Question

The maximum value of $5\sin x+12\cos x$ is-

• A. $5$
• B. $12$
• C. $13$
• D. none of these

Answer 1

Answer: (C)

$13$

$f\left(x\right)=5sinx+12cosx$

Now

$\sqrt{5^2+{12}^2}=\sqrt{169}=13$

Multiplying and dividing by 13, we get

$=13(\frac{5}{13}sinx+\frac{12}{13}cosx)$

$=13(sinx.cos\theta +cosx.sin\theta )$

$=13sin(x+\theta )$ where $\theta =co{s}^{-1}\left(\frac{5}{13}\right)=si{n}^{-1}\left(\frac{12}{13}\right)$

Hence

$f\left(x\right)=13sin(x+\theta )$. Therefore maximum value of $f\left(x\right)$ will be

$f(x{)}_{max}=13$. as $si{n}_{max}(x+\theta )=1$