The area bounded by the x-axis and the part of graph of y= cos x between x=(-π/2) and x=(π/2) is separated into two regions by the line x=k. If the area of the region for (-π/2) ≤ x ≤ k is three times the area of the region for k ≤ x ≤ (π/2), then k is equal to

Question

The area bounded by the x-axis and the part of graph of y= cos x between x=(-π/2) and x=(π/2) is separated into two regions by the line x=k. If the area of the region for (-π/2) ≤ x ≤ k is three times the area of the region for k ≤ x ≤ (π/2), then k is equal to (A)  arcsin ((1/4)) (B)  arcsin ((1/3)) (C) (π/6) (D) (π/3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0ca4d73b47f9bb9f679dc13849b32df9-.png" /> ∫ limits-π / 2 k cos xdx =3 ∫ limits k π / 2 cos xdx ; sin k - sin (-(π/2))=3( sin (π/2)- sin k ) sin k +1=3-3 sin k ; 4 sin k =2 ⇒ k =(π/6)

$arcsin (\frac{1}{4})$

$arcsin (\frac{1}{3})$

$\frac{π}{6}$

$\frac{π}{3}$

$- π /2 \int k cos x d x = 3 k \int π /2 cos x d x; sin k - sin (- \frac{π}{2}) = 3 (sin \frac{π}{2} - sin k)$
$sin k + 1 = 3 - 3 sin k; 4 sin k = 2 \Rightarrow k = \frac{π}{6}$

Q. The area bounded by the x-axis and the part of graph of y=cosx between x=2−π​ and x=2π​ is separated into two regions by the line x=k. If the area of the region for 2−π​≤x≤k is three times the area of the region for k≤x≤2π​, then k is equal to

arcsin(41​)

arcsin(31​)

6π​

3π​

−π/2∫k​cosxdx=3k∫π/2​cosxdx;sink−sin(−2π​)=3(sin2π​−sink) sink+1=3−3sink;4sink=2⇒k=6π​

$arcsin (\frac{1}{4})$

$arcsin (\frac{1}{3})$

$\frac{π}{6}$

$\frac{π}{3}$

$- π /2 \int k cos x d x = 3 k \int π /2 cos x d x; sin k - sin (- \frac{π}{2}) = 3 (sin \frac{π}{2} - sin k)$
$sin k + 1 = 3 - 3 sin k; 4 sin k = 2 \Rightarrow k = \frac{π}{6}$