If I = ∫ dx x^ 2 - 2x + 5 = (1)/(2)tan ^ - 1 (f(x)) + C (where, C is the constant of integration) and f(2) = (1)/(2) , then the maximum value of y = f(sin x) x ∈ ℝ is

Correct answer: C. 0 The given integral is I = ∫ dx (x - 1) ^ 2 + 4 = (1)/(2) ^ - 1 ((x - 1)/(2)) + c ∴ f(x) = (x - 1)/(2) i.e. f(sin x) = (sin x - 1)/(2) Hence, max (f(sin x)) = 0

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If I = ∫ dx x^ 2 - 2x + 5 = (1)/(2)tan ^ - 1 (f(x)) + C (where, C...

I = ∫ \dfrac{dx}{x^{2} - 2x + 5}= \dfrac{1}{2}tan{}^{- 1}\left(f\left(x\right)\right) + C

(where,

C

is the constant of integration)

and f\left(2\right) = \dfrac{1}{2}

, then the maximum value of

y = f\left(sin x\right) \forall x \in \mathbb{R}

Options

A $4$
B $2$
C $0$
D $- 1$

Correct answer

C. $0$

Step-by-step solution

The given integral is

I = ∫ \dfrac{dx}{{\left(x - 1\right)}^{2} + 4}

= \dfrac{1}{2}\tan^{- 1} \left(\dfrac{x - 1}{2}\right) + c

∴ f\left(x\right) = \dfrac{x - 1}{2}

i.e.

f\left(sin x\right) = \dfrac{sin x - 1}{2}

Hence, max \left(f\left(sin x\right)\right) = 0

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