If ∫0π/3(tan⁡θ)/(√(2k sec⁡θ))dθ=1-1/(√2), (k>0) , then the value of k is

Correct answer: (C) 2. ∫0π/3(tan⁡x)/(√(2ksec⁡x))dx =1/(√2k) ∫0π/3(sin⁡x)/(√(cos⁡x))dx Put t=cos⁡x dt= -sin⁡x dx, also x=0, t=1;x=(π)/3, t=1/2 ∴ 1/(√2k) ∫11/2(-dt)/(√t) =1/(√2k)∫1/2_(1t^(-12)dt=12k×2 t^121/21) =(√2)/(√k)1-1/(√2)=(√2-1)/(√k) Given (√2-1)/(√k)=1-1/(√2)=(√2-1)/(√2) ∴k=2

Q40 / 90 · JEE Main 2019 (09 Jan Shift 2) · Mathematics

If $\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)$ , then the value of $k$ is

A
$\frac{1}{2}$
B
$1$
Correct answer
C
$2$
D
$4$

Solution

\int_{0}^{π / 3} \frac{\tan x}{\sqrt{2 k \sec x}} d x

= \frac{1}{\sqrt{2 k}} \int_{0}^{π / 3} \frac{\sin x}{\sqrt{\cos x}} d x

Put

t = \cos x

d t = - \sin x d x,

also

x = 0, t = 1; x = \frac{π}{3}, t = \frac{1}{2}

∴ \frac{1}{\sqrt{2 k}} \int_{1}^{1 / 2} \frac{- d t}{\sqrt{t}}

= \frac{1}{\sqrt{2 k}} {\int_{1 / 2}^{1} t^{- \frac{1}{2}} d t = \frac{1}{\sqrt{2 k}} \times 2 {[t^{\frac{1}{2}}]}_{1 / 2}^{1}}_{}

= \frac{\sqrt{2}}{\sqrt{k}} (1 - \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} - 1}{\sqrt{k}}

Given

\frac{\sqrt{2} - 1}{\sqrt{k}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}

∴ k = 2

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

If ∫0π/3tan⁡θ2k sec⁡θdθ=1-12, (k>0) , then the value of k is

Solution

Relevant Questions

If $\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)$ , then the value of $k$ is