Sets, Relations & Functions
Functions
nta_abhyas_2025
Grade 11
Question:
If $f(\tan z) = \cos 2z, z \neq \left(2n+1\right)\frac{\pi}{2}, n \in \mathbb{Z}$, then incorrect statement is
f(z) is an even function
f(z) is an odd function
Range of f(z) is [-1,1]
Domain of f(z) is R
Step-by-Step Solution
Key Concept: An odd function satisfies $f(-x) = -f(x)$; using tangent substitution simplifies trigonometric expressions.
Let $\tan x = t$. Using the identity $\sin x = \frac{2t}{1+t^2}$ and $\cos 2x = \frac{1-t^2}{1+t^2}$, we substitute into the given equation. For $f(x)$ to be odd, we compute $f(-x) = -f(x)$. After simplification using trigonometric identities and the odd function property, $f(x) = \frac{1+t^2}{1-t^2}$ where $t = \tan x$.
Correct Answer: $\frac{1+t^2}{1-t^2}$