NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.2

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Class 11 Maths NCERT Solutions Chapter 6 Linear Inequalities Exercise 6.2 is based on the topic Graphical Solution of Linear Inequalities in Two Variables.

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CBSE CLASS XII Related Questions

1.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]
View Solution
2.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :
- \( \frac{7\pi}{12} \)
- \( \frac{5\pi}{12} \)
- \( -\frac{5\pi}{12} \)
- \( \frac{\pi}{3} \)
View Solution
3.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).
View Solution
4.
\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :
- \( \log \cos 2x + C \)
- \( \sec^{2} 2x + C \)
- \( e^{x}\tan x + C \)
- \( e^{x}\cot x + C \)
View Solution
5.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).
View Solution
6.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :
- \( -6 \) unit/sec
- \( 0 \) unit/sec
- \( 18 \) unit/sec
- \( 6 \) unit/sec
View Solution

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NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.2

CBSE CLASS XII Related Questions

1.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

2.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

3.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

4.
\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :

5.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

6.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

Similar Mathematics Concepts

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NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.2

CBSE CLASS XII Related Questions

1.If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

2.The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

3.Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

4.\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :

5.Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

6.The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

2.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

3.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

4.
\( \int e^{x}\left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx \) is equal to :

5.
Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

6.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :