SATHEE: Definite Integration Question 14

Definite Integration Question 14

Question 14

If $g (x) = \int_{\sin x}^{\sin (2 x)} \sin^{- 1} (t) d t$ , then (a) $g^{'} - \frac{π}{2} = 2 π$ (b) $g^{'} - \frac{π}{2} = - 2 π$ (c) $g^{'} \frac{π}{2} = 2 π$ (d) $g^{'} \frac{π}{2} = - 2 π$

Passage Based Questions

Let $f (x) = (1 - x)^{2} \sin^{2} x + x^{2}, \forall x \in R$ and

$g (x) = \int_{1}^{x} \frac{2 (t - 1)}{t + 1} - \ln t f (t) d t \forall x \in (1, \infty)$ .

Show Answer

Answer:

Correct Answer: 14. $(^{*})$

Solution:

$g (x) = \int_{\sin x}^{\sin 2 x} \sin^{- 1} (t) d t$

$\begin{aligned} g^{'} (x) & = 2 \cos 2 x \sin^{- 1} (\sin 2 x) - \cos x \sin^{- 1} (\sin x) g^{'} \frac{π}{2} & = - 2 \sin^{- 1} (0) = 0 g^{'} - \frac{π}{2} & = - 2 \sin^{- 1} (0) = 0 \end{aligned}$

No option is matching.

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Definite Integration Question 14

Question 14

Passage Based Questions

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu