If the range of functions fx-cos-1x32-sin-1x3-sin-1x32-24x2-2x-1 is -a-b- then the value of ba is

Let sin^ 1(x3)=t and g(x)=(cos^ 1x3)^2+πsin^ 1x3 (sin^ 1x3)^2 ⇒ g(x)=(π2 t)^2+π t t^2 ⇒ g(x)=π^24 ⇒ f(x)=π^24((x+1)^2+1) Now nbsp; 1≤x3≤1 ⇒ 3≤ x ≤3 ⇒ 2≤ x ...

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Byju's Answer

Standard XII

Mathematics

Inverse Function

If the range ...

Question

If the range of functions $f (x) = {({cos}^{- 1} \frac{x}{3})}^{2} + π {sin}^{- 1} \frac{x}{3} - {({sin}^{- 1} \frac{x}{3})}^{2} + \frac{π^{2}}{4} (x^{2} + 2 x + 1)$ is $[a, b]$ . then the value of $\frac{b}{a}$ is

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Solution

Let ${sin}^{- 1} (\frac{x}{3}) = t$ and $g (x) = {({cos}^{- 1} \frac{x}{3})}^{2} + π {sin}^{- 1} \frac{x}{3} - {({sin}^{- 1} \frac{x}{3})}^{2}$
$\Rightarrow g (x) = {(\frac{π}{2} - t)}^{2} + π t - t^{2} \Rightarrow g (x) = \frac{π^{2}}{4} \Rightarrow f (x) = \frac{π^{2}}{4} ((x + 1)^{2} + 1)$

Now
$- 1 \leq \frac{x}{3} \leq 1 \Rightarrow - 3 \leq x \leq 3 \Rightarrow - 2 \leq x + 1 \leq 4 \Rightarrow 0 \leq (x + 1)^{2} \leq 16 \Rightarrow 1 \leq (x + 1)^{2} + 1 \leq 17 \Rightarrow \frac{π^{2}}{4} \leq \frac{π^{2}}{4} ((x + 1)^{2} + 1) \leq 17 \frac{π^{2}}{4}$

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Similar questions

Q. If the range of functions

f (x) = {({cos}^{- 1} \frac{x}{3})}^{2} + π {sin}^{- 1} \frac{x}{3} - {({sin}^{- 1} \frac{x}{3})}^{2} + \frac{π^{2}}{4} (x^{2} + 2 x + 1)

[a, b]

. then the value of

\frac{b}{a}

Q. If the range of

f (x) = {sin}^{- 1} x + {tan}^{- 1} x + {cos}^{- 1} x

= [\frac{π}{a}, \frac{b π}{c}]

,
Find the value of

a + b + c

Q. Range of

f (x) = {tan}^{- 1} [\frac{2}{π} (2 {tan}^{- 1} x - {sin}^{- 1} x + {cot}^{- 1} x - {cos}^{- 1} x)]

contains

Q. Assertion :Statement 1 Range of

f (x) = {tan}^{- 1} x + {sin}^{- 1} x + {cos}^{- 1} x

(0, π)

Reason: Statement 2

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

for

x ϵ (- 1, 1]

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

(b)

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

where

0 \leq x \leq 1

then

(a, b) =

(a)

(0, π)

(b)

(0, π / 4)

(c)

(- π / 4, π / 4)

(d)

(π / 4, π / 2)

(d) If

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

then (a,b) is equal to

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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If the range of functions f(x)=(cos−1x3)2+πsin−1x3−(sin−1x3)2+π24(x2+2x+1) is [a,b]. then the value of ba is

Let sin−1(x3)=t and g(x)=(cos−1x3)2+πsin−1x3−(sin−1x3)2 ⇒g(x)=(π2−t)2+πt−t2⇒g(x)=π24⇒f(x)=π24((x+1)2+1) Now −1≤x3≤1⇒−3≤x≤3⇒−2≤x+1≤4⇒0≤(x+1)2≤16⇒1≤(x+1)2+1≤17⇒π24≤π24((x+1)2+1)≤17π24

If the range of functions $f (x) = {({cos}^{- 1} \frac{x}{3})}^{2} + π {sin}^{- 1} \frac{x}{3} - {({sin}^{- 1} \frac{x}{3})}^{2} + \frac{π^{2}}{4} (x^{2} + 2 x + 1)$ is $[a, b]$ . then the value of $\frac{b}{a}$ is