Test Index

CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

Show Para  Hide Para 
Section - C
Q. Nos. 27 to 32 carry 4 marks each.
© examsnet.com
Question : 27 of 36
Marks: +1, -0
Solve for x:sin⁡−1(1−x)−2sin⁡−1(x)=π2x: \sin^{-1}(1-x)-2\sin^{-1}(x)=\frac{\pi}{2}.
Solution:     👈: Video Solution
Given that sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2}
let x=sin⁡yx=\sin y
∴sin⁡−1(1−sin⁡y)−2y=π2\therefore \sin^{-1}(1-\sin y)-2y=\frac{\pi}{2}
⇒sin⁡−1(1−sin⁡y)=π2+2y\Rightarrow \sin^{-1}(1-\sin y)=\frac{\pi}{2}+2y
⇒1−sin⁡y=sin⁡(π2+2y)\Rightarrow 1-\sin y=\sin\left(\frac{\pi}{2}+2y\right)
⇒1−sin⁡y=cos⁡2y\Rightarrow 1-\sin y=\cos 2y
⇒1−sin⁡y=1−2sin⁡2y as cos⁡2y=1−2sin⁡2y\Rightarrow 1-\sin y=1-2\sin^2 y \text{ as } \cos 2y=1-2\sin^2 y
⇒2sin⁡2y−sin⁡y=0\Rightarrow 2\sin^2 y-\sin y=0
⇒2x2−x=0\Rightarrow 2x^2-x=0
⇒x(2x−1)=0\Rightarrow x(2x-1)=0
⇒x=0,  2x−1=0\Rightarrow x=0,\;2x-1=0
⇒x=0,  2x=1\Rightarrow x=0,\;2x=1
⇒x=0,  x=12\Rightarrow x=0,\;x=\frac{1}{2}
But x=12x=\frac{1}{2} does not satisfy the given equation
∴x=0\therefore x=0 is the solution of the given equation.
© examsnet.com
Go to Question:
Prev Question
Next Question