NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2

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Class 11 Maths NCERT Solutions Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2 is based on following concepts:

Argand Plane
Polar representation of a complex number

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CBSE CLASS XII Related Questions

1.
Find the foot of the perpendicular drawn from point $(2, -1, 5)$ to the line \[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \] Also, find the length of the perpendicular.
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2.

A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum.
On the basis of the above information, answer the following questions :
Find a relation between $ x $ and $ y $ such that the surface area $ S $ is minimum.
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3.
Find the absolute maximum and absolute minimum of the function $ f(x) = 2x^3 - 15x^2 + 36x + 1 $ on $ [1, 5] $.
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4.
The integrating factor of the differential equation $ (x + 2y^3) \frac{dy}{dx} = 2y $ is:
- $ e^{y^2} $
- $ \frac{1}{\sqrt{y}} $
- $ e^{-\frac{1}{y^2}} $
- $ e^{y^2} $
View Solution
5.
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.
View Solution
6.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
View Solution

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Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5: Complex Numbers and Quadratic Equations

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NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2

CBSE CLASS XII Related Questions

1.
Find the foot of the perpendicular drawn from point $(2, -1, 5)$ to the line \[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \] Also, find the length of the perpendicular.

3.
Find the absolute maximum and absolute minimum of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \).

4.
The integrating factor of the differential equation \( (x + 2y^3) \frac{dy}{dx} = 2y \) is:

5.
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.

6.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

Similar Mathematics Concepts

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NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2

CBSE CLASS XII Related Questions

1.Find the foot of the perpendicular drawn from point $(2, -1, 5)$ to the line \[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \] Also, find the length of the perpendicular.

3.Find the absolute maximum and absolute minimum of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \).

4.The integrating factor of the differential equation \( (x + 2y^3) \frac{dy}{dx} = 2y \) is:

5.If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.

6.Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the foot of the perpendicular drawn from point $(2, -1, 5)$ to the line \[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \] Also, find the length of the perpendicular.

3.
Find the absolute maximum and absolute minimum of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \).

4.
The integrating factor of the differential equation \( (x + 2y^3) \frac{dy}{dx} = 2y \) is:

5.
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.

6.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.