Let the function \( f:\left[0,\displaystyle\frac{\pi}{2}\right]\to\mathrm{R} \) given by \( f(x)=\sin x \) and \( g:\left[0,\displaystyle\frac{\pi}{2}\right]\to\mathrm{R} \) given by \( g(x)=\cos x \). Then
Which of following statements are TRUE?
(A) \( (f+g)(x)=\sin x+\cos x \)
(B) \( f+g \) is one-one
(C) \( f+g \) is not one-one
(D) \( f+g \) is one-one and onto
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
By definition \( (f+g)(x)=f(x)+g(x)=\sin x+\cos x \). Its values at the two end points are
\( \displaystyle (f+g)(0)=0+1=1 \)
\( \displaystyle (f+g)\left(\frac{\pi}{2}\right)=1+0=1 \)
Two different inputs give the same output, so f + g is not one-one. Also
\( \displaystyle \sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \)
which rises to \( \sqrt{2} \) at \( x=\frac{\pi}{4} \) and then falls back to 1, so its values lie only in \( [1,\sqrt{2}] \).
So \( (f+g)(x)=\sin x+\cos x \), \( (f+g)(0)=(f+g)\left(\frac{\pi}{2}\right)=1 \), and the values of f + g fill only \( [1,\sqrt{2}] \), not R.
Checking the statements
(A) \( (f+g)(x)=\sin x+\cos x \) — Correct. This is the definition of the sum of two functions.
(B) \( f+g \) is one-one — Incorrect. \( (f+g)(0)=(f+g)\left(\frac{\pi}{2}\right)=1 \).
(C) \( f+g \) is not one-one — Correct. 0 and \( \frac{\pi}{2} \) both give the value 1.
(D) \( f+g \) is one-one and onto — Incorrect. It is not one-one, and not onto R, since its values lie only in \( [1,\sqrt{2}] \).
NCERT (2026-27 edition):
Composition of Functions and Invertible Function
