Question

Let $f\left(x\right)=x\sqrt{4ax-x^2},(a>0).$. Then $f\left(x\right)$ is decreasing in:

• A. $(5a,\infty )$
• B. $(-\infty ,0)U(4a,\infty )$
• C. Always increasing
• D. None of the above.

Answer 1

Answer: (A), (B)

The correct options are
A $(5a,\infty )$
B $(-\infty ,0)U(4a,\infty )$

$f\left(x\right)=x\sqrt{4ax-x^2}$

Differntiate w.r.t 'x'

$f^1\left(x\right)=\frac{x(4x-2a)}{\alpha \sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$

$=\frac{\alpha x(2a-x)}{\alpha \sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$

$=\frac{x(2a-x)}{\sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$

Equate $f^1\left(x\right)=0$

$0=\frac{(2a-x)x}{\sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$

$-\sqrt{4ax-x^2}=\frac{(2a-x)x}{\sqrt{4ax-x^2}}$

$4ax-x^2=(2a-x)x$

So the factor is : $x\sqrt{x(4a-x)}=0$

$x=0or\sqrt{x(4a-x)}=0$

$4ax-x^2=0$

$x^2=4axx=4a$

So the function is increasing in the range (0,4a)

Therefore the decreasing range would be

$(-\infty ,0)\cup (4a,\infty )$

Hence the given option $(5a,\infty )and(-\infty ,0)\cup (4a,\infty )$ fall in this range

Hence option A and B are correct answers