Question

The minimum value of $\sqrt{(3sinx-4cosx-10)(3sinx+4cosx-10)}$ is

• A. 9
• B. 7
• C. 5
• D. 6

Answer 1

Answer: (B)

7

Given,

$\sqrt{(3sinx-4cosx-10)(3sinx+4cosx-10)}$

Simplifying, we get

$5\sqrt{(\frac{3}{5}sinx-\frac{4}{5}cosx-2)(\frac{3}{5}sinx+\frac{4}{5}cosx-2)}$

Put $sin\theta =\frac{4}{5}⟹\cos \theta =\frac{3}{5}$

$=5\sqrt{(sinx\cos \theta -\sin \theta cosx-2)(sinx\cos \theta +\sin \theta cosx-2)}$

$=5\sqrt{\left(sin\right(x-\theta )-2)\left(sin\right(\theta +x)-2)}$

..................... $\sin (A-B)=\sin A\cos B-\sin B\cos A$ and $\sin (A+B)=\sin A\cos B+\sin B\cos A$

$=5\sqrt{(2-sin(x-\theta \left)\right)(2-sin(\theta +x\left)\right)}$
$=5\sqrt{4-2\left[sin\right(x-\theta )+sin(x+\theta \left)\right]+sin(x-\theta ).sin(x+\theta )}$

$=5\sqrt{4-4sin\left(x\right).cos\left(\theta \right)+\frac{cos2\theta -cos2x}{2}}$

Hence the minimum value is achieved when
$x=\frac{\pi }{2}$.

Thus
$f\left(\frac{\pi }{2}\right)$$=\sqrt{\left(3sin\right(\frac{\pi }{2})-4cos(\frac{\pi }{2})-10)\left(3sin\right(\frac{\pi }{2})+4cos(\frac{\pi }{2})-10)}$

$=\sqrt{(3-10)\times (3-10)}=\sqrt{-7\times -7}=\sqrt{49}=7$.