Category: Class 12

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 = Ae^5x + Be^-5x
Solution:
y = Ae^5x + 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 = c₁e^2x + c₂e^5x
Solution:
y = c₁e^2x + c₂e^5x ………(1)
Differentiating twice w.r.t. x, we get
\(\frac{d y}{d x}\) = c₁e^2x × 2 + c₂e^5x × 5

The equations (1), (2) and (3) are consistent in c₁e^2x and c₂e^5x
∴ determinant of their consistency is zero.

This is the required D.E.

Alternative Method:
y = c₁e^2x + c₂e^5x
Dividing both sides by e^5x, 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)
∴ e^2x . 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.

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.

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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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 = \(\int_{1}^{4} x d y\)

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\([latex]\int_{0}^{4} x d x\)[/latex]
= 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

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) \(\log \left(\frac{189}{208}\right)\)
Answer:
(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) \(\frac{4+\pi}{2}\)
Answer:
(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
Answer:
(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) \(\frac{-5 \pi}{256}\)
Answer:
(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
Answer:
(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) \(\frac{\sqrt{e}-1}{e}\)
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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:

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:

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) = e^x, 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

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.5

Question 1.
Find the second order derivatives of the following:
(i) 2x⁵ – 4x³ – \(\frac{2}{x^{2}}\) – 9
Solution:
Let y = 2x⁵ – 4x³ – \(\frac{2}{x^{2}}\) – 9

(ii) e^2x . tan x
Solution:
Let y = e^2x . tan x

(iii) e^4x . cos 5x
Solution:
Let y = e^4x . cos 5x

(iv) x³ . log x
Solution:
Let y = x³ . log x

(v) log(log x)
Solution:
Let y = log(log x)

(vi) x^x
Solution:
y = x^x
log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

Question 2.
Find \(\frac{d^{2} y}{d x^{2}}\) of the following:
(i) x = a(θ – sin θ), y = a (1 – cos θ)
Solution:
x = a(θ – sin θ), y = a (1 – cos θ)
Differentiating x and y w.r.t. θ, we get
\(\frac{d x}{d \theta}=a \frac{d}{d \theta}(\theta-\sin \theta)\) = a(1 – cos θ) …….(1)

(ii) x = 2at², y = 4at
Solution:
x = 2at², y = 4at
Differentiating x and y w.r.t. t, we get

(iii) x = sin θ, y = sin³θ at θ = \(\frac{\pi}{2}\)
Solution:
x = sin θ, y = sin³θ
Differentiating x and y w.r.t. θ, we get,

(iv) x = a cos θ, y = b sin θ at θ = \(\frac{\pi}{4}\)
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

Question 3.
(i) If x = at² and y = 2at, then show that \(x y \frac{d^{2} y}{d x^{2}}+a=0\)
Solution:
x = at², y = 2at ………(1)
Differentiating x and y w.r.t. t, we get

(ii) If y = \(e^{m \tan ^{-1} x}\), show that \(\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-m) \frac{d y}{d x}=0\)
Solution:
y = \(e^{m \tan ^{-1} x}\) ……..(1)

(iii) If x = cos t, y = e^mt, show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-m^{2} y=0\)
Solution:
x = cos t, y = e^mt

(iv) If y = x + tan x, show that \(\cos ^{2} x \cdot \frac{d^{2} y}{d x^{2}}-2 y+2 x=0\)
Solution:
y = x + tan x

(v) If y = e^ax . sin (bx), show that y₂ – 2ay₁ + (a² + b²)y = 0.
Solution:
y = e^ax . sin (bx) ………(1)

(vi) If \(\sec ^{-1}\left(\frac{7 x^{3}-5 y^{3}}{7 x^{3}+5 y^{3}}\right)=m\), show that \(\frac{d^{2} y}{d x^{2}}=0\)
Solution:

(vii) If 2y = \(\sqrt{x+1}+\sqrt{x-1}\), show that 4(x² – 1)y₂ + 4xy₁ – y = 0.
Solution:
2y = \(\sqrt{x+1}+\sqrt{x-1}\) …… (1)
Differentiating both sides w.r.t. x, we get

(viii) If y = \(\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{m}\), show that \(\left(x^{2}+a^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=0\)
Solution:
y = \(\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{m}\) = \(m \log \left(x+\sqrt{x^{2}+a^{2}}\right)\)

(ix) If y = sin(m cos^-1x), then show that \(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+m^{2} y=0\)
Solution:
y = sin(m cos^-1x)
sin^-1y = m cos^-1x
Differentiating both sides w.r.t. x, we get

(x) If y = log(log 2x), show that xy₂ + y₁(1 + xy₁) = 0.
Solution:
y = log(log 2x)

(xi) If x² + 6xy + y² = 10, show that \(\frac{d^{2} y}{d x^{2}}=\frac{80}{(3 x+y)^{3}}\)
Solution:
x² + 6xy + y² = 10 …… (1)
Differentiating both sides w.r.t. x, we get

(xii) If x = a sin t – b cos t, y = a cos t + b sin t, Show that \(\frac{d^{2} y}{d x^{2}}=-\frac{x^{2}+y^{2}}{y^{3}}\)
Solution:
x = a sin t – b cos t, y = a cos t + b sin t
Differentiating x and y w.r.t. t, we get

Question 4.
Find the nth derivative of the following:
(i) (ax + b)^m
Solution:
Let y = (ax + b)^m

(ii) \(\frac{1}{x}\)
Solution:
Let y = \(\frac{1}{x}\)

(iii) e^ax+b
Solution:
Let y = e^ax+b

(iv) a^px+q
Solution:
Let y = a^px+q

(v) log(ax + b)
Solution:
Let y = log(ax + b)
Then \(\frac{d y}{d x}=\frac{d}{d x}[\log (a x+b)]\)

(vi) cos x
Solution:
Let y = cos x

(vii) sin(ax + b)
Solution:
Let y = sin(ax + b)

(viii) cos(3 – 2x)
Solution:

(ix) log(2x + 3)
Solution:

(x) \(\frac{1}{3 x-5}\)
Solution:
Let y = \(\frac{1}{3 x-5}\)

(xi) y = e^ax . cos (bx + c)
Solution:
y = e^ax . cos (bx + c)

(xii) y = e^8x . cos (6x + 7)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.4

Question 1.
Find \(\frac{d y}{d x}\) if
(i) x = at², y = 2at
Solution:
x = at², y = 2at
Differentiating x and y w.r.t. t, we get

(ii) x = a cot θ, y = b cosec θ
Solution:
x = a cot θ, y = b cosec θ
Differentiating x and y w.r.t. θ, we get

(iii) x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Solution:
x = \(\sqrt{a^{2}+m^{2}}\), y = log (a² + m²)
Differentiating x and y w.r.t. m, we get
\(\frac{d x}{d m}=\frac{d}{d m}\left(\sqrt{a^{2}+m^{2}}\right)\)

(iv) x = sin θ, y = tan θ
Solution:
x = sin θ, y = tan θ
Differentiating x and y w.r.t. θ, we get

(v) x = a(1 – cos θ), y = b(θ – sin θ)
Solution:
x = a(1 – cos θ), y = b(θ – sin θ)
Differentiating x and y w.r.t. θ, we get

(vi) x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\), where a > 0, a ≠ 1 and t ≠ 0
Solution:
x = \(\left(t+\frac{1}{t}\right)^{a}\), y = \(a^{t+\frac{1}{t}}\) ………(1)
Differentiating x and y w.r.t. t, we get

(vii) x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Solution:
x = \(\cos ^{-1}\left(\frac{2 t}{1+t^{2}}\right)\), y = \(\sec ^{-1}\left(\sqrt{1+t^{2}}\right)\)
Put t = tan θ Then θ = tan^-1t

(viii) x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Solution:
x = cos^-1(4t³ – 3t), y = \(\tan ^{-1}\left(\frac{\sqrt{1-t^{2}}}{t}\right)\)
Put t = cos θ. Then θ = cos^-1t
x = cos^-1(4cos³θ – 3cos θ)

Question 2.
Find \(\frac{d y}{d x}\), if
(i) x = cosec²θ, y = cot³θ at θ = \(\frac{\pi}{6}\)
Solution:
x = cosec²θ, y = cot³θ
Differentiating x and y w.r.t. θ, we get

(ii) x = a cos³θ, y = a sin³θ at θ = \(\frac{\pi}{3}\)
Solution:
x = a cos³θ, y = a sin³θ
Differentiating x and y w.r.t. θ, we get

