NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7 is covered in this article. Exercise 5.7 emphasizes the topic, Second Order Derivative. NCERT Solutions for Class 12 Maths Chapter 5 will carry a weightage of around 8-17 marks in the CBSE Term 2 Exam 2022. NCERT has provided a total of 17 problems and solutions based on the important topic covered in this exercise.

Download PDF NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

NCERT Solutions for Class 12 Maths Chapter 5: Important Topics

Important topics covered in the Continuity and Differentiability chapter are:

Mean Value Theorem
Rolle’s Theorem
Limits
Euler’s Number
Quotient Rule

Also check: NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

CBSE CLASS XII Related Questions

1.
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
2.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.
View Solution
3.
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
4.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:
- \( e^{-3} \)
- \( -1 \)
- \( 1 \)
- \( -e^3 \)
View Solution
5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]
View Solution
6.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.
View Solution

View All

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2

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You can also check

Exercise 5.1 Solutions	34 Questions (Short Answers)
Exercise 5.2 Solutions	10 Questions(Short Answers)
Exercise 5.3 Solutions	15 Questions ( Short Answers)
Exercise 5.4 Solutions	10 Questions (Short Answers)
Exercise 5.5 Solutions	18 Questions ( Short Answers)
Exercise 5.6 Solutions	11 Questions (Short Answers)
Exercise 5.7 Solutions	17 Questions (Short Answers)
Exercise 5.8 Solutions	6 Questions (Short Answers)
Miscellaneous Exercise Solutions	23 Questions (6 Long Answers, 17 Short Answers)

Continuous Function	Mean Value Theorem	Second Order Derivative
Exponential growth formula	Limits and Continuity	Slope of secant line formula
Rolle’s Theorem	Mean value theorem formula	Difference quotient formula

NCERT Solutions for Class 12 Maths	Maths MCQs	Maths Syllabus
Maths Study Material	Trigonometry	Arithmetic
Mathematics Notes	Permutation and Combination	Algebra

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

Download PDF NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

NCERT Solutions for Class 12 Maths Chapter 5: Important Topics

Other Exercise Solutions of Class 12 Maths Chapter 5 Continuity and Differentiability

CBSE CLASS XII Related Questions

1.
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 \[ 3q(x)-\pi. \]

2.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

3.
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): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

4.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

Download PDF NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7

NCERT Solutions for Class 12 Maths Chapter 5: Important Topics

Other Exercise Solutions of Class 12 Maths Chapter 5 Continuity and Differentiability

CBSE CLASS XII Related Questions

1.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 \[ 3q(x)-\pi. \]

2. A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

3.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): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

4. The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

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

6.Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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 \[ 3q(x)-\pi. \]

2.
A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

3.
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): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

4.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.
Check whether the function \[ f(x)= \begin{cases} \dfrac{|x-3|}{2(x-3)}, & x<3,\\[6pt] \dfrac{x-6}{6}, & x\geq 3 \end{cases} \] is continuous at \(x=3\) or not.