NCERT Solutions Class 12 Chapter 9 Differential Equations Exercise 9.3 Solutions

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Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.3 is provided in the article. These exercises focus on two key concepts:

Formation of a Differential Equation whose General Solution is given.
Procedure to form a differential equation that will represent a given family of curves

Download PDF: NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3

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
Algebra	Arithmetic	Calculus
Mensuration	Trigonometry	Geometry
Permutation and Combination	Math Study Notes	Math MCQs
NCERT Solutions for Class 12 Maths	Math Formula	Difference between in Maths

CBSE CLASS XII Related Questions

1.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]
View Solution
2.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]
View Solution
3.
Find the domain of \[ q(x)=\cos^{-1}(4x^2-3). \] Hence, find the value of \(x\) for which \[ q(x)=0. \] Also, write the range of \[ 3q(x)-\pi. \]
View Solution
4.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]
- Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
5.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:
- \(\frac{1}{3}\)
- \(\frac{1}{9}\)
- \(3\)
- \(9\)
View Solution
6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]
- local maximum value is 2
- local minimum value is \( -2 \)
- local maximum value is \( -2 \)
- local minimum value \( < \) local maximum value
View Solution

View All

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You can also check

NCERT Solutions Class 12 Chapter 9 Differential Equations Exercise 9.3 Solutions

CBSE CLASS XII Related Questions

1.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

2.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]

3.
Find the domain of \[ q(x)=\cos^{-1}(4x^2-3). \] Hence, find the value of \(x\) for which \[ q(x)=0. \] Also, write the range of \[ 3q(x)-\pi. \]

4.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

5.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

Similar Mathematics Concepts

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NCERT Solutions Class 12 Chapter 9 Differential Equations Exercise 9.3 Solutions

CBSE CLASS XII Related Questions

1. Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

2. If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]

3.Find the domain of \[ q(x)=\cos^{-1}(4x^2-3). \] Hence, find the value of \(x\) for which \[ q(x)=0. \] Also, write the range of \[ 3q(x)-\pi. \]

4.For two vectors \(\vec{a}\) and \(\vec{b}\): Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

5.If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

6. For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

2.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]

3.
Find the domain of \[ q(x)=\cos^{-1}(4x^2-3). \] Hence, find the value of \(x\) for which \[ q(x)=0. \] Also, write the range of \[ 3q(x)-\pi. \]

4.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

5.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:

6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]