The maximum value of 12 sin x - 5 cos x + 3 i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = 12 \sin x - 5 \cos x + 3$. We know that for any expression of the form $a \sin x + b \cos x$,the range is $[-\sqrt{a^2 + b^2}, \sqrt{a

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Let $f(x) = 12 \sin x - 5 \cos x + 3$. We know that for any expression of the form $a \sin x + b \cos x$,the range is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. Here,$a = 12$ and $b = -5$. So,$\sqrt{a^2 + b^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$. Thus,$-13 \leq 12 \sin x - 5 \cos x \leq 13$. Adding $3$ to all parts,we get: $-13 + 3 \leq 12 \sin x - 5 \cos x + 3 \leq 13 + 3$. $-10 \leq f(x) \leq 16$. Therefore,the maximum value of $f(x)$ is $16$.

The maximum value of $12 \sin x - 5 \cos x + 3$ is

Similar Questions

If $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $,then the difference between the maximum and minimum values of ${u^2}$ is given by

The range of $f(x) = \cos x - \sin x$ is

The minimum value of $27^{\cos 2x} 81^{\sin 2x}$ is

If $A + B = \frac{\pi}{2},$ the maximum value of $\cos A \cos B$ is

If $A + B + C = 180^\circ$,then $\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module