Consider a function \( f:\left[0,\displaystyle\frac{\pi}{2}\right]\to R \), given by \( f(x)=\sin x \) and \( g:\left[0,\displaystyle\frac{\pi}{2}\right]\to R \), given by \( g(x)=\cos x \). Where \( R \) is a set of real numbers, which of the following statements are correct?
(A) \( f \) and \( g \) are one-one
(B) \( f \) and \( g \) are not one-one
(C) \( f+g \) is not one-one
(D) \( f+g \) is one-one
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
On \( \left[0,\frac{\pi}{2}\right] \), \( \sin x \) increases strictly from 0 to 1 and \( \cos x \) decreases strictly from 1 to 0, so neither takes any value twice. A strictly monotonic function is one-one.
For f + g, two different points give the same value:
\( \displaystyle (f+g)(0)=\sin 0+\cos 0=1 \)
\( \displaystyle (f+g)\left(\frac{\pi}{2}\right)=\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1 \)
So f and g are one-one, and f + g is not one-one, since \( (f+g)(0)=(f+g)\left(\frac{\pi}{2}\right)=1 \).
Checking the statements
(A) \( f \) and \( g \) are one-one — Correct. \( \sin x \) is strictly increasing and \( \cos x \) strictly decreasing on \( \left[0,\frac{\pi}{2}\right] \).
(B) \( f \) and \( g \) are not one-one — Incorrect. Both are strictly monotonic on \( \left[0,\frac{\pi}{2}\right] \), so both are one-one.
(C) \( f+g \) is not one-one — Correct. \( (f+g)(0)=(f+g)\left(\frac{\pi}{2}\right)=1 \).
(D) \( f+g \) is one-one — Incorrect. The points 0 and \( \frac{\pi}{2} \) both give the value 1.
NCERT (2026-27 edition):
Relations and Functions
