NCERT Solutions for Class 12 Chapter 9 Differential Equations Exercise 9.1 Solutions

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Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.1 is provided in the article. Class 12 Chapter 9 Differential Equations Exercises include questions on Order and Degree of Differential Equations, Formation of Differential Equation, Methods of Solving First Order, First Degree Differential Equations.

Download PDF: NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.1

Also check other Exercise Solutions of Class 12 Maths Chapter 9 Differential Equations

Also Read:

Chapter 9 Differential Equations Important Topics
Differential Equations Applications	Difference between parametric and nonparametric	Partial Differential Equation
Ordinary differential equations	Differential Equations Formula	Differential Calculus
Partial Derivative	How to Solve Linear Differential Equation	Homogeneous Differential Equation

Also Read:

CBSE Class 12 Mathematics Study Guides
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CBSE CLASS XII Related Questions

1.
If $M$ and $N$ are square matrices of order 3 such that $\det(M) = m$ and $MN = mI$, then $\det(N)$ is equal to :
- $-1$
- 1
- $-m^2$
- $m^2$
View Solution
2.
If $ \mathbf{a} $ and $ \mathbf{b} $ are position vectors of two points $ P $ and $ Q $ respectively, then find the position vector of a point $ R $ in $ QP $ produced such that \[ QR = \frac{3}{2} QP. \]
View Solution
3.
If $ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 $, $ |\overrightarrow{a}| = \sqrt{37} $, $ |\overrightarrow{b}| = 3 $, and $ |\overrightarrow{c}| = 4 $, then the angle between $ \overrightarrow{b} $ and $ \overrightarrow{c} $ is:
- $ \frac{\pi}{6} $
- $ \frac{\pi}{4} $
- $ \frac{\pi}{3} $
- $ \frac{\pi}{2} $
View Solution
4.
If $ \int \frac{1}{2x^2} \, dx = k \cdot 2x + C $, then $ k $ is equal to:
- $ -1 $
- $ \log 2 $
- $ -\log 2 $
- $ 1/2 $
View Solution
5.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$
View Solution
6.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
View Solution

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You can also check

NCERT Solutions for Class 12 Chapter 9 Differential Equations Exercise 9.1 Solutions

CBSE CLASS XII Related Questions

1.
If $M$ and $N$ are square matrices of order 3 such that $\det(M) = m$ and $MN = mI$, then $\det(N)$ is equal to :

2.
If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

3.
If \( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \), \( |\overrightarrow{a}| = \sqrt{37} \), \( |\overrightarrow{b}| = 3 \), and \( |\overrightarrow{c}| = 4 \), then the angle between \( \overrightarrow{b} \) and \( \overrightarrow{c} \) is:

4.
If \( \int \frac{1}{2x^2} \, dx = k \cdot 2x + C \), then \( k \) is equal to:

5.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

6.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

Similar Mathematics Concepts

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NCERT Solutions for Class 12 Chapter 9 Differential Equations Exercise 9.1 Solutions

CBSE CLASS XII Related Questions

1.If $M$ and $N$ are square matrices of order 3 such that $\det(M) = m$ and $MN = mI$, then $\det(N)$ is equal to :

2.If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

3.If \( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \), \( |\overrightarrow{a}| = \sqrt{37} \), \( |\overrightarrow{b}| = 3 \), and \( |\overrightarrow{c}| = 4 \), then the angle between \( \overrightarrow{b} \) and \( \overrightarrow{c} \) is:

4.If \( \int \frac{1}{2x^2} \, dx = k \cdot 2x + C \), then \( k \) is equal to:

5.Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

6.Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If $M$ and $N$ are square matrices of order 3 such that $\det(M) = m$ and $MN = mI$, then $\det(N)$ is equal to :

2.
If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

3.
If \( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \), \( |\overrightarrow{a}| = \sqrt{37} \), \( |\overrightarrow{b}| = 3 \), and \( |\overrightarrow{c}| = 4 \), then the angle between \( \overrightarrow{b} \) and \( \overrightarrow{c} \) is:

4.
If \( \int \frac{1}{2x^2} \, dx = k \cdot 2x + C \), then \( k \) is equal to:

5.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

6.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.