## Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
(i) xy = log y + c; $$\frac{d y}{d x}=\frac{y^{2}}{1-x y}$$
Solution:
xy = log y + c
Differentiating w.r.t. x, we get

Hence, xy = log y + c is a solution of the D.E.
$$\frac{d y}{d x}=\frac{y^{2}}{1-x y^{\prime}}, x y \neq 1$$

(ii) y = (sin-1x)2 + c; (1 – x2) $$\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2$$
Solution:
y = (sin-1 x)2 + c …….(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

(iii) y = e-x + Ax + B; $$e^{x} \frac{d^{2} y}{d x^{2}}=1$$
Solution:
y = e-x + Ax + B
Differentiating w.r.t. x, we get

∴ $$e^{x} \frac{d^{2} y}{d x^{2}}=1$$
Hence, y = e-x + Ax + B is a solution of the D.E.
$$e^{x} \frac{d^{2} y}{d x^{2}}=1$$

(iv) y = xm; $$x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0$$
Solution:
y = xm
Differentiating twice w.r.t. x, we get

This shows that y = xm is a solution of the D.E.
$$x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0$$

(v) y = a + $$\frac{b}{x}$$; $$x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0$$
Solution:
y = a + $$\frac{b}{x}$$
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

Hence, y = a + $$\frac{b}{x}$$ is a solution of the D.E.
$$x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0$$

(vi) y = eax; x $$\frac{d y}{d x}$$ = y log y
Solution:
y = eax
log y = log eax = ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
$$\frac{1}{y} \cdot \frac{d y}{d x}$$ = a × 1
∴ $$\frac{d y}{d x}$$ = ay
∴ x $$\frac{d y}{d x}$$ = (ax)y
∴ x $$\frac{d y}{d x}$$ = y log y ………[By (1)]
Hence, y = eax is a solution of the D.E.
x $$\frac{d y}{d x}$$ = y log y.

Question 2.
Solve the following differential equations.
(i) $$\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$$
Solution:

(ii) log($$\frac{d y}{d x}$$) = 2x + 3y
Solution:

(iii) y – x $$\frac{d y}{d x}$$ = 0
Solution:
y – x $$\frac{d y}{d x}$$ = 0
∴ x $$\frac{d y}{d x}$$ = y
∴ $$\frac{1}{x} d x=\frac{1}{y} d y$$
Integrating both sides, we get
$$\int \frac{1}{x} d x=\int \frac{1}{y} d y$$
∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.

(iv) sec2x . tan y dx + sec2y . tan x dy = 0
Solution:
sec2x . tan y dx + sec2y . tan x dy = 0
∴ $$\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0$$
Integrating both sides, we get
$$\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=c_{1}$$
Each of these integrals is of the type
$$\int \frac{f^{\prime}(x)}{f(x)} d x$$ = log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c1 = log c
∴ log |tan x . tan y| = log c
∴ tan x . tan y = c
This is the general solution.

(v) cos x . cos y dy – sin x . sin y dx = 0
Solution:
cos x . cos y dy – sin x . sin y dx = 0
$$\frac{\cos y}{\sin y} d y-\frac{\sin x}{\cos x} d x=0$$
Integrating both sides, we get
∫cot y dy – ∫tan x dx = c1
∴ log|sin y| – [-log|cos x|] = log c, where c1 = log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y . cos x| = log c
∴ sin y . cos x = c
This is the general solution.

(vi) $$\frac{d y}{d x}$$ = -k, where k is a constant.
Solution:
$$\frac{d y}{d x}$$ = -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.

(vii) $$\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0$$
Solution:
$$\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0$$
∴ y cos2y dy + x cos2x dx = 0
∴ $$x\left(\frac{1+\cos 2 x}{2}\right) d x+y\left(1+\frac{\cos 2 y}{2}\right) d y=0$$
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c1 ……..(1)
Using integration by parts

Multiplying throughout by 4, this becomes
2x2 + 2y2 + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c1
∴ 2(x2 + y2) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c1
This is the general solution.

(viii) $$y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}$$
Solution:

(ix) 2ex+2y dx – 3 dy = 0
Solution:

(x) $$\frac{d y}{d x}$$ = ex+y + x2 ey
Solution:

∴ 3ex + 3e-y + x3 = -3c1
∴ 3ex + 3e-y + x3 = c, where c = -3c1
This is the general solution.

Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
(i) 3ex tan y dx + (1 + ex) sec2y dy = 0, when x = 0, y = π
Solution:
3ex tan y dx + (1 + ex) sec2y dy = 0

(ii) (x – y2x) dx – (y + x2y) dy = 0, when x = 2, y = 0
Solution:
(x – y2x) dx – (y + x2y) dy = 0
∴ x(1 – y2) dx – y(1 + x2) dy = 0

When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x2)(1 – y2) = 5.

(iii) y(1 + log x) $$\frac{d x}{d y}$$ – x log x = 0, y = e2, when x = e
Solution:
y(1 + log x) $$\frac{d x}{d y}$$ – x log x = 0

(iv) (ey + 1) cos x + ey sin x $$\frac{d y}{d x}$$ = 0, when x = $$\frac{\pi}{6}$$, y = 0
Solution:
(ey + 1) cos x + ey sin x $$\frac{d y}{d x}$$ = 0

$$\int \frac{f^{\prime}(x)}{f(x)} d x$$ = log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|ey + 1| = log c, where c1 = log c
∴ log|sin x . (ey + 1)| = log c
∴ sin x . (ey + 1) = c
When x = $$\frac{\pi}{4}$$, y = 0, we get
$$\left(\sin \frac{\pi}{4}\right)\left(e^{0}+1\right)=c$$
∴ c = $$\frac{1}{\sqrt{2}}$$(1 + 1) = √2
∴ the particular solution is sin x . (ey + 1) = √2

(v) (x + 1) $$\frac{d y}{d x}$$ – 1 = 2e-y, y = 0, when x = 1
Solution:

This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e0 = c(1 + 1)
∴ 3 = 2c
∴ c = $$\frac{3}{2}$$
∴ the particular solution is 2 + ey = $$\frac{3}{2}$$ (x + 1)
∴ 2(2 + ey) = 3(x + 1).

(vi) cos($$\frac{d y}{d x}$$) = a, a ∈ R, y (0) = 2
Solution:
cos($$\frac{d y}{d x}$$) = a
∴ $$\frac{d y}{d x}$$ = cos-1 a
∴ dy = (cos-1 a) dx
Integrating both sides, we get
∫dy = (cos-1 a) ∫dx
∴ y = (cos-1 a) x + c
∴ y = x cos-1 a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos-1 a + 2
∴ y – 2 = x cos-1 a
∴ $$\frac{y-2}{x}$$ = cos-1a
∴ cos($$\frac{y-2}{x}$$) = a

Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
(i) $$\frac{d y}{d x}$$ = cos(x + y)
Solution:

(ii) (x – y)2 $$\frac{d y}{d x}$$ = a2
Solution:

(iii) x + y $$\frac{d y}{d x}$$ = sec(x2 + y2)
Solution:

Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x2 + y2) = 2x + c
This is the general solution.

(iv) cos2(x – 2y) = 1 – 2 $$\frac{d y}{d x}$$
Solution:

Integrating both sides, we get
∫dx = ∫sec2u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.

(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ $$\frac{d y}{d x}=\frac{2(x-y)+3}{(x-y)+1}$$ ………(1)
Put x – y = u, Then $$1-\frac{d y}{d x}=\frac{d u}{d x}$$

∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0

## Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.2 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 1.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) x3 + y3 = 4ax
Solution:
x3 + y3 = 4ax ……..(1)
Differentiating both sides w.r.t. x, we get
3x2 + 3y2 $$\frac{d y}{d x}$$ = 4a × 1
∴ 3x2 + 3y2 $$\frac{d y}{d x}$$ = 4a
Substituting the value of 4a in (1), we get
x3 + y3 = (3x2 + 3y2 $$\frac{d y}{d x}$$) x
∴ x3 + y3 = 3x3 + 3xy2 $$\frac{d y}{d x}$$
∴ 2x3 + 3xy2 $$\frac{d y}{d x}$$ – y3 = 0
This is the required D.E.

(ii) Ax2 + By2 = 1
Solution:
Ax2 + By2 = 1
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y $$\frac{d y}{d x}$$ = 0
∴ Ax + By $$\frac{d y}{d x}$$ = 0 ……..(1)
Differentiating again w.r.t. x, we get

Substituting the value of A in (1), we get

This is the required D.E.

Alternative Method:
Ax2 + By2 = 1 ……..(1)
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y $$\frac{d y}{d x}$$ = 0
∴ Ax + By $$\frac{d y}{d x}$$ = 0 ……….(2)
Differentiating again w.r.t. x, we get,

The equations (1), (2) and (3) are consistent in A and B.
∴ determinant of their consistency is zero.

This is the required D.E.

(iii) y = A cos(log x) + B sin(log x)
Solution:
y = A cos(log x) + B sin (log x) ……. (1)
Differentiating w.r.t. x, we get

(iv) y2 = (x + c)3
Solution:
y2 = (x + c)3
Differentiating w.r.t. x, we get

This is the required D.E.

(v) y = Ae5x + Be-5x
Solution:
y = Ae5x + Be-5x ……….(1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

(vi) (y – a)2 = 4(x – b)
Solution:
(y – a)2 = 4(x – b)
Differentiating both sides w.r.t. x, we get
2(y – a) . $$\frac{d}{d x}$$(y – a) = 4 $$\frac{d}{d x}$$(x – b)
∴ 2(y – a) . ($$\frac{d y}{d x}$$ – 0) = 4(1 – 0)
∴ 2(y – a) $$\frac{d y}{d x}$$ = 4
∴ (y – a) $$\frac{d y}{d x}$$ = 2 ……..(1)
Differentiating w.r.t. x, we get

This is the required D.E.

(vii) y = a + $$\frac{a}{x}$$
Solution:
y = a + $$\frac{a}{x}$$
Differentiating w.r.t. x, we get

Substituting the value of a in (1), we get

This is the required D.E.

(viii) y = c1e2x + c2e5x
Solution:
y = c1e2x + c2e5x ………(1)
Differentiating twice w.r.t. x, we get
$$\frac{d y}{d x}$$ = c1e2x × 2 + c2e5x × 5

The equations (1), (2) and (3) are consistent in c1e2x and c2e5x
∴ determinant of their consistency is zero.

This is the required D.E.

Alternative Method:
y = c1e2x + c2e5x
Dividing both sides by e5x, we get

This is the required D.E.

(ix) c1x3 + c2y2 = 5.
Solution:
c1x3 + c2y2 = 5 ……….(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

The equations (1), (2) and (3) in c1, c2 are consistent.
∴ determinant of their consistency is zero.

This is the required D.E.

(x) y = e-2x(A cos x + B sin x)
Solution:
y = e-2x(A cos x + B sin x)
∴ e2x . y = A cos x + B sin x ………(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

This is the required D.E.

Question 2.
Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Solution:
The equation of the line having intercepts a and b on the coordinate axes respectively, is
$$\frac{x}{a}+\frac{y}{b}=1$$ ……….(1)
where a and b are arbitrary constants.
[For different values of a and b, we get, different lines. Hence (1) is the equation of family of lines.]
Differentiating (1) w.r.t. x, we get

Differentiating again w.r.t. x, we get $$\frac{d^{2} y}{d x^{2}}=0$$
This is the required D.E.

Question 3.
Find the differential equation all parabolas having length of latus rectum 4a and axis is parallel to the X-axis.
Solution:

Let A(h, k) be the vertex of the parabola whose length of latus rectum is 4a.
Then the equation of the parabola is (y – k)2 = 4a (x – h), where h and k are arbitrary constants.
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

This is the required D.E.

Question 4.
Find the differential equation of the ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$
i.e., $$\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1$$
∴ $$\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1$$
∴ x2 + 4y2 = 4b2
Differentiating w.r.t. x, we get
2x + 4 × 2y $$\frac{d y}{d x}$$ = 0
∴ x + 4y $$\frac{d y}{d x}$$ = 0
This is the required D.E.

Question 5.
Form the differential equation of family of lines parallel to the line 2x + 3y + 4 = 0.
Solution:
The equation of the line parallel to the line 2x + 3y + 4 = 0 is 2x + 3y + c = 0, where c is an arbitrary constant.
Differentiating w.r.t. x, we get
2 × 1 + 3 $$\frac{d y}{d x}$$ + 0 = 0
∴ 3 $$\frac{d y}{d x}$$ + 2 = 0
This is the required D.E.

Question 6.
Find the differential equation of all circles having radius 9 and centre at point (h, k).
Solution:
Equation of the circle having radius 9 and centre at point (h, k) is
(x – h)2 + (y – k)2 = 81 …… (1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get

Differentiating again w.r.t. x, we get

From (2), x – h = -(y – k) $$\frac{d y}{d x}$$
Substituting the value of (x – h) in (1), we get

This is the required D.E.

Question 7.
Form the differential equation of all parabolas whose axis is the X-axis.
Solution:

The equation of the parbola whose axis is the X-axis is
y2 = 4a(x – h) …… (1)
where a and h are arbitrary constants.
Differentiating (1) w.r.t. x, we get
2y $$\frac{d y}{d x}$$ = 4a(1 – 0)
∴ y $$\frac{d y}{d x}$$ = 2a
Differentiating again w.r.t. x, we get

This is the required D.E.

## Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

1. Determine the order and degree of each of the following differential equations:

Question (i).
$$\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x$$
Solution:
The given D.E. is $$\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 1.
∴ the given D.E. is of order 2 and degree 1.

Question (ii).
$$\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}$$
Solution:
The given D.E. is $$\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}$$
On cubing both sides, we get
$$1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 3.
∴ the given D.E. is of order 2 and degree 3.

Question (iii).
$$\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}$$
Solution:
The given D.E. is $$\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}$$
∴ $$\left(\frac{d y}{d x}\right)^{2}$$ = 2 sin x + 3
This D.E. has highest order derivative $$\frac{d y}{d x}$$ with power 2.
∴ the given D.E. is of order 1 and degree 2.

Question (iv).
$$\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}$$
Solution:
The given D.E. is $$\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}$$
On squaring both sides, we get
$$\left(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x\right)^{2}=1+\frac{d^{3} y}{d x^{3}}$$
This D.E. has highest order derivative $$\frac{d^{3} y}{d x^{3}}$$ with power 1.
∴ the given D.E. has order 3 and degree 1.

Question (v).
$$\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0$$
Solution:
The given D.E. is $$\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 1.
∴ the given D.E. has order 2 and degree 1.

Question (vi).
(y”‘)2 + 3y” + 3xy’ + 5y = 0
Solution:
The given D.E. is (y”‘)2 + 3y” + 3xy’ + 5y = 0
This can be written as:
$$\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+3 \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+5 y=0$$
This D.E. has highest order derivative $$\frac{d^{3} y}{d x^{3}}$$ with power 2.
∴ The given D.E. has order 3 and degree 2.

Question (vii).
$$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$$
Solution:
The given D.E. is $$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$
∴ order = 2
Since this D.E. cannot be expressed as a polynomial in differential coefficients, the degree is not defined.

Question (viii).
$$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}$$
Solution:
The given D.E. is $$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}$$
On squaring both sides, we get
$$\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=8^{2} \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$$
This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 2.
∴ the given D.E. has order 2 and degree 2.

Question (ix).
$$\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20$$
Solution:
The given D.E. is $$\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20$$
∴ $$\left(\frac{d^{3} y}{d x^{3}}\right)^{3} \cdot\left(\frac{d y}{d x}\right)^{2}=20^{6}$$
This D.E. has highest order derivative $$\frac{d^{3} y}{d x^{3}}$$ with power 3.
∴ the given D.E. has order 3 and degree 3.

Question (x).
$$x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}$$
Solution:
The given D.E. is $$x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}$$
On squaring both sides, we get

This D.E. has highest order derivative $$\frac{d^{2} y}{d x^{2}}$$ with power 1.
∴ the given D.E. has order 2 and degree 1.

## Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
(a) 12 sq units

Question 2.
The area of the region enclosed by the curve y = $$\frac{1}{x}$$, and the lines x = e, x = e2 is given by
(a) 1 sq unit
(b) $$\frac{1}{2}$$ sq units
(c) $$\frac{3}{2}$$ sq units
(d) $$\frac{5}{2}$$ sq units
(a) 1 sq unit

Question 3.
The area bounded by the curve y = x3, the X-axis and the lines x = -2 and x = 1 is
(a) -9 sq units
(b) $$-\frac{15}{4}$$ sq units
(c) $$\frac{15}{4}$$ sq units
(d) $$\frac{17}{4}$$ sq units
(c) $$\frac{15}{4}$$ sq units

Question 4.
The area enclosed between the parabola y2 = 4x and line y = 2x is
(a) $$\frac{2}{3}$$ sq units
(b) $$\frac{1}{3}$$ sq units
(c) $$\frac{1}{4}$$ sq units
(d) $$\frac{3}{4}$$ sq units
(b) $$\frac{1}{3}$$ sq units

Question 5.
The area of the region bounded between the line x = 4 and the parabola y2 = 16x is
(a) $$\frac{128}{3}$$ sq units
(b) $$\frac{108}{3}$$ sq units
(c) $$\frac{118}{3}$$ sq units
(d) $$\frac{218}{3}$$ sq units
(a) $$\frac{128}{3}$$ sq units

Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
(d) 4 sq units

Question 7.
The area bounded by the parabola y2 = 8x, the X-axis and the latus rectum is
(a) $$\frac{31}{3}$$ sq units
(b) $$\frac{32}{3}$$ sq units
(c) $$\frac{32 \sqrt{2}}{3}$$ sq units
(d) $$\frac{16}{3}$$ sq units
(b) $$\frac{32}{3}$$ sq units

Question 8.
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is
(a) 4 sq units
(b) $$\frac{3}{4}$$ sq units
(c) $$\frac{2}{3}$$ sq units
(d) $$\frac{4}{3}$$ sq units
(d) $$\frac{4}{3}$$ sq units

Question 9.
The area of the circle x2 + y2 = 25 in first quadrant is
(a) $$\frac{25 \pi}{3}$$ sq units
(b) 5π sq units
(c) 5 sq units
(d) 3 sq units
(a) $$\frac{25 \pi}{3}$$ sq units

Question 10.
The area of the region bounded by the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ is
(a) ab sq units
(b) πab sq units
(c) $$\frac{\pi}{a b}$$ sq units ab
(d) πa2 sq units
(b) πab sq units

Question 11.
The area bounded by the parabola y2 = x and the line 2y = x is
(a) $$\frac{4}{3}$$ sq units
(b) 1 sq unit
(c) $$\frac{2}{3}$$ sq unit
(d) $$\frac{1}{3}$$ sq unit
(a) $$\frac{4}{3}$$ sq units

Question 12.
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ $$\frac{\pi}{6}$$ and the X-axis is
(a) $$\frac{1}{2}$$ sq unit
(b) 1 sq unit
(c) $$\frac{2}{3}$$ sq unit
(d) $$\frac{1}{3}$$ sq unit
(d) $$\frac{1}{3}$$ sq unit

Question 13.
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is
(a) 2√3 sq units
(b) 9 sq units
(c) $$\frac{34}{3}$$ sq units
(d) 18 sq units
(b) 9 sq units

Question 14.
The area bounded by the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ and the line $$\frac{x}{a}+\frac{y}{b}=1$$ is
(a) (πab – 2ab) sq units
(b) $$\frac{\pi a b}{4}-\frac{a b}{2}$$ sq units
(c) (πab – ab) sq units
(d) πab sq units
(b) $$\frac{\pi a b}{4}-\frac{a b}{2}$$ sq units

Question 15.
The area bounded by the parabola y = x2 and the line y = x is
(a) $$\frac{1}{2}$$ sq unit
(b) $$\frac{1}{3}$$ sq unit
(c) $$\frac{1}{6}$$ sq unit
(d) $$\frac{1}{12}$$ sq unit
(c) $$\frac{1}{6}$$ sq unit

Question 16.
The area enclosed between the two parabolas y2 = 4x and y = x is
(a) $$\frac{8}{3}$$ sq units
(b) $$\frac{32}{3}$$ sq units
(c) $$\frac{16}{3}$$ sq units
(d) $$\frac{4}{3}$$ sq units
(c) $$\frac{16}{3}$$ sq units

Question 17.
The area bounded by the curve y = tan x, X-axis and the line x = $$\frac{\pi}{4}$$ is
(a) $$\frac{1}{3}$$ log 2 sq units
(b) log 2 sq units
(c) 2 log 2 sq units
(d) 3 log 2 sq units
(a) $$\frac{1}{3}$$ log 2 sq units

Question 18.
The area of the region bounded by x2 = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is
(a) $$\frac{7}{3}$$ sq units
(b) $$\frac{8}{3}$$ sq units
(c) $$\frac{64}{3}$$ sq units
(d) $$\frac{56}{3}$$ sq units
(d) $$\frac{56}{3}$$ sq units

Question 19.
The area of the region included between the parabolas y2 = 4ax and x2 = 4ay, (a > 0) is given by
(a) $$\frac{16 a^{2}}{3}$$ sq units
(b) $$\frac{8 a^{2}}{3}$$ sq units
(c) $$\frac{4 a^{2}}{3}$$ sq units
(d) $$\frac{32 a^{2}}{3}$$ sq units
(a) $$\frac{16 a^{2}}{3}$$ sq units

Question 20.
The area of the region included between the line x + y = 1 and the circle x2 + y2 = 1 is
(a) $$\frac{\pi}{2}-1$$ sq units
(b) π – 2 sq units
(c) $$\frac{\pi}{4}-\frac{1}{2}$$ sq units
(d) π – $$\frac{1}{2}$$ sq units
(c) $$\frac{\pi}{4}-\frac{1}{2}$$ sq units

(II) Solve the following:

Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = $$\frac{\pi}{3}$$
Solution:
(i) Required area = $$\int_{0}^{5} y d x$$, where y = 2
= $$\int_{0}^{5} 2 d x$$
= $$[2 x]_{0}^{5}$$
= 2 × 5 – 0
= 10 sq units.

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.

Two bounded regions A1 and A2 are obtained. Both the regions have equal areas.
∴ required area = A1 + A2 = 2A1

(iii) Required area = $$\int_{0}^{\pi / 3} y d x$$, where y = sin x

Question 2.
Find the area of the circle x2 + y2 = 9, using integration.
Solution:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.

From the equation of the circle, y2 = 9 – x2.
In the first quadrant, y > 0
∴ y = $$\sqrt{9-x^{2}}$$
∴ area of the circle = 4 (area of the region OABO)

Question 3.
Find the area of the ellipse $$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$$ using integration.
Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
$$\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}$$
∴ y2 = $$\frac{16}{25}$$ (25 – x2)
In the first quadrant y > 0
∴ y = $$\frac{4}{5} \sqrt{25-x^{2}}$$
∴ area of the ellipse = 4(area of the region OABO)

Question 4.
Find the area of the region lying between the parabolas:
(i) y2 = 4x and x2 = 4y
(ii) 4y2 = 9x and 3x2 = 16y
(iii) y2 = x and x2 = y.
Solution:
(i)

For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations.
From the equation x2 = 4y, y = $$\frac{x^{2}}{4}$$
y = $$\frac{x^{4}}{16}$$
$$\frac{x^{4}}{16}$$ = 4x
∴ x4 – 64x = 0
∴ x(x3 – 64) = 0
∴ x = 0 or x3 = 64 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = $$\frac{4^{2}}{4}$$ = 4
∴ the points of intersection are 0(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x, i.e. y = 2√x between x = 0 and x = 4

(ii)

For finding the points of intersection of the two parabolas, we equate the values of 4y2 from their equations.
From the equation 3x2 = 16y, y = $$\frac{3 x^{2}}{16}$$
∴ y = $$\frac{3 x^{4}}{256}$$
∴ $$\frac{3 x^{4}}{256}$$ = 9x
∴ 3x4 – 2304x = 0
∴ x(x3 – 2304) = 0
∴ x = 0 or x3 = 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = $$\frac{4^{2}}{4}$$
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area of the region ODABO = area under the rabola x2 = 4y,
i.e. y = $$\frac{x^{2}}{4}$$ between x = 0 and x = 4

(iii)

For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations.
From the equation x2 = y, y = $$\frac{x^{2}}{y}$$
∴ y = $$\frac{x^{2}}{y}$$
∴ $$\frac{x^{2}}{y}$$ = x
∴ x2 – y = 0
∴ x(x3 – y) = 0
∴ x = 0 or x3 = y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = $$\frac{4^{2}}{4}$$ = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area ofthe region ODABO = area under the rabola x2 = 4y,
i.e. y = $$\frac{x^{2}}{3}$$ between x = 0 and x = 4

Question 5.
Find the area of the region in the first quadrant bounded by the circle x2 + y2 = 4 and the X-axis and the line x = y√3.
Solution:

For finding the points of intersection of the circle and the line, we solve
x2 + y2 = 4 ………(1)
and x = y√3 ……..(2)
From (2), x2 = 3y2
From (1), x2 = 4 – y2
3y2 = 4 – y2
4y2 = 4
y2 = 1
y = 1 in the first quadrant.
When y = 1, r = 1 × √3 = √3
∴ the circle and the line intersect at A(√3, 1) in the first quadrant
Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED
Now, area of the region OCADO = area under the line x = y√3, i.e. y = $$\frac{x}{\sqrt{3}}$$ between x = 0
and x = √3

Question 6.
Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant.
Solution:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.

∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola y2 = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1

Area of the region OBADO = area under the line y = x between x = 0 and x = 1

Question 7.
Find the area enclosed between the circle x2 + y2 = 1 and the line x + y = 1, lying in the first quadrant.
Solution:

Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle x2 + y2 = 1 between x = 0 and x = 1

Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1

∴ required area = $$\left(\frac{\pi}{4}-\frac{1}{2}\right)$$ sq units.

Question 8.
Find the area of the region bounded by the curve (y – 1)2 = 4(x + 1) and the line y = (x – 1).
Solution:
The equation of the curve is (y – 1)2 = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)2 = 4(x + 1)
∴ x2 – 4x + 4 = 4x + 4
∴ x2 – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).

To find the points where the parabola (y – 1)2 = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)2 = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1)2 = 4(x + 1), Y-axis from y = -1 to y = 3

Since, the area cannot be negative,
Area of the region BFAB = $$\left|-\frac{8}{3}\right|=\frac{8}{3}$$ sq units.
Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG

Since, area cannot be negative,
area of the region = $$\left|-\frac{1}{2}\right|=\frac{1}{2}$$ sq units.
∴ required area = $$\frac{8}{3}+\frac{109}{6}+\frac{1}{2}$$
= $$\frac{16+109+3}{6}$$
= $$\frac{128}{6}$$
= $$\frac{64}{3}$$ sq units.

Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:
The equation of the line is
2y = 5x + 7, i.e., y = $$\frac{5}{2} x+\frac{7}{2}$$
Required area = area of the region ABCDA = area under the line y = $$\frac{5}{2} x+\frac{7}{2}$$ between x = 2 and x = 5

Question 10.
Find the area of the region bounded by the curve y = 4x2, Y-axis and the lines y = 1, y = 4.
Solution:

By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
From the equation of the parabola, x2 = $$\frac{y}{4}$$
In the first quadrant, x > 0
∴ x = $$\frac{1}{2} \sqrt{y}$$
∴ required area = $$\int_{1}^{4} x d y$$

## Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Ex 5.1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

1. Find the area of the region bounded by the following curves, X-axis, and the given lines:

(i) y = 2x, x = 0, x = 5.
Solution:
Required area = $$\int_{0}^{5} y d x$$, where y = 2x
= $$\int_{0}^{5} 2x d x$$
= $$\left[\frac{2 x^{2}}{2}\right]_{0}^{5}$$
= 25 – 0
= 25 sq units.

(ii) x = 2y, y = 0, y = 4.
Solution:
Required area = $$\int_{0}^{4} x d y$$, where x = 2y
= $$\int_{0}^{4} 2 y d y$$
= $$\left[\frac{2 y^{2}}{2}\right]_{0}^{4}$$
= 16 – 0
= 16 sq units.

(iii) x = 0, x = 5, y = 0, y = 4.
Solution:
Required area = $$\int_{0}^{5} y d x$$, where y = 4
= $$\int_{0}^{5} 4 d x$$
= $$[4 x]_{0}^{5}$$
= 20 – 0
= 20 sq units.

(iv) y = sin x, x = 0, x = $$\frac{\pi}{2}$$
Solution:
Required area = $$\int_{0}^{\pi / 2} y d x$$, where y = sin x
= $$\int_{0}^{\pi / 2} \sin x d x$$
= $$[-\cos x]_{0}^{\pi / 2}$$
= -cos $$\frac{\pi}{2}$$ + cos 0
= 0 + 1
= 1 sq unit.

(v) xy = 2, x = 1, x = 4.
Solution:
For xy = 2, y = $$\frac{2}{x}$$
Required area = $$\int_{1}^{4} y d x$$, where y = $$\frac{2}{x}$$
= $$\int_{1}^{4} \frac{2}{x} d x$$
= $$[2 \log |x|]_{1}^{4}$$
= 2 log 4 – 2 log 1
= 2 log 4 – 0
= 2 log 4 sq units.

(vi) y2 = x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y2 = x, y = √x
Required area = A1 + A2 = 2A1

(vii) y2 = 16x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y2 = x, y = √x
Required area = A1 + A2 = 2A1

2. Find the area of the region bounded by the parabola:

(i) y2 = 16x and its latus rectum.
Solution:
Comparing y2 = 16x with y2 = 4ax, we get
4a = 16
∴ a = 4
∴ focus is S(a, 0) = (4, 0)

For y2 = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]

(ii) y = 4 – x2 and the X-axis.
Solution:
The equation of the parabola is y = 4 – x2
∴ x2 = 4 – y
i.e. (x – 0)2 = -(y – 4)
It has vertex at P(0, 4)
For points of intersection of the parabola with X-axis,
we put y = 0 in its equation.
∴ 0 = 4 – x2
∴ x2 = 4
∴ x = ± 2
∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)

Required area = area of the region APBOA
= 2[area of the region OPBO]

3. Find the area of the region included between:

(i) y2 = 2x and y = 2x.
Solution:
The vertex of the parabola y2 = 2x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,
y2 = y
∴ y2 – y = 0
∴ y = 0 or y = 1
When y = 0, x = $$\frac{0}{2}$$ = 0
When y = 1, x = $$\frac{1}{2}$$
∴ the points of intersection are 0(0, 0) and B($$\frac{1}{2}$$, 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y2 = 2x between x = 0 and x = $$\frac{1}{2}$$

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = $$\frac{1}{2}$$

(ii) y2 = 4x and y = x.
Solution:
The vertex of the parabola y2 = 4x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = $$\frac{0}{2}$$ = 0
When y = 1, x = $$\frac{1}{2}$$
∴ the points of intersection are O(0, 0) and B($$\frac{1}{2}$$, 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y2 = 4x between x = 0 and x = $$\frac{1}{2}$$

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = $$\frac{1}{2}$$

(iii) y = x2 and the line y = 4x.
Solution:
The vertex of the parabola y = x2 is at the origin 0(0, 0)
To find the points of the intersection of a line and the parabola.

Equating the values of y from the two equations, we get
x2 = 4x
∴ x2 – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of intersection are 0(0, 0) and B(4, 16)
Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4
= $$\int_{0}^{4} y d x$$, where y = 4x
= $$\int_{0}^{4} 4 x d x$$
= 4$$\int_{0}^{4} x d x$$
= 4$$$\int_{0}^{4} x d x$$$
= 2(16 – 0)
= 32
Area of the region ODBAO = area under the parabola y = x2 between x = 0 and x = 4
= $$\int_{0}^{4} y d x$$, where y = x2
= $$\int_{0}^{4} x^{2} d x$$
= $$\left[\frac{x^{3}}{3}\right]_{0}^{4}$$
= $$\frac{1}{3}$$ (64 – 0)
= $$\frac{64}{3}$$
∴ required area = 32 – $$\frac{64}{3}$$ = $$\frac{32}{3}$$ sq units.

(iv) y2 = 4ax and y = x.
Solution:
The vertex of the parabola y2 = 4ax is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = $$\frac{0}{2}$$ = 0
When y = 1, x = $$\frac{1}{2}$$
∴ the points of intersection are O(0, 0) and B($$\frac{1}{2}$$, 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO
= area under the parabola y2 = 4ax between x = 0 and x = $$\frac{1}{2}$$

Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = $$\frac{1}{4 a x}$$

(v) y = x2 + 3 and y = x + 3.
Solution:
The given parabola is y = x2 + 3, i.e. (x – 0)2 = y – 3
∴ its vertex is P(0, 3).

To find the points of intersection of the line and the parabola.
Equating the values of y from both the equations, we get
x2 + 3 = x + 3
∴ x2 – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
∴ the points of intersection are P(0, 3) and B(1, 4)
Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO
Now, area of the region OPABDO
= area under the line y = x + 3 between x = 0 and x = 1

Area of the region OPCBDO = area under the parabola y = x2 + 3 between x = 0 and x = 1

## Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Miscellaneous Exercise 4 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

I. Choose the correct option from the given alternatives:

Question 1.
$$\int_{2}^{3} \frac{d x}{x\left(x^{3}-1\right)}=$$
(a) $$\frac{1}{3} \log \left(\frac{208}{189}\right)$$
(b) $$\frac{1}{3} \log \left(\frac{189}{208}\right)$$
(c) $$\log \left(\frac{208}{189}\right)$$
(d) $$\log \left(\frac{189}{208}\right)$$
(a) $$\frac{1}{3} \log \left(\frac{208}{189}\right)$$

Question 2.
$$\int_{0}^{\pi / 2} \frac{\sin ^{2} x \cdot d x}{(1+\cos x)^{2}}=$$
(a) $$\frac{4-\pi}{2}$$
(b) $$\frac{\pi-4}{2}$$
(c) 4 – $$\frac{\pi}{2}$$
(d) $$\frac{4+\pi}{2}$$
(a) $$\frac{4-\pi}{2}$$

Question 3.
$$\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} \cdot d x=$$
(a) 3 + 2π
(b) 4 – π
(c) 2 + π
(d) 4 + π
(b) 4 – π

Question 4.
$$\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{2} x \cdot d x=$$
(a) $$\frac{7 \pi}{256}$$
(b) $$\frac{3 \pi}{256}$$
(c) $$\frac{5 \pi}{256}$$
(d) $$\frac{-5 \pi}{256}$$
(c) $$\frac{5 \pi}{256}$$

Question 5.
If $$\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{X}}=\frac{k}{3}$$, then k is equal to
(a) √2(2√2 – 2)
(b) $$\frac{\sqrt{2}}{3}$$(2 – 2√2)
(c) $$\frac{2 \sqrt{2}-2}{3}$$
(d) 4√2
(d) 4√2

Question 6.
$$\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{1}{x}} \cdot d x=$$
(a) √e + 1
(b) √e − 1
(c) √e(√e − 1)
(d) $$\frac{\sqrt{e}-1}{e}$$
(c) √e(√e − 1)

Question 7.
If $$\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] \cdot d x=a+\frac{b}{\log 2}$$, then
(a) a = e, b = -2
(b) a = e, b = 2
(c) a = -e, b = 2
(d) a = -e, b = -2
(a) a = e, b = -2

Question 8.
Let $$\mathrm{I}_{1}=\int_{e}^{e^{2}} \frac{d x}{\log x}$$ and $$\mathrm{I}_{2}=\int_{1}^{2} \frac{e^{x}}{\boldsymbol{X}} \cdot d x$$, then
(a) I1 = $$\frac{1}{3}$$ I2
(b) I1 + I2 = 0
(c) I1 = 2I2
(d) I1 = I2
(d) I1 = I2

Question 9.
$$\int_{0}^{9} \frac{\sqrt{X}}{\sqrt{X}+\sqrt{9-X}} \cdot d x=$$
(a) 9
(b) $$\frac{9}{2}$$
(c) 0
(d) 1
(b) $$\frac{9}{2}$$

Question 10.
The value of $$\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right) \cdot d \theta$$ is
(a) 0
(b) 1
(c) 2
(d) π
(a) 0

II. Evaluate the following:

Question 1.
$$\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x$$
Solution:
Let I = $$\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x$$
Put Numerator = A(Denominator) + B[$$\frac{d}{d x}$$(Denominator)]
∴ cos x = A(3 cos x + sin x) + B[$$\frac{d}{d x}$$(3 cos x + sin x)]
= A(3 cos x + sin x) + B(-3 sin x + cos x)
∴ cos x + 0 . sin x = (3A + B) cos x + (A – 3B) sin x
Comparing the coefficients of sinx and cos x on both the sides, we get
3A + B = 1 ………. (1)
A – 3B = 0 ………. (2)
Multiplying equation (1) by 3, we get
9A + 3B = 3 ………(3)
Adding (2) and (3), we get
10A = 3
∴ A = $$\frac{3}{10}$$

Question 2.
$$\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^{3}} d \theta$$
Solution:

Question 3.
$$\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x$$
Solution:
Let I = $$\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x$$
Put √x = t
∴ x = t2 and dx = 2t . dt
When x = 0, t = 0
When x = 1, t = 1

Question 4.
$$\int_{0}^{\pi / 4} \frac{\tan ^{3} x}{1+\cos 2 x} d x$$
Solution:

Question 5.
$$\int_{0}^{1} t^{5} \sqrt{1-t^{2}} d t$$
Solution:

Question 6.
$$\int_{0}^{1}\left(\cos ^{-1} x\right)^{2} d x$$
Solution:

Question 7.
$$\int_{-1}^{1} \frac{1+x^{3}}{9-x^{2}} d x$$
Solution:

Question 8.
$$\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x$$
Solution:

Question 9.
$$\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x} d x$$
Solution:

Question 10.
$$\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x$$
Solution:

III. Evaluate the following:

Question 1.
$$\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right) \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$
Solution:

Question 2.
$$\int_{0}^{\pi / 2} \frac{1}{6-\cos x} d x$$
Solution:

Question 3.
$$\int_{0}^{a} \frac{1}{a^{2}+a x-x^{2}} d x$$
Solution:

Question 4.
$$\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x$$
Solution:

Question 5.
$$\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$
Solution:
Let I = $$\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$

Question 6.
$$\int_{0}^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x$$
Solution:

Question 7.
$$\int_{0}^{\pi / 2}[2 \log (\sin x)-\log (\sin 2 x)] d x$$
Solution:

Question 8.
$$\int_{0}^{\pi}\left(\sin ^{-1} x+\cos ^{-1} x\right)^{3} \sin ^{3} x d x$$
Solution:

Question 9.
$$\int_{0}^{4}\left[\sqrt{x^{2}+2 x+3}\right]^{-1} d x$$
Solution:

Question 10.
$$\int_{-2}^{3}|x-2| d x$$
Solution:
|x – 2|= 2 – x, if x < 2
= x – 2, if x ≥ 2

IV. Evaluate the following:

Question 1.
If $$\int_{a}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x$$, find the value of $$\int_{a}^{a+1} x d x$$.
Solution:

Question 2.
If $$\int_{0}^{k} \frac{1}{2+8 x^{2}} \cdot d x=\frac{\pi}{16}$$, find k.
Solution:

Question 3.
If f(x) = a + bx + cx2, show that $$\int_{0}^{1} f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]$$
Solution:

## Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.2 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

I. Evaluate:

Question 1.
$$\int_{1}^{9} \frac{x+1}{\sqrt{x}} \cdot d x$$
Solution:

Question 2.
$$\int_{2}^{3} \frac{1}{x^{2}+5 x+6} \cdot d x$$
Solution:

Question 3.
$$\int_{0}^{\pi / 4} \cot ^{2} \cdot d x$$
Solution:

The integral does not exist since cot 0 is not defined.

Question 4.
$$\int_{-\pi / 4}^{\pi / 4} \frac{1}{1-\sin x} \cdot d x$$
Solution:

Question 5.
$$\int_{3}^{5} \frac{1}{\sqrt{2 x+3}-\sqrt{2 x-3}} \cdot d x$$
Solution:

Question 6.
$$\int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} \cdot d x$$
Solution:

Question 7.
$$\int_{0}^{\pi / 4} \sin 4 x \sin 3 x \cdot d x$$
Solution:

Question 8.
$$\int_{0}^{\pi / 4} \sqrt{1+\sin 2 x} \cdot d x$$
Solution:

Question 9.
$$\int_{0}^{\pi / 4} \sin ^{4} x \cdot d x$$
Solution:

Question 10.
$$\int_{-4}^{2} \frac{1}{x^{2}+4 x+13} \cdot d x$$
Solution:

Question 11.
$$\int_{0}^{4} \frac{1}{\sqrt{4 x-x^{2}}} \cdot d x$$
Solution:

Question 12.
$$\int_{0}^{1} \frac{1}{\sqrt{3+2 x-x^{2}}} \cdot d x$$
Solution:

Question 13.
$$\int_{0}^{\pi / 2} x \cdot \sin x \cdot d x$$
Solution:

Question 14.
$$\int_{0}^{1} x \cdot \tan ^{-1} x \cdot d x$$
Solution:

Question 15.
$$\int_{0}^{\infty} x \cdot e^{-x} \cdot d x$$
Solution:

II. Evaluate:

Question 1.
$$\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} \cdot d x$$
Solution:

Question 2.
$$\int_{0}^{\pi / 4} \frac{\sec ^{2} x}{3 \tan ^{2} x+4 \tan x+1} \cdot d x$$
Solution:

Question 3.
$$\int_{0}^{4 \pi} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \cdot d x$$
Solution:

Question 4.
$$\int_{0}^{2 \pi} \sqrt{\cos x} \cdot \sin ^{3} x \cdot d x$$
Solution:

Question 5.
$$\int_{0}^{\pi / 2} \frac{1}{5+4 \cos x} \cdot d x$$
Solution:

Question 6.
$$\int_{0}^{\pi / 4} \frac{\cos x}{4-\sin ^{2} x} \cdot d x$$
Solution:

Question 7.
$$\int_{0}^{\pi / 2} \frac{\cos X}{(1+\sin x)(2+\sin x)} \cdot d x$$
Solution:

Question 8.
$$\int_{-1}^{1} \frac{1}{a^{2} e^{x}+b^{2} e^{-x}} \cdot d x$$
Solution:

Question 9.
$$\int_{0}^{\pi} \frac{1}{3+2 \sin x+\cos x} \cdot d x$$
Solution:

Question 10.
$$\int_{0}^{\pi / 4} \sec ^{4} x \cdot d x$$
Solution:

Question 11.
$$\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \cdot d x$$
Solution:

Question 12.
$$\int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} \cdot d x$$
Solution:

Question 13.
$$\int_{0}^{\pi / 2} \sin 2 x \cdot \tan ^{-1}(\sin x) \cdot d x$$
Solution:

Question 14.
$$\int_{\frac{1}{\sqrt{2}}}^{1} \frac{\left(e^{\cos ^{-1} x}\right)\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \cdot d x$$
Solution:

Question 15.
$$\int_{2}^{3} \frac{\cos (\log x)}{x} \cdot d x$$
Solution:

III. Evaluate:

Question 1.
$$\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} \cdot d x$$
Solution:

Question 2.
$$\int_{0}^{\pi / 2} \log \tan x \cdot d x$$
Solution:

Question 3.
$$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) \cdot d x$$
Solution:

Question 4.
$$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cdot \cos x} \cdot d x$$
Solution:

Question 5.
$$\int_{0}^{3} x^{2}(3-x)^{\frac{5}{2}} \cdot d x$$
Solution:

Question 6.
$$\int_{-3}^{3} \frac{x^{3}}{9-x^{2}} \cdot d x$$
Solution:

Question 7.
$$\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2+\sin x}{2-\sin x}\right) \cdot d x$$
Solution:

Question 8.
$$\int_{-\pi / 4}^{\pi / 4} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} \cdot d x$$
Solution:

Question 9.
$$\int_{-\pi / 4}^{\pi / 4} x^{3} \cdot \sin ^{4} x \cdot d x$$
Solution:

Question 10.
$$\int_{0}^{1} \frac{\log (x+1)}{x^{2}+1} \cdot d x$$
Solution:

Question 11.
$$\int_{-1}^{1} \frac{x^{3}+2}{\sqrt{x^{2}+4}} \cdot d x$$
Solution:

Question 12.
$$\int_{-a}^{a} \frac{x+x^{3}}{16-x^{2}} \cdot d x$$
Solution:

Question 13.
$$\int_{0}^{1} t^{2} \sqrt{1-t} \cdot d t$$
Solution:

Question 14.
$$\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x$$
Solution:

Question 15.
$$\int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \cdot d x$$
Solution:

## Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

I. Evaluate the following integrals as a limit of a sum.

Question 1.
$$\int_{1}^{3}(3 x-4) \cdot d x$$
Solution:
Let f(x) = 3x – 4, for 1 ≤ x ≤ 3
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3
∴ nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here, a = 1
∴ f(a + rh) = f(1 + rh)
= 3(1 + rh) – 4
= 3rh – 1

Question 2.
$$\int_{0}^{4} x^{2} d x$$
Solution:
Let f(x) = x2, for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4
i.e. 0, h, 2h, ….., rh, ….., nh = 4
∴ h = $$\frac{4}{n}$$ as n → ∞, h → 0
Here, a = 0

Question 3.
$$\int_{0}^{2} e^{x} d x$$
Solution:
Let f(x) = ex, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n equal subntervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here, a = 0

Question 4.
$$\int_{0}^{2}\left(3 x^{2}-1\right) d x$$
Solution:
Let f(x) = 3x2 – 1, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n subintervals each of length h at the points.
0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh)
= f(rh)
= 3(rh)2 – 1
= 3r2h2 – 1

Question 5.
$$\int_{1}^{3} x^{3} d x$$
Solution:
Let f(x) = x3, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points
1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3
∴ nh = 2
∴ h = $$\frac{2}{n}$$ and as n → ∞, h → 0
Here a = 1

## Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Miscellaneous Exercise 1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(I) Choose the correct option from the given alternatives:

Question 1.
Let f(1) = 3, f'(1) = $$-\frac{1}{3}$$, g(1) = -4 and g'(1) = $$-\frac{8}{3}$$. The derivative of $$\sqrt{[f(x)]^{2}+[g(x)]^{2}}$$ w.r.t. x at x = 1 is
(a) $$-\frac{29}{15}$$
(b) $$\frac{7}{3}$$
(c) $$\frac{31}{15}$$
(d) $$\frac{29}{15}$$
(d) $$\frac{29}{15}$$

Question 2.
If y = sec(tan-1 x), then $$\frac{d y}{d x}$$ at x = 1, is equal to
(a) $$\frac{1}{2}$$
(b) 1
(c) $$\frac{1}{\sqrt{2}}$$
(d) 2
(c) $$\frac{1}{\sqrt{2}}$$

Question 3.
If f(x) = $$\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)$$, which of the following is not the derivative of f(x)?
(a) $$\frac{2 \cdot 4^{x} \log 4}{1+4^{2 x}}$$
(b) $$\frac{4^{x+1} \log 2}{1+4^{2 x}}$$
(c) $$\frac{4^{x+1} \log 4}{1+4^{4 x}}$$
(d) $$\frac{2^{2(x+1)} \log 2}{1+2^{4 x}}$$
(c) $$\frac{4^{x+1} \log 4}{1+4^{4 x}}$$

Question 4.
If xy = yx, then$$\frac{d y}{d x}$$ = _______
(a) $$\frac{x(x \log y-y)}{y(y \log x-x)}$$
(b) $$\frac{y(y \log x-x)}{x(x \log y-y)}$$
(c) $$\frac{y^{2}(1-\log x)}{x^{2}(1-\log y)}$$
(d) $$\frac{y(1-\log x)}{x(1-\log y)}$$
(b) $$\frac{y(y \log x-x)}{x(x \log y-y)}$$

Question 5.
If y = sin (2 sin-1 x), then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}$$
(b) $$\frac{2+4 x^{2}}{\sqrt{1-x^{2}}}$$
(c) $$\frac{4 x^{2}-1}{\sqrt{1-x^{2}}}$$
(d) $$\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$$
(a) $$\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}$$

Question 6.
If y = $$\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)+\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{x}{\sqrt{1-x^{2}}}$$
(b) $$\frac{1-2 x}{\sqrt{1-x^{2}}}$$
(c) $$\frac{1-2 x}{2 \sqrt{1-x^{2}}}$$
(d) $$\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$$
(c) $$\frac{1-2 x}{2 \sqrt{1-x^{2}}}$$

Question 7.
If y is a function of x and log(x + y) = 2xy, then the value of y'(0) = _______
(a) 2
(b) 0
(c) -1
(d) 1
(d) 1

Question 8.
If g is the inverse of function f and f'(x) = $$\frac{1}{1+x^{7}}$$, then the value of g'(x) is equal to:
(a) 1 + x7
(b) $$\frac{1}{1+[g(x)]^{7}}$$
(c) 1 + [g(x)]7
(d) 7x6
(c) 1 + [g(x)]7

Question 9.
If $$x \sqrt{y+1}+y \sqrt{x+1}=0$$ and x ≠ y, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{1}{(1+x)^{2}}$$
(b) $$-\frac{1}{(1+x)^{2}}$$
(c) (1 + x)2
(d) $$-\frac{x}{x+1}$$
(b) $$-\frac{1}{(1+x)^{2}}$$

Question 10.
If y = $$\tan ^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)$$, where -a < x < a, then $$\frac{d y}{d x}$$ = _______
(a) $$\frac{x}{\sqrt{a^{2}-x^{2}}}$$
(b) $$\frac{a}{\sqrt{a^{2}-x^{2}}}$$
(c) $$-\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
(d) $$\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
(c) $$-\frac{1}{2 \sqrt{a^{2}-x^{2}}}$$
[Hint: Put x = a cos θ]

Question 11.
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), then $$\left[\frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}$$ = _______
(a) $$\frac{8 \sqrt{2}}{a \pi}$$
(b) $$-\frac{8 \sqrt{2}}{a \pi}$$
(c) $$\frac{a \pi}{8 \sqrt{2}}$$
(d) $$\frac{4 \sqrt{2}}{a \pi}$$
(a) $$\frac{8 \sqrt{2}}{a \pi}$$

Question 12.
If y = a cos (log x) and $$A \frac{d^{2} y}{d x^{2}}+B \frac{d y}{d x}+C=0$$, then the values of A, B, C are _______
(a) x2, -x, -y
(b) x2, x, y
(c) x2, x, -y
(d) x2, -x, y
(b) x2, x, y

(II) Solve the following:

Question 1.

Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
Solution:

Question 2.
The values of f(x), g(x), f'(x) and g'(x) are given in the following table:

Match the following:

Solution:

Question 3.
Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.

(i) The derivative of f[g(x)] w.r.t. x at x = 0 is _______
(ii) The derivative of g[f(x)] w.r.t. x at x = 0 is _______
(iii) The value of $$\left[\frac{d}{d x}\left[x^{10}+f(x)\right]^{-2}\right]_{x=1}$$ is _______
(iv) The derivative of f[(x+g(x))] w.r.t. x at x = 0 is _______
Solution:

Question 4.
Differentiate the following w.r.t. x:
(i) $$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$
Solution:
Let y = $$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$$
Put x = cos θ, Then θ = cos-1 x and

(ii) $$\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]$$
Solution:
Let y = $$\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]$$
Put x = cos θ. Then θ = cos-1 x and

(iii) $$\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]$$
Solution:

(iv) $$\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$$
Solution:
Let y = $$\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)$$
Put x = cos θ. Then θ = cos-1 x and

(v) $$\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)+\cot ^{-1}\left(\frac{1-10 x^{2}}{7 x}\right)$$
Solution:

(vi) $$\tan ^{-1}\left[\sqrt{\frac{\sqrt{1+x^{2}+x}}{\sqrt{1+x^{2}}-x}}\right]$$
Solution:

Question 5.
(i) If $$\sqrt{y+x}+\sqrt{y-x}=c$$, show that $$\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$$
Solution:
$$\sqrt{y+x}+\sqrt{y-x}=c$$
Differentiating both sides w.r.t. x, we get

(ii) If $$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$, then show that $$\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$$
Solution:
$$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$$
$$y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$$
Differentiating both sides w.r.t. x, we get

(iii) If x sin(a + y) + sin a cos(a + y) = 0, then show $$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$$
Solution:
x sin(a + y) + sin a . cos (a + y) = 0 ….. (1)
Differentiating w.r.t. x, we get

(iv) If sin y = x sin(a + y), then show that $$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$$
Solution:

(v) If x = $$e^{\frac{x}{y}}$$, then show that $$\frac{d y}{d x}=\frac{x-y}{x \log x}$$
Solution:
x = $$e^{\frac{x}{y}}$$
$$\frac{x}{y}$$ = log x …..(1)
y = $$\frac{x}{\log x}$$

(vi) If y = f(x) is a differentiable function of x, then show that $$\frac{d^{2} x}{d y^{2}}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^{2} y}{d x^{2}}$$
Solution:
If y = f(x) is a differentiable function of x such that inverse function x = f-1(y) exists,

Question 6.
(i) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$$
Solution:

(ii) Differentiate $$\log \left[\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}-x}\right]$$ w.r.t. cos(log x)
Solution:

(iii) Differentiate $$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$$ w.r.t. $$\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^{2}}}{2 \sqrt{1+x^{2}}}}\right)$$
Solution:

Question 7.
(i) If y2 = a2 cos2x + b2 sin2x, show that $$y+\frac{d^{2} y}{d x^{2}}=\frac{a^{2} b^{2}}{y^{3}}$$
Solution:
y2 = a2 cos2x + b2 sin2x …… (1)
Differentiating both sides w.r.t. x, we get

(ii) If log y = log(sin x) – x2, show that $$\frac{d^{2} y}{d x^{2}}+4 x \frac{d y}{d x}+\left(4 x^{2}+3\right) y=0$$
Solution:

(iii) If x = a cos θ, y = b sin θ, show that $$a^{2}\left[y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]+b^{2}=0$$
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

(iv) If y = A cos(log x) + B sin(log x), show that x2y2 + xy1 + y = o.
Solution:
y = A cos (log x) + B sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(v) If y = A emx + B enx, show that y2 – (m + n) y1 + (mn) y = 0.
Solution:
y = A emx + B enx
Differentiating w.r.t. x, we get

## Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.4 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4

Question 1.
Maximize : z = 11x + 8y subject to x ≤ 4, y ≤ 6,
x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.

The feasible region is shaded portion OAPDO in the graph.
The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)
P is point of intersection of lines x + y = 6 and x = 4.
Substituting x = 4 in x + y = 6, we get
4 + y = 6 ∴ y = 2 ∴ P is (4, 2).
∴ the corner points of feasible region are O (0, 0), A (4, 0), P(4, 2) and D(0 ,6).
The values of the objective function z = 11x + 8y at these vertices are
z (O) = 11(0) + 8(0) = 0 + 0 = 0
z(a) = 11(4) + 8(0) = 44 + 0 = 44
z (P) = 11(4) + 8(2) = 44 + 16 = 60
z (D) = 11(0) + 8(2) = 0 + 16 = 16
∴ z has maximum value 60, when x = 4 and y = 2.

Question 2.
Maximize : z = 4x + 6y subject to 3x + 2y ≤ 12,
x + y ≥ 4, x, y ≥ 0.
Solution:
First we draw the lines AB and AC whose equations are 3x + 2y = 12 and x + y = 4 respectively.

The feasible region is the ∆ABC which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A (4, 0), B (0, 6) and C (0, 4).
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(4) + 6(0) = 16 + 0 = 16
z(B) = 4(0)+ 6(6) = 0 + 36 = 36
z(C) = 4(0) + 6(4) = 0 + 24 = 24
∴ has maximum value 36, when x = 0, y = 6.

Question 3.
Maximize : z = 7x + 11y subject to 3x + 5y ≤ 26
5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 26 and 5x + 3y = 30 respectively.

The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (6, 0), p and B(0, $$\frac{26}{5}$$)
The vertex P is the point of intersection of the lines
3x + 5y = 26 … (1)
and 5x + 3y = 30 … (2)
Multiplying equation (1) by 3 and equation (2) by 5, we get
9x + 15y = 78
and 25x + 15y = 150
On subtracting, we get
16x = 72 ∴ x = $$\frac{72}{16}=\frac{9}{2}$$ = 4.5
Substituting x = 4.5 in equation (2), we get
5(4.5) + 3y = 30
22.5 + 3y = 30
∴ 3y = 7.5 ∴ y = 2.5
∴ P is (4.5, 2.5)
The values of the objective function z = 7x + 11y at these corner points are
z (O) = 7(0) + 11(0) = 0 + 0 = 0
z (C) = 7(6) + 11(0) = 42 + 0 = 42
z (P) = 7(4.5) + 11 (2.5) = 31.5 + 27.5 = 59.0 = 59
z(B) = 7(0) + 11$$\left(\frac{26}{5}\right)=\frac{286}{5}$$ = 57.2
∴ z has maximum value 59, when x = 4.5 and y = 2.5.

Question 4.
Maximize : z = 10x + 25y subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3, x + y ≤ 5 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.

The feasible region is OAPQDO which is shaded in the i graph.
The vertices of the feasible region are O (0, 0), A (3, 0), P, Q and D(0, 3).
t P is the point of intersection of the lines x + y = 5 and x = 3.
Substituting x = 3 in x + y = 5, we get
3 + y = 5 ∴ y = 2
∴ P is (3, 2)
Q is the point of intersection of the lines x + y = 5 and y = 3
Substituting y = 3 in x + y = 5, we get
x + 3 = 5 ∴ x = 2
∴ Q is (2, 3)
The values of the objective function z = 10x + 25y at these vertices are
z(O) = 10(0) + 25(0) = 0 + 0 = 0
z(a) = 10(3) + 25(0) = 30 + 0 = 30
z(P) = 10(3) + 25(2) = 30 + 50 = 80
z(Q) = 10(2) + 25(3) = 20 + 75 = 95
z(D) = 10(0)+ 25(3) = 0 + 75 = 75
∴ z has maximum value 95, when x = 2 and y = 3.

Question 5.
Maximize : z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21,
x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.

The feasible region is OCPQBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (7, 0), P, Q and B (0, 6).
P is the point of intersection of the lines
3x + y = 21 … (1)
and x + y = 9 … (2)
On subtracting, we get 2x = 12 ∴ x = 6
Substituting x = 6 in equation (2), we get
6 + y = 9 ∴ y = 3
∴ P = (6, 3)
Q is the point of intersection of the lines
x + 4y = 24 … (3)
and x + y = 9 … (2)
On subtracting, we get
3y = 15 ∴ y = 5
Substituting y = 5 in equation (2), we get
x + 5= 9 ∴ x = 4
∴ Q = (4, 5)
∴ the corner points of the feasible region are 0(0,0), C(7, 0), P (6, 3), Q (4, 5) and B (0, 6).
The values of the objective function 2 = 3x + 5y at these corner points are
z(O) = 3(0)+ 5(0) = 0 + 0 = 0
z(C) = 3(7) + 5(0) = 21 + 0 = 21
z(P) = 3(6) + 5(3) = 18 + 15 = 33
z(Q) = 3(4) + 5(5) = 12 + 25 = 37
z(B) = 3(0)+ 5(6) = 0 + 30 = 30
∴ z has maximum value 37, when x = 4 and y = 5.

Question 6.
Minimize : z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3,
x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 5x + y = 5 and x + y = 3 respectively.

The feasible region is XCPBY which is shaded in the graph.
The vertices of the feasible region are C (3, 0), P and B (0, 5).
P is the point of the intersection of the lines
5x + y = 5
and x + y = 3
On subtracting, we get
4x = 2 ∴ x = $$\frac{1}{2}$$
Substituting x = $$\frac{1}{2}$$ in x + y = 3, we get
$$\frac{1}{2}$$ + y = 3
∴ y = $$\frac{5}{2}$$ ∴ P = $$\left(\frac{1}{2}, \frac{5}{2}\right)$$
The values of the objective function z = 7x + y at these vertices are
z(C) = 7(3) + 0 = 21
z(B) = 7(0) + 5 = 5
∴ z has minimum value 5, when x = 0 and y = 5.

Question 7.
Minimize : z = 8x + 10y subject to 2x + y ≥ 7, 2x + 3y ≥ 15,
y ≥ 2, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.

The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0,7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.
Substituting y – 2 in 2x + 3y = 15, we get 2x + 3(2) = 15
∴ 2x = 9 ∴ x = 4.5 ∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 … (1)
and 2x + y = 7 … (2)
On subtracting, we get
2y = 8 ∴ y = 4
∴ from (2), 2x + 4 = 7
∴ 2x = 3 ∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are
z(P) = 8(4.5) + 10(2) = 36 + 20 = 56
z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52
z(B) = 8(0) +10(7) = 70
∴ z has minimum value 52, when x = 1.5 and y = 4

Question 8.
Minimize : z = 6x + 21y subject to x + 2y ≥ 3, x + 4y ≥ 4,
3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 2y = 3, x + 4y = 4 and 3x + y = 3 respectively.

The feasible region is XCPQFY which is shaded in the graph.
The vertices of the feasible region are C (4, 0), P, Q and F(0, 3).
P is the point of intersection of the lines x + 4y = 4 and x + 2y = 3
On subtracting, we get
2y = 1 ∴ y = $$\frac{1}{2}$$
Substituting y = $$\frac{1}{2}$$ in x + 2y = 3, we get
x + 2$$\left(\frac{1}{2}\right)$$ = 3
∴ x = 2
∴ P = (2, $$\frac{1}{2}$$)
Q is the point of intersection of the lines
x + 2y = 3 … (1)
and 3x + y = 3 ….(2)
Multiplying equation (1) by 3, we get 3x + 6y = 9
Subtracting equation (2) from this equation, we get
5y = 6
∴ y = $$\frac{6}{5}$$
∴ from (1), x + 2$$\left(\frac{6}{5}\right)$$ = 3
∴ x = 3 – $$\frac{12}{5}=\frac{3}{5}$$
Q ≡ $$\left(\frac{3}{5}, \frac{6}{5}\right)$$
The values of the objective function z = 6x + 21y at these vertices are
z(C) = 6(4) + 21(0) = 24
z(P) = 6(2) + 21$$\left(\frac{1}{2}\right)$$
= 12 + 10.5 = 22.5
z(Q)= 6$$\left(\frac{3}{5}\right)$$ + 21$$\left(\frac{6}{5}\right)$$
= $$\frac{18}{5}+\frac{126}{5}=\frac{144}{5}$$ = 28.8
2 (F) = 6(0) + 21(3) = 63
∴ z has minimum value 22.5, when x = 2 and y = $$\frac{1}{2}$$.