Question

The least value of the function $F(x) =$ $\int_{5\pi/4}^{x}{(3\sin u + 4\cos u)du}$ on the interval $\left\lbrack \frac{5\pi}{4},\frac{4\pi}{3} \right\rbrack$ is

• A. $\sqrt{3} + \frac{3}{2}$
• B. $- 2\sqrt{3} + \frac{3}{2} + \frac{1}{\sqrt{2}}$
• C. $\frac{3}{2} + \frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (B)

$- 2\sqrt{3} + \frac{3}{2} + \frac{1}{\sqrt{2}}$

Answer 2

We have $F'(x) = 3\sin x + 4\cos x$

Since in $\left\lbrack \frac{5\pi}{4},\frac{4\pi}{3} \right\rbrack,F'(x) < 0,$ so assume the least value at the point $x = \frac{4\pi}{3}.$ Thus, $f\left( \frac{4\pi}{3} \right) = \int_{5\pi/4}^{4\pi/3}{(3\sin u + 4\cos u)du}$

$= \frac{3}{2} - 2\sqrt{3} + \frac{1}{\sqrt{2}}$.