Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

$cos^{-1}\begin{Bmatrix}\frac{1}{2}x^2 + \sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}\end{Bmatrix}= cos^{-1}\frac{x}{2}-cos^{-1}x $ holds for

Options:

x ∈[-1,1]

x ∈ R

x ∈[0,1]

x ∈[-1,0]

Correct Answer:

x ∈[0,1]

Explanation:

We observe that

$\frac{x^2}{2}+\sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}$ is positive and defined for all x ∈ [-1, 1]

$∴cos^{-1}\begin{Bmatrix}\frac{x^2}{2} + \sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}\end{Bmatrix}∈\left[0, \frac{\pi}{2}\right] $

Now, RHS of the given relation is defined for

$\left|\frac{x}{2}\right|$ and $|x| ≤ 1 $ i.e. for x ∈ [-1,1].

Also, LHS ≥ 0

$⇒ cos^{-1}\frac{x}{2}-cos^{-1} x ≥ 0 ⇒ cos^{-1}\frac{x}{2} ≥cos^{-1}x ⇒x ∈[0, 1]$