Let f:A→B be an onto function defined as fx=sin−1x+tan−1xcos−1x+cot−1x . Then which of the following is/are TRUE?

We have f (x) =π2−cos−1x+π2−cot−1xcos−1x+cot−1x=πcos−1x+cot−1x−1Clearly A=−1,1 . So in −1,1 , cos−1x∈0,πand cot−1x∈π4,3π4 . ∴cos−1x+cot−1x=π4,7π4∀ x∈AAlso cos−1x+cot−1x is strictly decreasing on A ∴m=fxmin=πcos−1x+cot−1xmax−1 =πcos−1−1+cot−1−1−1=−37And M=fxmax=πcos−1x+cos−1xMin.−1 =πcos−11+cot−11−1=3Hence Mm=3−37=−7

Let f:A→B be an onto function defined as fx=sin−1x+tan−1xcos−1x+cot−1x . Then which of the following is/are TRUE?

We have f (x) =π2−cos−1x+π2−cot−1xcos−1x+cot−1x=πcos−1x+cot−1x−1Clearly A=−1,1 . So in −1,1 , cos−1x∈0,πand cot−1x∈π4,3π4 . ∴cos−1x+cot−1x=π4,7π4∀ x∈AAlso cos−1x+cot−1x is strictly decreasing on A ∴m=fxmin=πcos−1x+cot−1xmax−1 =πcos−1−1+cot−1−1−1=−37And M=fxmax=πcos−1x+cos−1xMin.−1 =πcos−11+cot−11−1=3Hence Mm=3−37=−7

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $f : A \to B$ be an onto function defined as $f (x) = \frac{\sin^{- 1} x + \tan^{- 1} x}{\cos^{- 1} x + \cot^{- 1} x}$ . Then which of the following is/are TRUE?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

If the minimum and maximum value of f (x) be m and M respectively then the value of $|\frac{M}{m}|$ is 8

The number of solutions of the equations $f (x^{3} + 14 x^{2} + 13 x - 5) = f (1 - x^{2} + x^{3})$ is 1

The number of solutions of the equations $f (x^{3} + 14 x^{2} + 13 x - 5) = f (1 - x^{2} + x^{3})$ is 2

If the minimum and maximum value of f (x) be m and M respectively then the value of $|\frac{M}{m}|$ is 7

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We have f (x)
$= \frac{\frac{π}{2} - \cos^{- 1} x + \frac{π}{2} - \cot^{- 1} x}{\cos^{- 1} x + \cot^{- 1} x} = \frac{π}{\cos^{- 1} x + \cot^{- 1} x} - 1$
Clearly $A = [- 1, 1]$ . So in $[- 1, 1]$ , $\cos^{- 1} x \in [0, π]$
and $\cot^{- 1} x \in [\frac{π}{4}, \frac{3 π}{4}]$ .
$∴ \cos^{- 1} x + \cot^{- 1} x = [\frac{π}{4}, \frac{7 π}{4}] \forall x \in A$
Also $\cos^{- 1} x + \cot^{- 1} x$ is strictly decreasing on A
$∴ m = f {(x)}_{\min} = \frac{π}{{(\cos^{- 1} x + \cot^{- 1} x)}_{\max}} - 1$
$= \frac{π}{\cos^{- 1} (- 1) + \cot^{- 1} (- 1)} - 1 = \frac{- 3}{7}$
And $M = f {(x)}_{\max} = \frac{π}{{(\cos^{- 1} x + \cos^{- 1} x)}_{M i n .}} - 1$
$= \frac{π}{\cos^{- 1} (1) + \cot^{- 1} (1)} - 1 = 3$
Hence $\frac{M}{m} = \frac{3}{(- \frac{3}{7})} = - 7$