Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-1/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.1

Question 1.
\(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } +({ sin }y^{ “‘ })\) = 0
Solution:
The highest order derivatives is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } }\)
∴ Order of the equation is 4
The differential equation is not a polynomial equation in its derivatives.
Hence its degree is not defined.

Question 2.
y’ + 5y = 0
Solution:
The highest order derivative is y’
∴ Order is 1
This is a polynomial equation in y’ and the
Hence its degree is 1.

Question 3.
\({ \left( \frac { ds }{ dt } \right) }^{ 4 }+3s{ \left( \frac { { d }^{ 2 }s }{ { dt }^{ 2 } } \right) }\) = 0
Solution:
The highest order derivative is \(\frac{d^{2} s}{d t^{2}}\)
∴ Order is 2
This is a polynomial equation in \(\frac{d^{2} s}{d t^{2}}\) and \(\frac{ds}{dt}\) and the power of \(\frac{d^{2} s}{d t^{2}}\) is one
Hence its degree is one.

Question 4.
\({ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }+cos\left( \frac { dy }{ dx } \right) =0\)
Solution:
The highest order derivatives is \(\frac{d^{2} y}{d x^{2}}\)
∴ Order of the equation is 2
The differential equation is not a polynomial equation in its derivatives.
Hence its degree is not defined.

Question 5.
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } }\) = cos3x + sin3x
Solution:
The highest order derivative is \(\frac{d^{2} y}{d x^{2}}\)
∴ Order is 2
This is a polynomial equation in \(\frac{d^{2} y}{d x^{2}}\) and the power of \(\frac{d^{2} y}{d x^{2}}\) is one
Hence its degree is one.

Question 6.
\({ { (y }^{ ”’ }) }^{ 2 }+{ { ( }y^{ ” }) }^{ 3 }+{ { (y }^{ ‘ }) }^{ 4 }+{ y }^{ 5 }\) = 0
Solution:
The highest order derivative is y”‘
∴ Order is 3
This is a polynomial equation in y”‘, y” and y’ the power of y” is two.
Hence its degree is 2.

Question 7.
y”‘ + 2y” + y’ = 0
Solution:
The highest order derivative is y”‘
∴ Order is 3
This is a polynomial equation in y”‘, y” and y’ and the power of y”‘ is one.
Hence its degree is 1.

Question 8.
y’ + y = e^x
Solution:
The highest order derivative is y’.
∴ Order is 1
This is a polynomial equation in y’ and the power of y’ is one.
Hence its degree is 1.

Question 9.
y” + (y’) +2y = 0
Solution:
The highest order derivative is y”
∴ Order is 2
This is a polynomial equation in y” and
y’ and the degree of y” is one.
Hence its degree is 1.

Question 10.
y” + 2y’ + sin y = 0
Solution:
The highest order derivative is y”.
∴ Order is 2
This is a polynomial equation in y”, y’ and the power of y” is one.
Hence its degree is 1.

Question 11.
The degree of the differential equation
\({ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 3 }{ +\left( \frac { dy }{ dx } \right) }^{ 2 }+sin{ \left( \frac { dy }{ dx } \right) }+1=0\)
(a) 3
(b) 2
(c) 1
(d) not defined
Solution:
The differential equation is not a polynomial equation in derivatives. Hence the degree is not defined.

Question 12.
The order of the differential equation
\({ 2x }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } +y=0\)
(a) 2
(b) 1
(c) 0
(d) not defined
Solution:
The highest order derivative is \(\frac{d^{2} y}{d x^{2}}\)
∴ Order is 2.

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-2/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.2

10.2 Class 12 Question 1.
Compute the magnitude of the following vectors:
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 2i } -\hat { 7j } -\hat { 3k } \)
\(\overrightarrow { c } =\frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } -\frac { 1 }{ \sqrt { 3 } } \hat { k } \)
Solution:

Question 2.
Write two different vectors having same magnitude.
Solution:
Let \(\overrightarrow { a } =\hat { i } +\hat { 2j } +\hat { 3k } ,\overrightarrow { b } =\hat { 3i } +\hat { 2j } +\hat { k } \)
\(|\vec{a}|=\sqrt{1+1+9}=\sqrt{11}\)
\(|\vec{b}|=\sqrt{9+1+1}=\sqrt{11}\)
∴ \(\overline{a}\) and \(\overline{b}\) are examples for two vectors having the same magnitude.
There are infinitely many Vectors having same magnitude.

