}, 0 < x < \frac{π}{2}, x = \frac{π}{2}$ is continuous at $x = \frac{π}{2}$ , is :

$\frac{17}{20}$

$\frac{2}{5}$

$\frac{3}{5}$

$- \frac{2}{5}$

$k + 2/5 = (4/5)$
$= k + \frac{2}{4} = 1$
$K = \frac{3}{5}$

Q. The value of k for which the function f(x)={(54​)tan 5xtan 4x​k+52​​,0<x<2π​,x=2π​​ is continuous at x=2π​, is :

2017​

52​

53​

−52​