SATHEE: Limit Continuity and Differentiability 7 Question 27

Limit Continuity and Differentiability 7 Question 27

27. Let $f : (0, π) \to R$ be a twice differentiable function such that $lim_{t \to x} \frac{f (x) \sin t - f (t) \sin x}{t - x} = \sin^{2} x$ for all $x \in (0, π)$ .

If $f \frac{π}{6} = - \frac{π}{12}$ , then which of the following statement(s) is (are) TRUE?

(2018 Adv.)

(a) $f \frac{π}{4} = \frac{π}{4 \sqrt{2}}$

(b) $f (x) < \frac{x^{4}}{6} - x^{2}$ for all $x \in (0, π)$

(d) $f^{''} \frac{π}{2} + f \frac{π}{2} = 0$

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Answer:

Correct Answer: 27. (a, c, d)

Solution:

Given, $y = f \frac{2 x - 1}{x^{2} + 1}$

and $f^{'} (x) = \sin^{2} x$

$\begin{aligned} ∴ \frac{d y}{d x} & = f^{'} \frac{2 x - 1}{x^{2} + 1} \cdot \frac{d}{d x} \frac{2 x - 1}{x^{2} + 1} \\ = \sin^{2} \frac{2 x - 1}{x^{2} + 1} \cdot \frac{(x^{2} + 1) \cdot 2 - (2 x - 1) (2 x)}{{(x^{2} + 1)}^{2}} \end{aligned}$

$\begin{aligned} = \sin^{2} \frac{2 x - 1}{x^{2} + 1} \cdot \frac{- 2 x^{2} + 2 x + 2}{{(x^{2} + 1)}^{2}} \\ = \frac{- 2 (x^{2} - x - 1)}{{(x^{2} + 1)}^{2}} \sin^{2} \frac{2 x - 1}{x^{2} + 1} \end{aligned}$

Limit Continuity and Differentiability 7 Question 27

27. Let $f : (0, π) \to R$ be a twice differentiable function such that $lim_{t \to x} \frac{f (x) \sin t - f (t) \sin x}{t - x} = \sin^{2} x$ for all $x \in (0, π)$ .

Answer:

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Limit Continuity and Differentiability 7 Question 27

27. Let f:(0,π)→R be a twice differentiable function such that limt→xf(x)sin⁡t−f(t)sin⁡xt−x=sin2⁡x for all x∈(0,π).

Answer:

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27. Let $f : (0, π) \to R$ be a twice differentiable function such that $lim_{t \to x} \frac{f (x) \sin t - f (t) \sin x}{t - x} = \sin^{2} x$ for all $x \in (0, π)$ .