(iii) x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\)) at t = 1
Solution:
x = t² + t + 1, y = sin(\(\frac{\pi t}{2}\)) + cos(\(\frac{\pi t}{2}\))
Differentiating x and y w.r.t. t, we get

(iv) x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = \(\frac{\pi}{4}\)
Solution:
x = 2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get

(v) x = t + 2 sin(πt), y = 3t – cos(πt) at t = \(\frac{1}{2}\)
Solution:
x = t + 2 sin(πt), y = 3t – cos(πt)
Differentiating x and y w.r.t. t, we get

Question 3.
(i) If x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\), then show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Solution:
x = \(a \sqrt{\sec \theta-\tan \theta}\), y = \(a \sqrt{\sec \theta+\tan \theta}\)

(ii) If x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\), then show that \(\frac{d y}{d x}=-\frac{y \log x}{x \log y}\)
Solution:
x = \(e^{\sin 3 t}\), y = \(e^{\cos 3 t}\)
log x = log \(e^{\sin 3 t}\), log y = log \(e^{\cos 3 t}\)
log x = (sin 3t)(log e), log y = (cos 3t)(log e)
log x = sin 3t, log y = cos 3t ….. (1) [∵ log e = 1]
Differentiating both sides w.r.t. t, we get

(iii) If x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\), then show that y2 – \(\frac{d y}{d x}\) = 0.
Solution:
x = \(\frac{t+1}{t-1}\), y = \(\frac{1-t}{t+1}\)

(iv) If x = a cos³t, y = a sin³t, then show that \(\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}\)
Solution:
x = a cos³t, y = a sin³t
Differentiating x and y w.r.t. t, we get

(v) If x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3), show that \(\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}\)
Solution:
x = 2 cos⁴(t + 3), y = 3 sin⁴(t + 3)

(vi) If x = log (1 + t²), y = t – tan^-1t, show that \(\frac{d y}{d x}=\frac{\sqrt{e^{x}-1}}{2}\)
Solution:
x = log (1 + t²), y = t – tan^-1t
Differentiating x and y w.r.t. t, we get

(vii) If x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\), show that sin x + \(\frac{d y}{d x}\) = 0
Solution:
x = \(\sin ^{-1}\left(e^{t}\right)\), y = \(\sqrt{1-e^{2 t}}\)
Differentiating x and y w.r.t. t, we get

(viii) If x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\), show that \(\frac{d x}{d y}=-\frac{b^{2} y}{a^{2} x}\)
Solution:
x = \(\frac{2 b t}{1+t^{2}}\), y = \(a\left(\frac{1-t^{2}}{1+t^{2}}\right)\)

Question 4.
(i) Differentiate x sin x w.r.t tan x.
Solution:
Let u = x sinx and v = tan x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get

(ii) Differentiate \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) w.r.t \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let u = \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) and v = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)

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

(iv) Differentiate \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) w.r.t. tan^-1x
Solution:
Let u = \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) and v = tan^-1x
Then we want to find \(\frac{d u}{d v}\)
Put x = tan θ. Then θ = tan^-1x.

(v) Differentiate 3x w.r.t. log_x3.
Solution:
Let u = 3x and v = log_x3.
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(3^{x}\right)=3^{x} \cdot \log 3\)

(vi) Differentiate \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) w.r.t. sec^-1x.
Solution:
Let u = \(\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)\) and v = sec^-1x
Then we want to find \(\frac{d u}{d v}\).
Differentiating u and v w.r.t. x, we get

(vii) Differentiate x^x w.r.t. x^{sin x}.
Solution:
Let u = x^x and v = x^{sin x}
Then we want to find \(\frac{d u}{d x}\).
Take, u = x^x
log u = log x^x = x log x
Differentiating both sides w.r.t. x, we get

(viii) 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:
Let u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) and v = \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Then we want to find \(\frac{d u}{d v}\)
u = \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\)
Put x = tan θ. Then θ = tan^-1x and

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.3

Question 1.
Differentiate the following w.r.t. x:
(i) \(\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}\)
Solution:
Let y = \(\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}\)
Then, log y = log [latex]\frac{(x+1)^{2}}{(x+2)^{3}(x+3)^{4}}[/latex]
= log (x + 1)² – log (x + 2)³ – log (x + 3)⁴
= 2 log (x +1) – 3 log (x + 2) – 4 log (x + 3)
Differentiating both sides w.r.t. x, we get

