NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.4

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Class 11 Maths NCERT Solutions Chapter 14 Mathematical Reasoning Exercise 14.4 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 14.4

Check out the solutions of Class 11 Maths NCERT Solutions Chapter 14 Mathematical Reasoning Exercise 14.4

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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.
Sports car racing is a form of motorsport which uses sports car prototypes.The competition is held on special tracks designed in various shapes.
The equation of a sports car racing track is given as: \[ f(x)= \begin{cases} x^4-4x^2+4, & 0\leq x<3,\\ x^2+40, & x\geq 3 \end{cases} \] Based on this information:
3.
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
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.
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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.
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

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NCERT Solutions for Class 11 Maths Chapter 14: Mathematical Reasoning

Important Questions for Class 11 Maths Chapter 14 Mathematical Reasoning

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 14.3

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 14.4

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.

3.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

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.
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.
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.

Similar Mathematics Concepts

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NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.4

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.

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

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.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. 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.

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.

3.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

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.
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.
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.