The maximum value of \(f(x) = 3\sin x + 4\cos x\), \(x \in R\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
Setting the derivative equal to zero:
\(f'(x) = 3\cos x - 4\sin x = 0\)
\( \displaystyle \tan x = \frac{3}{4} \)
Where f is largest, \(\sin x\) and \(\cos x\) are both positive, so
\( \displaystyle \sin x = \frac{3}{5},\ \cos x = \frac{4}{5} \)
\( \displaystyle f(x) = \frac{9}{5} + \frac{16}{5} = 5 \)
This agrees with the result that the maximum of \(a\sin x + b\cos x\) is \(\sqrt{a^2 + b^2} = \sqrt{9 + 16} = 5\).
Why the other options are incorrect
Option (1) 7 — Incorrect. 3 + 4 would need \(\sin x = 1\) and \(\cos x = 1\) at the same x, which never happens.
Option (2) 4 — Incorrect. This is only f(0); f(x) is larger at \(\tan x = \frac{3}{4}\).
Option (4) 1 — Incorrect. f(x) reaches 5 where \(\tan x = \frac{3}{4}\), so 1 is not the maximum.