(ii) \(\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}\)
Solution:
Let y = \(\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}\)
Then log y = log [latex]\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^{2}}}[/latex]

Differentiating both sides w.r.t. x, we get

(iii) \(\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}\)
Solution:
Let y = \(\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}\)
Then log y = log [latex]\left(x^{2}+3\right)^{\frac{3}{2}} \cdot \sin ^{3} 2 x \cdot 2^{x^{2}}[/latex]

Differentiating both sides w.r.t. x, we get

(iv) \(\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}\)
Solution:
Let y = \(\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}\)
Then log y = log [latex]\frac{\left(x^{2}+2 x+2\right)^{\frac{3}{2}}}{(\sqrt{x}+3)^{3}(\cos x)^{x}}[/latex]

Differentiating both sides w.r.t. x, we get

(v) \(\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}\)
Solution:
Let y = \(\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}\)
Then log y = log [latex]\frac{x^{5} \cdot \tan ^{3} 4 x}{\sin ^{2} 3 x}[/latex]
= log x⁵ + log tan³4x – log sin²3x
= 5 log x+ 3 log (tan 4x) – 2 log (sin 3x)
Differentiating both sides w.r.t. x, we get

(vi) \(x^{\tan ^{-1} x}\)
Solution:
Let y = \(x^{\tan ^{-1} x}\)
Then log y = log (\(x^{\tan ^{-1} x}\)) = (tan^-1 x)(log x)
Differentiating both sides w.r.t. x, we get

(vii) (sin x)^x
Solution:
Let y = (sin x)^x
Then log y = log (sin x)^x = x . log (sin x)
Differentiating both sides w.r.t. x, we get

(viii) sin x^x
Solution:
Let y = (sin x^x)
Then \(\frac{d y}{d x}=\frac{d}{d x}\left[\left(\sin x^{x}\right)\right]\)
\(\frac{d y}{d x}=\cos \left(x^{x}\right) \cdot \frac{d}{d x}\left(x^{x}\right)\) ……. (1)
Let u = x^x
Then log u = log x^x = x . log x
Differentiating both sides w.r.t. x, we get

Question 2.
Differentiate the following w.r.t. x:
(i) x^e + x^x + e^x + e^e
Solution:
Let y = x^e + x^x + e^x + e^e
Let u = x^x
Then log u = log x^x = x log x
Differentiating both sides w.r.t. x, we get
\(\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(x \log x)\)

(ii) \(x^{x^{x}}+e^{x^{x}}\)
Solution:
Let y = \(x^{x^{x}}+e^{x^{x}}\)
Put u = \(x^{x^{x}}\) and v = \(e^{x^{x}}\)
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Take u = \(x^{x^{x}}\)
log u = log \(x^{x^{x}}\) = x^x . log x
Differentiating both sides w.r.t. x, we get

(iii) (log x)^x – (cos x)^{cot x}
Solution:
Let y = (log x)^x – (cos x)^{cot x}
Put u = (log x)^x and v = (cos x)^{cot x}
Then y = u – v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\) ……..(1)
Take u = (log x)^x
∴ log u = log (log x)^x = x . log (log x)
Differentiating both sides w.r.t. x, we get

(iv) \(x^{e^{x}}+(\log x)^{\sin x}\)
Solution:
Let y = \(x^{e^{x}}+(\log x)^{\sin x}\)
Put u = \(x^{e^{x}}\) and v = (log x)^{sin x}
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……….(1)
Take u = \(x^{e^{x}}\)
∴ log u = log \(x^{e^{x}}\) = e^x . log x
Differentiating both sides w.r.t. x, we get

Also, v = (log x)^{sin x}
∴ log v = log (log x)^{sin x} = (sin x) . (log log x)
Differentiating both sides w.r.t. x, we get
\(\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}[(\sin x) \cdot(\log \log x)]\)
= \((\sin x) \cdot \frac{d}{d x}\left[(\log \log x)+(\log \log x) \cdot \frac{d}{d x}(\sin x)\right]\)

(v) \(e^{\tan x}+(\log x)^{\tan x}\)
Solution:
Let y = \(e^{\tan x}+(\log x)^{\tan x}\)
Put u = (log x)^{tan x}
∴ log u =log(log x)^{tan x} = (tan x).(log log x)
Differentiating both sides w.r.t. x, we get