Question 3.
Write two different vectors having same direction.
Solution:
\(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{3i}+\hat{3j}+\hat{3k}\) are examples for two vectors having the same direction. Generally, if \(\vec{a}\) is a nonzero vector, then \(\vec{a}\) and λ\(\vec{a}\) have the same direction whenever λ is positive.

Question 4.
Find the values of x and y so that the vectors \(2 \hat{i}+3 \hat{j} \text { and } x \hat{i}+y \hat{j}\) are equal.
Solution:
We are given \(2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}\)
If vectors are equal, then their respective components are equal. Hence x = 2, y = 3.

Question 5.
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Solution:
Let A(2, 1) be the initial point and B(-5,7) be the terminal point \(\overrightarrow { AB } =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \hat { i } +\left( { y }_{ 2 }-{ y }_{ 1 } \right) \hat { j } =-\hat { 7i } +\hat { 6j } \)
∴ The vector components are \(\vec{-7i}\), \(\vec{6j}\) and scalar components are – 7 and 6.

Question 6.
Find the sum of three vectors:
\(\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,\)
Solution:
\(\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k }\)

Question 7.
Find the unit vector in the direction of the vector \(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k } \).
Solution:

Question 8.
Find the unit vector in the direction of vector \(\overrightarrow { PQ }\), where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.
Solution:

Question 9.
For given vectors \(\overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k }\) find the unit vector in the direction of the vector \(\overrightarrow { a } +\overrightarrow { b }\)
Solution:

Question 10.
Find a vector in the direction of \(5\hat { i } -\hat { j } +2\hat { k }\) which has magnitude 8 units.
Solution:

Question 11.
Show that the vector \(2 \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are collinear.
Solution:
\(\overrightarrow { a } =2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { b } =-4\hat { i } +6\hat { j } -8\hat { k } \)
\(=-2(2\hat { i } -3\hat { j } +4\hat { k } ) \)
vector \(\vec{a}\) and \(\vec{b}\) have the same direction they are collinear.

Question 12.
Find the direction cosines of the vector \(\hat { i } +2\hat { j } +3\hat { k }\)
Solution:

Question 13.
Find the direction cosines of the vector joining the points A (1, 2, – 3) and B(- 1, – 2, 1), directed from A to B.
Solution:

∴ Direction cosines of \(\overline{AB}\) = Scalar components of \(\overline{AB}\)
= \(\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}\)

Question 14.
Show that the vector \(\hat { i } +\hat { j } +\hat { k }\) are equally inclined to the axes OX, OY, OZ.
Solution:
Let \(\vec{r}\) = \(\hat { i } +\hat { j } +\hat { k }\)
The direction ratios of \(\vec{r}\) are 1, 1, 1.
The direction cosines of \(\vec{r}\) are
\(\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}\)
\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
∴ \(\vec{r}\) is equally inclined to the axes.

Question 15.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \(\hat{i}+2 \hat{j}-\hat{k}\) and \(-\hat{i}+\hat{j}+\hat{k}\) respectively, in the ratio 2:1
i. internally
ii. externally
Solution:
i. Internal division
Let \(\vec{a}\) = position vectorof P = \(\hat{i}+2 \hat{j}-\hat{k}\)
Let \(\vec{b}\) = position vectorof Q = \(-\hat{i}+\hat{j}+\hat{k}\)
Let R divides PQ internally in the ratio 2 : 1

ii. External division
Let S divides PQ externally in the ratio 2 : 1

Question 16.
Find position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).
Solution:
Let R be the midpoint of PQ

Question 17.
Show that the points A, B and C with position vector \(\overrightarrow { a } =3\hat { i } -4\hat { j } -4\hat { k } ,\overrightarrow { b } =2\hat { i } -\hat { j } +\hat { k } and\quad \overrightarrow { c } =\hat { i } -3\hat { j } -5\hat { k }\) respectively form the vertices of a right angled triangle.
Solution:

Question 18.
In triangle ABC (fig.), which of the following is not

(a) \(\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 } \)
(b) \(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 } \)
(c) \(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { CA } =\overrightarrow { 0 } \)
(d) \(\overrightarrow { AB } -\overrightarrow { CB } +\overrightarrow { CA } =\overrightarrow { 0 } \)
Solution:
By the triangle law of vector addition,
\(\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 } \)
\(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 } \), is not true.

Question 19.
If \(\overrightarrow { a } ,\overrightarrow { b } \) are two collinear vectors then which of the following are incorrect:
(a) \(\overrightarrow { b } =\lambda \overrightarrow { a } \), for some scalar λ.
(b) \(\overrightarrow { a } =\pm \overrightarrow { b } \)
(c) the respective components of \(\overrightarrow { a } ,\overrightarrow { b } \) are proportional.
(d) both the vectors \(\overrightarrow { a } ,\overrightarrow { b } \) have same direction, but different magnitudes.
Solution:
If \(\vec{a}\) and \(\vec{b}\) are collinear, then they need not be in the same direction, d and b may have opposite directions.

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-8-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise

Question 1.
Find the area under the given curves and given lines:
i. y = x², x = 1, x = 2 and x – axis
ii. y = x⁴, x = 1, x = 5 and x-axis
Solution:

i. The given curves are y = x² , x = 1 and x = 2. The required area lies between the lines x = 1 and x = 2. Hence the limits of integration are 1 and 2.

ii. The given curves are y = x⁴, x = 1 and x = 5. The required area lies between the lines x = 1 and x = 5. Hence the limits of integration are 1 and 5.

Question 2.
Find the area between the curves y = x and y = x².
Solution:

y = x … (1)
y = x² … (2)
(2) → x = x²
⇒ x² – x = 0 ⇒ x(x – 1) = 0
⇒ x = 0 and x = 1
Hence the limits of integration are 0 and 1. Required area = (Area under the line y = x) – (Area under the parabola y = x²)

Question 3.
Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.
Solution:
The given curves are y = 4x²
⇒ x² = \(\frac { y }{ 4 }\) ⇒ x = \(\frac{1}{2} \sqrt{y}\)
x = 0, y= 1, y = 4
The required area lies in the first quadrant between the lines y= 1 and y = 4

Hence the limits of integration are 1 and 4.
Required area = Shaded area

Question 4.
Sketch the graph of y = |x + 3| and evaluate \(\int_{-6}^{0}|x+3| d x\).
Solution:
y = |x + 3| is redefined as follows,

Question 5.
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Solution:
The area bounded by the curve y = sinx between x = 0 and x = 2π is shaded in the figure.1

The area bounded by the curve between x = 0 and x = π is same as the area bounded by it between x = π and x = 2π
Required area 2 x Area of the shaded
region above x – axis between x = 0
and π = 2\(2 \int_{0}^{\pi} y d x=2 \int_{0}^{\pi} \sin x d x\)
= 2\([-\cos x]_{0}^{\pi}=-2[\cos x]_{0}^{\pi}\)
= – 2(cos π – cos0) = – 2(- 1 – 1) = 4 sq.unit

Question 6.
Find the area enclosed between the parabola y² = 4ax and the line y = mx.
Solution:

The curves are y² = 4ax … (1)
and y = mx … (2)
Solving (1) and (2), we get
m²x² = 4ax ⇒ m²x² – 4ax = 0
i.e., x(m²x – 4a) = 0 ⇒ x = 0 or x = \(\frac{4 a}{m^{2}}\)
The x coordinate of the point of intersection
are x = 0 and x = \(\frac{4 a}{m^{2}}\)
Required area = (Area under the parabola y² = 4ax) – (Area under the line y = mx)

Question 7.
Find the area enclosed by the parabola 4y = 3x² and the line 2y = 3x+ 12.
Solution:

The curves are 4y = 3x²

∴ The points of intersection are (4, 12) and (-2, 3)
Required area is shaded in the figure

Question 8.
Find the area of the smaller region bounded by the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 and the line \(\frac{x}{3}+\frac{y}{2}=1\)
Solution:

\(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 is the ellipse with centre at origin intersecting the positive x-axis at A(3, 0) and positive y-axis at B(0, 2)

The required area is shaded in the figure.

Question 9.
Find the area of the smaller region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) and the line \(\frac{x}{a}+\frac{y}{b}\) = 1.
Solution:

The ellipse intersects the positive x-axis at A (a, 0) and y-axis at B(0, b).

Required area is shaded in the figure.
Required area = Area under the ellipse in the first quadrant – Area under the line AB

Question 10.
Find the area of the region enclosed by the parabola x² = y, the line y = x + 2 and the x-axis.
Solution:
The given curves are x² = y … (1)
and y = x + 2 … (2)

(1) → x² = x + 2 ⇒ x² – x – 2 = 0
⇒ (x – 2)(x + 1) = 0 ⇒ x = 2, x = – 1
When x = 2, y = 4
When x = – 1, y = 1
The points of intersection are (-1, 1) and (2,4)
Required area is shaded in the figure

Question 11.
Using the method of integration find the area bounded by the curve |x| + |y| = 1.
Solution:

The given curve is |x| + |y| = 1
x + y = 1 … (1)
x – y = 1 … (2)
– x + y = 1 … (3)
– x – y = 1 … (4)
Required area is symmetric w.r.t. both the axes.
Required area = 4 x Area of shaded region
= \(4 \int_{0}^{1}(1-x) d x=4\left[x-\frac{x^{2}}{2}\right]_{0}^{1}\)
= \(4\left(1-\frac{1}{2}\right)=4\left(\frac{1}{2}\right)\)
= 2 sq.units

Question 12.
Find the area bounded by curves {(x, y) : y ≥ x² and y = |x|}
Solution:
The area bounded by the curves y ≥ x² and y = |x| is same as the area bounded by y = x², y = |x| are same

The point B is obtained by solving y = x
and y = x²
i.e., x = x² ⇒ x² – x = 0
x(x – 1) = 0 ⇒ x = 0 or x = 1
When x = 0, y = 0 and when x = 1, y = 1
∴ B is(1, 1)
Required area = 2 x (Area of the shaded portion in the first quadrant)
= 2[(Area under the line y = x) – (Area under the parabola y = x²)]
= \(2\left[\int_{0}^{1} x d x-\int_{0}^{1} x^{2} d x\right]=2\left[\left[\frac{x^{2}}{2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}\right]\)
= \(2\left(\frac{1}{2}-\frac{1}{3}\right)=2\left(\frac{1}{6}\right)=\frac{1}{3} \text { sq.unit }\)

Question 13.
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6,3).
Solution:

Equation of BC: y – 5 = \(\left(\frac{3-5}{6-4}\right)\)(x – 4)
y – 5 = – (x – 4) ⇒ y = 9 – x
Equation of AC : y – 0 = \(\left(\frac{3-0}{6-2}\right)\)(x – 2)
y = \(\frac{3}{4}\)(x – 2)
Area of ∆ABC = Area under AB + Area under BC – Area under AC

= 5 + 8 – 6 = 7 sq. unit

Question 14.
Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
Solution:

The given lines are 2x + y = 4 … (1)
3x – 2y = 6 … (2)
x – 3y + 5 = 0 … (3)
Solving (1) and (2), we get x = 2, y = 0
Let A be the point (2, 0)
Solving (2) and (3), we get x = 4, y = 3
Let B be the point (4, 3)
Solving (1) and (3), we get x = 1, y = 2
Let C be the point (1, 2)
Equation of AB is y = \(\frac{3x-6}{2}\)
Equation of BC is y = \(\frac{x+5}{3}\)
Equation of AC is y = 4 – 2x
Area of ∆ABC Area of trapezium DCBE – Area of ∆DCA – Area of ∆ABE

Question 15.
Find the area of the region \(\left\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\right\}\)
Solution:

circle with centre at origin and intersecting positive x-axis at C(\(\frac{3}{2}\), 0)
Solving 4x² + 4y² = 9 and y² = 4x, we get 4x² + 16x = 9 ⇒ 4x² + 16x – 9 = 0

Area of the shaded region = Area under OA + Area under AC

The required region is symmetric with respect to jt-axis.
∴ Required area = 2 x shaded area
= \(2\left(\frac{\sqrt{2}}{12}+\frac{9 \pi}{16}-\frac{9}{8} \sin ^{-1}\left(\frac{1}{3}\right)\right)\)
= \(\frac{\sqrt{2}}{6}+\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right) \text { sq.unit }\)

Question 16.
Area bounded by the curve y = x³, the x- axis and the ordinates x = – 2 and x = 1 is
a. – 9
b. \(\frac{-15}{4}\)
c. \(\frac{15}{4}\)
d. \(\frac{17}{4}\)
Solution:

Question 17.
The area bounded by the curve y = x|x|, x-axis and the ordinates x = – 1 and x = 1 is given by
a. 0
b. \(\frac{1}{3}\)
c. \(\frac{2}{3}\)
d. \(\frac{4}{3}\)
Solution:
y = x|x| is redefined as

Question 18.
The area of the circle x² + y² = 16 exterior to the parabola y² = 6x is
a. \(\frac{4}{3}(4 \pi-\sqrt{3})\)
b. \(\frac{4}{3}(4 \pi+\sqrt{3})\)
c. \(\frac{4}{3}(8 \pi-\sqrt{3})\)
d. \(\frac{4}{3}(8 \pi+\sqrt{3})\)
Solution:
c. \(\frac{4}{3}(8 \pi-\sqrt{3})\)

x² + y² = 16 is a circle with centre at origin
and intersecting positive x-axis at C(4,0).
Solving x² + y² = 16 and y² = 6x, we get
x² + 6x = 16
x² + 6x – 16 = 0 ⇒ (x + 8)(x – 2) = 0
∴ x = – 8 and x = 2
But x = – 8 is not possible ∴ x = 2
When x = 2, y² = 12 ∴ y = ±2\(\sqrt{3}\)
The points of intersection are
A(2, \(\sqrt{3}\)) and B(2, -2\(\sqrt{3}\))
The required area is shaded in the figure.
The nonshaded region OACBO is symmetric w.r.t. x-axis and lies between x = 0 and x = 4.
∴ Area of the nonshaded region = 2 x Area of the nonshaded region in the first quadrant.
= 2 [Area under OA + Area under AC]

Question 19.
The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ \(\frac { π }{ 2 }\) is
a. 2(\(\sqrt{2}\) – 1)
b. \(\sqrt{2}\) – 1
c. \(\sqrt{2}\) + 1
d. \(\sqrt{2}\)
Solution:
b. \(\sqrt{2}\) – 1

The curves are y = sinx … (1)
and y = cos x … (2)
From (1) and (2), we get sin x = cos x
⇒ x = \(\frac { π }{ 4 }\)
∴ y = sin\(\frac { π }{ 4 }\) = \(\frac{1}{\sqrt{2}}\)
Point of intersection of curves is (\(\frac { π }{ 4 }\), \(\frac{1}{\sqrt{2}}\))
Required area = Area of the shaded region = (Area under the curve y = cos x) – (Area above the curve y = sinx)

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-6/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.6

Question 1.
\(\frac { dy }{ dx }\) + 2y = sin x
Solution:
\(\frac { dy }{ dx }\) + 2y = sin x is a linear differential equation.
∴ p = 2 and Q = sin x

is the general solution.

Question 2.
\(\frac { dy }{ dx } +3y={ e }^{ -2x }\)
Solution:
\(\frac { dy }{ dx } +3y={ e }^{ -2x }\) is a linear differential equation.
∴ p = 3 and Q = e^-2x

y = e^-2x + Ce^-3x is the general solution.

Question 3.
\(\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }\)
Solution:

Question 4.
\(\frac { dy }{ dx } +(secx)y=tanx\left( 0\le x<\frac { \pi }{ 2 } \right) \)
Solution:
Here, P = secx, Q = tanx; \(IF={ e }^{ \int { p.dx } }={ e }^{ \int { secx.dx } }\)
\(={ e }^{ log|secx+tanx| }\)
= sec x + tan x
i.e., The solu. is y.× I.F. = ∫Q × I.F. dx + c
or y × (secx+tanx) = ∫tanx(secx+tanx)dx+c
Required solution is
∴ y(secx + tanx) = (secx + tanx) – x + c

Question 5.
\(\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)\)
Solution:
\(\cos ^{2} x \frac{d y}{d x}+y=\tan x\)
i.e., \(\frac{d y}{d x}+y \sec ^{2} x=\sec ^{2} x \tan x\) is a linear differential equation.

y = (tan x – 1) + Cr^-tanx is the general solution.

Question 6.
\(x \frac{d y}{d x}+2 y=x^{2} \log x\)
Solution:

y = \(\frac{x^{2}}{16}(4 \log x-1)+C x^{-2}\) is the general solution.

Question 7.
\(xlogx\frac { dy }{ dx } + y=\frac { 2 }{ x } logx\)
Solution:

Question 8.
(1 + x²)dy + 2xy dx = cotx dx(x ≠ 0)
Solution:

Question 9.
\(x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)\)
Solution:

Question 10.
\((x+y)\frac { dy }{ dx }\) = 1
Solution:

Question 11.
y dx + (x – y²)dy = 0
Solution:

is the general solution.

Question 12.
\(\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)\)
Solution:

Question 13.
\(\frac { dy }{ dx } +2ytanx=sinx,y=0\quad when\quad x=\frac { \pi }{ 3 } \)
Solution:
\(\frac { dy }{ dx }\) + 2y tanx = sin x is a linear differential equation.
∴ p = 2 tanx, Q = sin x

Question 14.
\(\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } +2xy=\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\quad when\quad x=1\)
Solution:

Question 15.
\(\frac { dy }{ dx } -3ycotx=sin2x,y=2\quad when\quad x=\frac { \pi }{ 2 } \)
Solution:
\(\frac { dy }{ dx }\) – 3y cot x = sin 2x is a linear differential equation.
∴ p = – 3 cot x, Q = sin 2x

y cosec³x = – 2 cosec + C … (1)
When x = \(\frac { π }{ 2 }\), y = 2
(1) → 2 x 1 = – 2 + C ∴ C = 4
Hence the particular solution is
y cosec³x = – 2 cosec + 4
y = \(\frac{-2}{cosec^{2} x}\) + \(\frac{4}{cosec^{3} x}\)
y = 4 sin³x – 2 sin²x

Question 16.
Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point
Solution:
\(\frac { dy }{ dx }\) = x + y
\(\frac { dy }{ dx }\) – y = x, is a linear differential equation
∴ P = – 1, Q = x

The curve is passing through the origin
∴ x = 0, y = 0
(1) → C = 1
i.e., x + y + 1 = e^x is the required equation of the curve.

Question 17.
Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5
Solution:
x + y = \(\frac { dy }{ dx }\)
\(\frac { dy }{ dx }\) – y = x – 5, is a linear differential equation
∴ P = – 1, Q = x – 5
∫P dx = ∫-1 dx = – x
I.F = e^∫pdx = e^-x
∴ Solution of the differential equation is

Given that the cure pass through the point (0, 2)
(1) → 2 = 4 + C ∴ C = – 2
y = 4 – x – 2e^x is the required equation of the curve.

Question 18.
The integrating factor of the differential equation \(x\frac { dy }{ dx } -y={ 2x }^{ 2 }\)
(a) \({ e }^{ -x }\)
(b) \({ e }^{ -y }\)
(c) \(\frac { 1 }{ x } \)
(d) x
Solution:

Question 19.
The integrating factor of the differential equation \(\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay\)(-1<y<1) is
(a) \(\frac { 1 }{ { y }^{ 2 }-1 } \)
(b) \(\frac { 1 }{ \sqrt { { y }^{ 2 }-1 } } \)
(c) \(\frac { 1 }{ 1-{ y }^{ 2 } } \)
(d) \(\frac { 1 }{ \sqrt { { 1-y }^{ 2 } } } \)
Solution:
(d) \(\frac { 1 }{ \sqrt { { 1-y }^{ 2 } } } \)

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-4/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.4

Class 12 Maths Chapter 9 Exercise 9.4  Question 1.
\(\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx } \)
Solution:

y = 2tan\(\frac { x }{ 2 }\) – x + C, is the required solution.

Exercise 9.4class 12 NCERT Solutions Question 2.
\(\frac{d y}{d x}=\sqrt{4-y^{2}} \quad(-2<y<2)\)
Solution:

∴ y = 2sin (x + C), is the general solution.

Exercise 9.4 Class 12 NCERT Solutions Question 3.
\(\frac { dy }{ dx } +y=1(y\neq 1)\)
Solution:

1 – y = Ce^-x ⇒ y = 1 – Ce^-x
⇒ y = + Ae^-x, where A = – C is the general solution.

Ex 9.4 Class 12 NCERT Solutions Question 4.
sec² x tany dx+sec² y tanx dy = 0
Solution:
sec² x tany dx+sec² y tanx dy = 0
sec² x tan x dy = – sec² x tan y dx

is the required general solution.

Ex 9.4 Class 12 Maths NCERT Solutions Question 5.
\(\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0\)
Solution:
\(\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right)\)dx = 0
(e^x + e^-x)dy = (e^x – e^-x)dx
i.e., dy = \(\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)\)dx
Integrating both sides we get,
\(\int d y=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x\)
i.e., y = log|e^x + e^-x| + C is the general solution.

Ex 9.4 Class 12 Learn Cbse NCERT Solutions Question 6.
\(\frac { dy }{ dx } =\left( { 1+x }^{ 2 } \right) \left( { 1+y }^{ 2 } \right) \)
Solution:
\(\frac { dy }{ { 1+y }^{ 2 } } =\left( { 1+x }^{ 2 } \right) dx \)
Integrating on both side we get
\({ tan }^{ -1 }y={ x+\frac { 1 }{ 3 } }x^{ 3 }+c \)
which is the required general solution.

Question 7.
y logy dx – x dy = 0
Solution:
y logy dx – x dy = 0
y logy dx = x dy
i.e., \(\frac{d x}{x}=\frac{d y}{y \log y}\)
Integrating both sides, we get

y = e^Cx is the general solution.

Question 8.
\({ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }\)
Solution:

Question 9.
\(\frac { dy }{ dx } ={ sin }^{ -1 }x\)
Solution:
\(\frac { dy }{ dx } ={ sin }^{ -1 }x\) ∴ dy = sin^-1x dx
Integrating both sides, we get

∴ y = x sin^-1x + \(\sqrt{1-x^{2}}\) + C, is the general solution.

Question 10.
\({ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0\)
Solution:

Question 11.
\(\left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x\) y = 1 when x = 0
Solution:

Put x = – 1, we get 2A = 1 ∴ A = \(\frac { 1 }{ 2 }\)
Equating coefficients of x², we get
A + B = 2
∴ B = \(\frac { 3 }{ 2 }\)
Put x = 0, we get A + C = 0 ∴ C = – \(\frac { 1 }{ 2 }\)

Question 12.
\(x\left(x^{2}-1\right) \frac{d y}{d x}=1\)y = 0 when x = 2
Solution:

1 = A(x + 1)(x – 1) + Bx(x – 1) + Cx(x + 1)
When x = 0, A = – 1
When x = 1, C = \(\frac { 1 }{ 2 }\)
When x = – 1, B = \(\frac { 1 }{ 2 }\)

Question 13.
cos(\(\frac { dy }{ dx }\)) = a(a ∈ R); y = 1, when x = 0.
Solution:
cos(\(\frac { dy }{ dx }\)) = a
∴ \(\frac { dy }{ dx }\) = cos^-1(a)
i.e., dy = cos^-1(a)dx
Integrating both sides we get
∫dy = ∫cos^-1(a)dx
i.e., y = cos^-1(a) x + C …. (1)
When x = 0, y = 1
∴ 1 = 0 + C
i.e., C = 1
Substituting the value of C in (1), we get
y = x cos^-1(a) + 1
y – 1 = x cos^-1(a)
cos(\(\frac { y-1 }{ x }\)) = a is the required particular solution.

Question 14.
\(\frac { dy }{ dx }\) = y tan x; y = 1, when x = 0.
Solution:
\(\frac { dy }{ dx }\) ∴\(\frac { dy }{ y }\) = tanx dx
Integrating, we get ∫\(\frac { dy }{ dx }\) = ∫tan x dx
⇒ log|y| = log|secx| + log|C|
When x = 0, y = 1
⇒ log 1 = log sec0 + C ⇒ 0 = log1 + C
⇒ C = 0
∴ logy = log sec x
⇒ y = sec x is the required particular solution.

Question 15.
Find the equation of the curve passing through the point (0,0) and whose differential equation y’ = e^x sin x.
Solution:
y’ = e^x sin x
i.e., \(\frac { dy }{ dx }\) = e^x sin x
dy = e^x sinx dx
Integrating both sides, we get
∫dy = ∫e^x sin x dx
i.e., y = \(\frac{e^{x}}{2}[\sin x-\cos x]+C\) … (1)
(1) passess through the point (0, 0)
we get 0 = \(\frac { 1 }{ 2 }\) [0 – 1] + C
∴ C = \(\frac { 1 }{ 2 }\)
∴ (1) → y = \(\frac{e^{x}}{2}[\sin x-\cos x]\) + \(\frac { 1 }{ 2 }\)
i.e., 2y = e^x[sinx – cosx] + 1
2y – 1 = e^x [sinx – cosx], is the equation of the curve.

Question 16.
For the differential equation xy \(\frac { dy }{ dx }\) = (x + 2) (y + 2), find the solution curve passing through the point (1, – 1)
Solution:

y – 2 log |y + 2| = x + 2 log |x| + C
y – x = log (y + 2)² + logx² + C
y – x = log (x(y + 2))² + C … (1)
(1) passess through (1, – 1), we get
– 1 – 1 = log 1 + C
– 2 = C ⇒ C = – 2
(1) → y – x = log x(x(y + 2))² – 2
i.e., y – x + 2 = log(x(y + 2))², is the equation of the curve.

Question 17.
Find the equation of a curve passing through the point (0, -2) given that at any point (pc, y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point
Solution:
According to the question \(y\frac { dy }{ dx } =x\)
\(\Rightarrow \int { ydy } =\int { xdx } \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c\)
0, – 2) lies on it.c = 2
∴ Equation of the curve is : x² – y² + 4 = 0.

Question 18.
At any point (x, y) of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3) find the equation of the curve given that it passes through (- 2,1).
Solution:
The slope of the line points (x, y) and (- 4, 3) is \(\frac { y+3 }{ x+4 }\)
Also the slope of the tangent at any point on the curve is \(\frac { dy }{ dx }\)

Since it passes through the point (-2, 1), we get log(1 + 3) = log (-2 + 4)² + C
log 4 = log 4 + C
⇒ C = 0
∴ log|y + 3| = log(x + 4)²
⇒ y + 3 = (x + 4)² is the equation of the curve.

Question 19.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution:

Question 20.
In a bank principal increases at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years
Solution:
Let P_t be the principal after t years.
The initial principal P₀ = ₹ 100
Rate of interest = r %, time t = 10 years
P₁₀ = 200
Given that \(\frac{d P}{d t}=\frac{r}{100} \mathrm{P} \Rightarrow \frac{d P}{P}=\frac{r}{100} d t\)
Integrating both sides, we get

Question 21.
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years
Solution:
Let P_t be the principal after t years.
The initial principal P₀ = 1000
Rate of interest = 5 %, time t = 10 years
Given that \(\frac{d P}{d t}=\frac{5}{100} \mathrm{P} \Rightarrow \frac{d P}{P}=\frac{1}{20} d t\)
Integrating both sides, we get

Question 22.
In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present
Solution:
Let N₁ be the count of bacteria at any time t. Initial count of bacteria be N₀.
Given that \(\frac { dN }{ dt }\) ∝ N
⇒ \(\frac { dN }{ dt }\) = kN
\(\frac { dN }{ N }\) = k dt
Integrating both sides, we get ∫\(\frac { dN }{ N }\) = ∫k dt
⇒ log N = kt + C … (1)
When t = 0, N = N₀ = 100000
i.e., C = log 100000
(1) ⇒ logN = kt + log 100000
log N – log 100000 = kt
log(\(\frac { N }{ 100000 }\)) = kt … (2)
When t = 2, N = N₂

Question 23.
The general solution of a differential equation \(\frac{d y}{d x}=e^{x+y} \text { is }\)
(a) \({ e }^{ x }+{ e }^{ -y }=c \)
(b) \({ e }^{ x }+{ e }^{ y }=c \)
(c) \({ e }^{ -x }+{ e }^{ y }=c \)
(d) \({ e }^{ -x }+{ e }^{ -y }=c \)
Solution:
(a) \(\frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }\Rightarrow \int { { e }^{ -y }dy } =\int { { e }^{ x }dx } \)
\(\Rightarrow { e }^{ -y }={ e }^{ x }+k\Rightarrow { e }^{ x }+{ e }^{ -y }=C \) is the general solution.

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-1/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.1

Ex 10.1 Class 12 NCERT Solutions Question 1.
Represent graphically a displacement of 40 km, 30° east of north.
Solution:

The vector \(\overrightarrow{OP}\) represents the required displacement of 40 km, 30° east of north.

Exercise 10.1 Class 12 NCERT Solutions Question 2.
Classify the following measures as scalars and vectors.
(i) 10 kg
(ii) 2 metres north- west
(iii) 40°
(iv) 40 watt
(v) 10^-19 coulomb
(vi) 20 m/sec².
Solution:
(i) Mass-scalar
(ii) Directed distance-vector
(iii) Temperature-scalar
(iv) Rate of electricity-scalar
(v) Electric charge-vector
(vi) Acceleration-vector

Question 3.
Classify the following as scalar and vector quantities
(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work.
Solution:
Scalar Quantity: (i) time period (ii) distance (v) work.
Vector Quantity: (iii) force (iv) velocity

Question 4.
In a square, identify the following vectors.

(i) Co-initial
(ii) Equal
(iii) collinear but not equal
Solution:
a. Coinitial
b. Equal
c. Collinear but not equal
(i) Co initial vectors are \(\overrightarrow { a } ,\overrightarrow { d } \)
(ii) Equal Vectors are \(\overrightarrow { b } ,\overrightarrow { d } \)
(iii) Collinear but not equal vectors are \(\overrightarrow { a } ,\overrightarrow { c } \)

Question 5.
Answer the following as true or false:
(i) \(\overrightarrow { a } ,\overrightarrow { -a } \) are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
Solution:
(i) True
(ii) False
(iii) False
(iv) False.