NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 Solutions

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Class 12 Maths NCERT Solutions Chapter 12 Linear Programming Exercise 12.1 is provided in the article. Class 12 Chapter 12 Linear Programming Exercises include questions on Mathematical formulation of the problem and Graphical method of solving linear programming problems.

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Chapter 12 Linear Programming Important Topics
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Class 12 Mathematics Notes	Math MCQs	Trigonometry
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Algebra	Difference Between Topics in Maths	Math Study Guides
Math Formula	NCERT Solutions for Class 12 Maths	Mensuration

CBSE CLASS XII Related Questions

1.
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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2.
For a square matrix \(A\), \[ (3A)^{-1}= \]

\( 3A^{-1} \)
\( 9A^{-1} \)
\( \frac{1}{3} A^{-1} \)
\( \frac{1}{9} A^{-1} \)

View Solution

3.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

\([-1, 1]\)
\([4, 6]\)
\([-7, -3]\)
\([2, 3]\)

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

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6.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

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

View Solution

NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 Solutions

CBSE CLASS XII Related Questions

1.
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.

2.
For a square matrix \(A\), \[ (3A)^{-1}= \]

3.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

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

6.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

Similar Mathematics Concepts

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NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 Solutions

CBSE CLASS XII Related Questions

1. 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.

2. For a square matrix \(A\), \[ (3A)^{-1}= \]

3.The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

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

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

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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.

2.
For a square matrix \(A\), \[ (3A)^{-1}= \]

3.
The domain of \[ f(x)=\cos^{-1}(2x-5) \] is:

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

6.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is: