SATHEE: Inverse Circular Functions 2 Question 1

Inverse Circular Functions 2 Question 1

1. Let $f (x) = \log_{e} (\sin x), (0 < x < π)$ and $g (x) = \sin^{- 1} (e^{- x})$ , $(x \geq 0)$ . If $α$ is a positive real number such that $a = (f \circ g)^{'} (α)$ and $b = (f \circ g) (α)$ , then (2019 Main, 10 April II)

(a) $a α^{2} - b α - a = 0$

(b) $a α^{2} - b α - a = 1$

(d) $a α^{2} + b α + a = 0$

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Answer:

Correct Answer: 1. (b)

Solution:

Given functions, $f (x) = \log_{e} (\sin x), (0 < x < π)$ and $g (x) = \sin^{- 1} (e^{- x}), x \geq 0$ . Now, $f \circ g (x) = f (g (x)) = f (\sin^{- 1} (e^{- x}))$

$= \log_{e} (\sin (\sin^{- 1} (e^{- x})))$

$= \log_{e} (e^{- x})$

$∵ \sin (\sin^{- 1} x) = x$ ,

if $x \in [- 1, 1]$

$= - x$

and

$(f \circ g)^{'} (x) = \frac{d}{d x} (- x) = - 1$

According to the question, $∵ a = (f \circ g)^{'} (α) = - 1$ [from Eq. (ii)] [from Eq. (i)]

and $b = (f \circ g) (α) = - (α)$

for a positive real value ’ $α$ ‘.

Since, the value of $a = - 1$ and $b = - α$ , satisfy the quadratic equation (from the given options)

$a α^{2} - b α - a = 1$

Inverse Circular Functions 2 Question 1

1. Let $f (x) = \log_{e} (\sin x), (0 < x < π)$ and $g (x) = \sin^{- 1} (e^{- x})$ , $(x \geq 0)$ . If $α$ is a positive real number such that $a = (f \circ g)^{'} (α)$ and $b = (f \circ g) (α)$ , then (2019 Main, 10 April II)

Answer:

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Inverse Circular Functions 2 Question 1

1. Let f(x)=loge⁡(sin⁡x),(0<x<π) and g(x)=sin−1⁡(e−x), (x≥0). If α is a positive real number such that a=(f∘g)′(α) and b=(f∘g)(α), then (2019 Main, 10 April II)

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. Let $f (x) = \log_{e} (\sin x), (0 < x < π)$ and $g (x) = \sin^{- 1} (e^{- x})$ , $(x \geq 0)$ . If $α$ is a positive real number such that $a = (f \circ g)^{'} (α)$ and $b = (f \circ g) (α)$ , then (2019 Main, 10 April II)