NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12 2 Solutions

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Class 12 Maths NCERT Solutions Chapter 12 Linear Programming Exercise 12.2 is provided in the article. Class 12 Chapter 12 Linear Programming Exercises include questions on Different Types of Linear Programming Problems:

Manufacturing problems
Diet problems
Transportation problems

Download PDF NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2

Check out the solutions of Class 12 Maths NCERT solutions chapter 12 Linear Programming Exercise 12.2

Also check other Exercise Solutions of Class 12 Maths Chapter 12 Linear Programming

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Chapter 12 Linear Programming Important Topics
Optimization	Class 12 Mathematics Chapter 12 Linear Programming	Graphical method linear programming

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Class 12 Mathematics Important Guides
Class 12 Mathematics Notes	Math MCQs	Trigonometry
Number Systems	Probability and Statistics	Arithmetic
Algebra	Difference Between Topics in Maths	Math Study Guides
Math Formula	NCERT Solutions for Class 12 Maths	Mensuration

CBSE CLASS XII Related Questions

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

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2.
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.

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

\( e^{-3} \)
\( -1 \)
\( 1 \)
\( -e^3 \)

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4.
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.

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5.
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

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6.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]

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NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.2 Solutions

CBSE CLASS XII Related Questions

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

2.
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.

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

4.
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.

5.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

6.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]

Similar Mathematics Concepts

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NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.2 Solutions

CBSE CLASS XII Related Questions

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

2.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.

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

4. 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.

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

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

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

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

2.
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.

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

4.
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.

5.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

6.
If \[ \frac{d}{dx}(F(x))=\frac{1}{e^x+1}, \] then find \(F(x)\), given that \[ F(0)=\log\left(\frac{1}{2}\right). \]