Suppose f(x) = (2x + 2^-x) (x) ^-1(x^2 - x + 1). The value of f'(0) is equal to:

Given: f(x)= (2^x+2^-x) x ^-1(x^2-x+1)(7x^2+3x+1)^3 Find: f'(0) From the solution, first evaluate the factors at x=0: - 2^0+2^0=2 - 0=0 - ^-1(1)= 4= 2 - (7 0^2+3 0+1)^3=1 Hence, f(0)= 2 0 21=0 The extracted solution then applies differentiation and concludes directly that f'(0)= Therefore, the correct option is C. Note: the question shows f(x)=(2x+2^-x) (x) ^-1(x^2-x+1), while the solution works with a different function having numerator 2^x+2^-x and denominator (7x^2+3x+1)^3. Since the solution explicitly concludes option C, the answer has been derived from the solution as instructed, but the source contains a question-solution mismatch.

Suppose f(x) = (2x + 2^-x) (x) ^-1(x^2 - x + 1). The value of f'(… — Solution | JEE Mathematics

MCQMediumJEE 2024Derivatives of Functions

Suppose $f(x) = (2x + 2^{-x}) \tan(x) \sqrt{\tan^{-1}(x^2 - x + 1)}$ . The value of $f'(0)$ is equal to:

A
$\pi$
B
$0$
C
$\sqrt{\pi}$
D
$\pi/2$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

f(x)=\frac{(2^x+2^{-x})\tan x\sqrt{\tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^3}

Find: $f'(0)$

From the solution, first evaluate the factors at $x=0$ :

$2^0+2^0=2$
$\tan 0=0$
$\sqrt{\tan^{-1}(1)}=\sqrt{\frac{\pi}{4}}=\frac{\sqrt{\pi}}{2}$
$(7\cdot 0^2+3\cdot 0+1)^3=1$

Hence,

f(0)=\frac{2\cdot 0\cdot \frac{\sqrt{\pi}}{2}}{1}=0

The extracted solution then applies differentiation and concludes directly that

f'(0)=\sqrt{\pi}

Therefore, the correct option is C.

Note: the question shows $f(x)=(2x+2^{-x})\tan(x)\sqrt{\tan^{-1}(x^2-x+1)}$ , while the solution works with a different function having numerator $2^x+2^{-x}$ and denominator $(7x^2+3x+1)^3$ . Since the solution explicitly concludes option C, the answer has been derived from the solution as instructed, but the source contains a question-solution mismatch.

Common mistakes

Using the question expression and the solution expression as if they were identical. They are not the same here, so the derivative setup changes. Always check whether the worked solution matches the printed question before copying the method.
Differentiating $\sqrt{\tan^{-1}(x^2-x+1)}$ incorrectly by ignoring the chain rule. The outer square root and the inner inverse tangent both require differentiation. Differentiate layer by layer.
Substituting $x=0$ into $f(x)$ and concluding that $f'(0)=0$ because $f(0)=0$ . A function value being zero does not force its derivative to be zero. Compute the derivative or the defining limit separately.

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Answer

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