If f(x) = \sin ( \cos ^ - 1( 1 - 2^2x \over 1 + 2^2x ) ) and its first derivative with respect to x is - b \over a\log _e2 when x = 1, where a and b are integers, then the minimum value of | a2 - b2 | is \underline\hspace2cm .

Differentiation PYQ — JEE Main 2021 · 17 Mar · Shift 1 · Q41 with Solution

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Correct answer: 481

Given function

f(x)=\sin\left(\cos^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)

We use the identity

\sin(\cos^{-1} t)=\sqrt{1-t^2}

since $\cos^{-1}t\in[0,\pi]$ , so sine is non-negative.

Thus,

f(x)=\sqrt{1-\left(\frac{1-2^{2x}}{1+2^{2x}}\right)^2}

Simplify the expression

Let

y=2^{2x}

Then

f(x)=\sqrt{1-\left(\frac{1-y}{1+y}\right)^2}

Now,

1-\left(\frac{1-y}{1+y}\right)^2 =\frac{(1+y)^2-(1-y)^2}{(1+y)^2}

Using

(a+b)^2-(a-b)^2=4ab

with $a=1, b=y$ , we get

(1+y)^2-(1-y)^2=4y

So,

f(x)=\sqrt{\frac{4y}{(1+y)^2}}=\frac{2\sqrt{y}}{1+y}

Since $y=2^{2x}>0$ , we have

\sqrt{y}=2^x

Hence,

f(x)=\frac{2\cdot 2^x}{1+2^{2x}}=\frac{2^{x+1}}{1+2^{2x}}

Differentiate

Let

u=2^{x+1}, \qquad v=1+2^{2x}

Then

u' = 2^{x+1}\ln 2

and

v' = 2^{2x}\cdot 2\ln 2 = 2^{2x+1}\ln 2

By quotient rule,

f'(x)=\frac{u'v-uv'}{v^2}

So,

f'(x)=\frac{2^{x+1}\ln 2\,(1+2^{2x})-2^{x+1}(2^{2x+1}\ln 2)}{(1+2^{2x})^2}

Factor out $2^{x+1}\ln 2$ :

f'(x)=\frac{2^{x+1}\ln 2\left[(1+2^{2x})-2^{2x+1}\right]}{(1+2^{2x})^2}

Since $2^{2x+1}=2\cdot 2^{2x}$ ,

(1+2^{2x})-2^{2x+1}=1-2^{2x}

Therefore,

f'(x)=\frac{2^{x+1}(1-2^{2x})}{(1+2^{2x})^2}\ln 2

Evaluate at $x=1$

At $x=1$ ,

2^{x+1}=2^2=4, \qquad 2^{2x}=2^2=4

Thus,

f'(1)=\frac{4(1-4)}{(1+4)^2}\ln 2

f'(1)=\frac{4(-3)}{25}\ln 2=-\frac{12}{25}\ln 2

Given that

f'(1)=-\frac{b}{a}\log_e 2

we compare and get

\frac{b}{a}=\frac{12}{25}

So we may take

a=25, \qquad b=12

Find $|a^2-b^2|$

|a^2-b^2|=|25^2-12^2|=|625-144|=481

Since $\frac{12}{25}$ is already in lowest terms, this gives the minimum value.

Final Answer

\boxed{481}

More from Differentiation

View written solutionFree

Correct answer: 481

Given function

f(x)=\sin\left(\cos^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)

We use the identity

\sin(\cos^{-1} t)=\sqrt{1-t^2}

since $\cos^{-1}t\in[0,\pi]$ , so sine is non-negative.

Thus,

f(x)=\sqrt{1-\left(\frac{1-2^{2x}}{1+2^{2x}}\right)^2}

Simplify the expression

Let

y=2^{2x}

Then

f(x)=\sqrt{1-\left(\frac{1-y}{1+y}\right)^2}

Now,

1-\left(\frac{1-y}{1+y}\right)^2 =\frac{(1+y)^2-(1-y)^2}{(1+y)^2}

Using

(a+b)^2-(a-b)^2=4ab

with $a=1, b=y$ , we get

(1+y)^2-(1-y)^2=4y

So,

f(x)=\sqrt{\frac{4y}{(1+y)^2}}=\frac{2\sqrt{y}}{1+y}

Since $y=2^{2x}>0$ , we have

\sqrt{y}=2^x

Hence,

f(x)=\frac{2\cdot 2^x}{1+2^{2x}}=\frac{2^{x+1}}{1+2^{2x}}

Differentiate

Let

u=2^{x+1}, \qquad v=1+2^{2x}

Then

u' = 2^{x+1}\ln 2

and

v' = 2^{2x}\cdot 2\ln 2 = 2^{2x+1}\ln 2

By quotient rule,

f'(x)=\frac{u'v-uv'}{v^2}

So,

f'(x)=\frac{2^{x+1}\ln 2\,(1+2^{2x})-2^{x+1}(2^{2x+1}\ln 2)}{(1+2^{2x})^2}

Factor out $2^{x+1}\ln 2$ :

f'(x)=\frac{2^{x+1}\ln 2\left[(1+2^{2x})-2^{2x+1}\right]}{(1+2^{2x})^2}

Since $2^{2x+1}=2\cdot 2^{2x}$ ,

(1+2^{2x})-2^{2x+1}=1-2^{2x}

Therefore,

f'(x)=\frac{2^{x+1}(1-2^{2x})}{(1+2^{2x})^2}\ln 2

Evaluate at $x=1$

At $x=1$ ,

2^{x+1}=2^2=4, \qquad 2^{2x}=2^2=4

Thus,

f'(1)=\frac{4(1-4)}{(1+4)^2}\ln 2

f'(1)=\frac{4(-3)}{25}\ln 2=-\frac{12}{25}\ln 2

Given that

f'(1)=-\frac{b}{a}\log_e 2

we compare and get

\frac{b}{a}=\frac{12}{25}

So we may take

a=25, \qquad b=12

Find $|a^2-b^2|$

|a^2-b^2|=|25^2-12^2|=|625-144|=481

Since $\frac{12}{25}$ is already in lowest terms, this gives the minimum value.

Final Answer

\boxed{481}

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