At what rate of interest will the present value of a perpetuity of Rs. 1000 payable at the end of every six months be Rs. 40000?

Given: $f(x) = \tan x - 4x$f(x)=tanx−4x where $x \in (0, \frac{\pi}{2})$x∈(0,2π​)

A function is strictly decreasing when its derivative is negative.

Taking the derivative:

$$f'(x) = \sec^2 x - 4$$f′(x)=sec2x−4

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For the function to be strictly decreasing:

$$f'(x) < 0$$f′(x)<0

$$\sec^2 x - 4 < 0$$sec2x−4<0

$$\sec^2 x < 4$$sec2x<4

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Since $\sec^2 x = \frac{1}{\cos^2 x}$sec2x=cos2x1​:

$$\frac{1}{\cos^2 x} < 4$$cos2x1​<4

$$1 < 4\cos^2 x$$1<4cos2x

$$\cos^2 x > \frac{1}{4}$$cos2x>41​

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Taking the square root (note that $\cos x$cosx is positive in $(0, \frac{\pi}{2})$(0,2π​)):

$$\cos x > \frac{1}{2}$$cosx>21​

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The value $\cos x = \frac{1}{2}$cosx=21​ at $x = \frac{\pi}{3}$x=3π​

Since cosine decreases from 1 to 0 as $x$x goes from 0 to $\frac{\pi}{2}$2π​, the inequality $\cos x > \frac{1}{2}$cosx>21​ holds when $x < \frac{\pi}{3}$x<3π​.

Combined with the given domain:

$$0 < x < \frac{\pi}{3}$$0<x<3π​

Therefore, the function is strictly decreasing in the interval $(0, \frac{\pi}{3})$(0,3π​).