If f x -sin -1 x -2 tan -1 x - x 2-4 x -1- then minimum value of f x isA- -2 B- -2C- -2D- 2-

The correct option is B ( π + 2)Domain = [ 1, 1] and f(x) is increasing function ∴ f(x) mini = f ( 1) = π/2 + 2 ( π/4) + 1 4 + 1 = π 2 = (π + 2) ...

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Range of a Function

If f x =sin -...

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If $f (x) = s i n^{- 1} x + 2 t a n^{- 1} x + x^{2} + 4 x + 1$ , then minimum value of f(x) is

π + 2

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2 - π

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(- π + 2)

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π - 2

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f (x) = {sin}^{- 1} x + 2 {tan}^{- 1} x + x^{5} - 5 x^{4} + 5 x^{3} + 1

π / 2 \int 2 / π \frac{{tan}^{- 1} x}{x} d x

	List I		List II
A.	Range of $f (x) = {sin}^{- 1} x + {cos}^{- 1} x + {cot}^{- 1} x$ is	1.	$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
B.	Range of $f (x) = cot - 1 x + {tan}^{- 1} x + c o s e c^{- 1} x$ is	2.	$[\frac{π}{2}, \frac{3 π}{2}]$
C.	Range of $f (x) = {cot}^{- 1} x + {tan}^{- 1} x + {cos}^{- 1} x$ is	3.	${0, π}$
D.	Range of $f (x) = {sec}^{- 1} x + c o s e c^{- 1} x + {sin}^{- 1} x$ is	4.	$(\frac{π}{2}, \frac{3 π}{2})$

x \geq 1 t h e n 2 t a n^{- 1} x + s i n^{- 1} (\frac{2 x}{1 + x^{2}})

If f (x) = 2 {tan}^{- 1} x + {sin}^{- 1} (\frac{2 x}{1 + x^{2}}), x > 1,

f (5)

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