If $ A=\left\{x:-\frac{2}{5} \leq x \leq \frac{\pi-2}{5}\right\}, B=\{y:-1 \leq y \leq 1\} $ and $ f(x)=\cos (5 x+2) $, then the mapping $ f: A \rightarrow B $ is

Question

If $ A=\left\{x:-\frac{2}{5} \leq x \leq \frac{\pi-2}{5}\right\}, B=\{y:-1 \leq y \leq 1\} $ and $ f(x)=\cos (5 x+2) $, then the mapping $ f: A \rightarrow B $ is, one-one but not onto, onto but not one-one, both one-one and onto, neither one-one nor onto

MARKS App · Accepted Answer

Let t = 5 x + 2 , then A = t : 0 ≤ t ≤ π A bijective function is both one-one and onto. ∴ f t = cos t , which is bijective in 0 , π since cos t is decreasing in 0 , π and Range of cos t is - 1 , 1 . Hence, f x is bijective.

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