NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.3

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Class 11 Maths NCERT Solutions Chapter 14 Mathematical Reasoning Exercise 14.3 are based on the following concepts:

New statements from old statements
Implications
Validating Statements
Conjunction

Download PDF NCERT Solutions for Class 11 Chapter 14 Mathematical Reasoning Exercise 14.3

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Chapter Related Topics
Irrational Number	Prime Number	Real Number

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CBSE CLASS XII Related Questions

1.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.
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.
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.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:
- \( e^{-3} \)
- \( -1 \)
- \( 1 \)
- \( -e^3 \)
View Solution
5.
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
6.
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

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NCERT Solutions for Class 11 Maths Chapter 14: Mathematical Reasoning

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NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.1

NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.2

NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.4

NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.5

NCERT Solutions for Class 11 Maths Chapter 14 Miscellaneous Exercises

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NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.3

CBSE CLASS XII Related Questions

1.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

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

5.
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.

6.
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): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

Similar Mathematics Concepts

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NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.3

CBSE CLASS XII Related Questions

1.Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

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

5.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.

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

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

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

5.
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.

6.
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): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]