(vi) (sin x)^{tan x} + (cos x)^{cot x}
Solution:
Let y = (sin x)^{tan x} + (cos x)^{cot x}
Put u = (sin x)^{tan x} and v = (cos x)^{cot x}
Then y = u + v
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ………(1)
Take u = (sin x)^{tan x}
∴ log u = log (sin x)^{tan x} = (tan x) . (log sin x)
Differentiating both sides w.r.t. x, we get

(vii) \(10^{x^{x}}+x^{x^{10}}+x^{10^{x}}\)
Solution:
Let y = \(10^{x^{x}}+x^{x^{10}}+x^{10^{x}}\)
Put u = \(10^{x^{x}}\), v = \(x^{x^{10}}\) and w = \(x^{10^{x}}\)
Then y = u + v + w
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}\) ………(1)

(viii) \(\left[(\tan x)^{\tan x}\right]^{\tan x}\) at x = \(\frac{\pi}{4}\)
Solution:
Let y = \(\left[(\tan x)^{\tan x}\right]^{\tan x}\)
∴ log y = log [latex]\left[(\tan x)^{\tan x}\right]^{\tan x}[/latex]
= tan x . log(tan x)^{tan x}
= tan x . tan x log (tan x)
= (tan x)² . log (tan x)
Differentiating both sides w.r.t. x, we get

Question 3.
Find \(\frac{d y}{d x}\) if
(i) √x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get

(ii) x√x + y√y = a√a
Solution:
x√x + y√y = a√a
∴ \(x^{\frac{3}{2}}+y^{\frac{3}{2}}=a^{\frac{3}{2}}\)
Differentiating both sides w.r.t. x, we get

(iii) x + √xy + y = 1
Solution:
x + √xy + y = 1
Differentiating both sides w.r.t. x, we get

(iv) x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get

(v) x²y² – tan^-1(\(\sqrt{x^{2}+y^{2}}\)) = cot^-1(\(\sqrt{x^{2}+y^{2}}\))
Solution:
x²y² – tan^-1(\(\sqrt{x^{2}+y^{2}}\)) = cot^-1(\(\sqrt{x^{2}+y^{2}}\))
∴ x²y² = tan^-1(\(\sqrt{x^{2}+y^{2}}\)) + cot^-1(\(\sqrt{x^{2}+y^{2}}\))
∴ x²y² = \(\frac{\pi}{2}\) …….[∵ \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)]
Differentiating both sides w.r.t. x, we get

(vi) xe^y + ye^x = 1
Solution:
xe^y + ye^x = 1
Differentiating both sides w.r.t. x, we get

(vii) e^x+y = cos (x – y)
Solution:
e^x+y = cos (x – y)
Differentiating both sides w.r.t. x, we get

(viii) cos (xy) = x + y
Solution:
cos (xy) = x + y
Differentiating both sides w.r.t. x, we get

(ix) \(e^{e^{x-y}}=\frac{x}{y}\)
Solution:
\(e^{e^{x-y}}=\frac{x}{y}\)
∴ e^x-y = log(\(\frac{x}{y}\)) …….[e^x = y ⇒ x = log y]
∴ e^x-y = log x – log y
Differentiating both sides w.r.t. x, we get

Question 4.
Show that \(\frac{d y}{d x}=\frac{y}{x}\) in the following, where a and p are constants.
(i) x⁷y⁵ = (x + y)¹²
Solution:
x⁷y⁵ = (x + y)¹²
(log x⁷y⁵) = log(x + y)¹²
log x⁷ + log y⁵ = log(x + y)¹²
7 log x + 5 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get

(ii) x^py⁴ = (x + y)^p+4, p∈N
Solution:
x^py⁴ = (x + y)^p+4
Taking log
log (x^py⁴) = log(x + y)^p+4
log x^p + log y⁴ = (p + 4) log(x + y)
p log x + 4 log y = (p + 4) log(x + y)
Differentiating both sides w.r.t. x, we get

(iii) \(\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}\)
Solution:
\(\sec \left(\frac{x^{5}+y^{5}}{x^{5}-y^{5}}\right)=a^{2}\)

Differentiating both sides w.r.t. x, we get


Differentiating both sides w.r.t. x, we get

(iv) \(\tan ^{-1}\left(\frac{3 x^{2}-4 y^{2}}{3 x^{2}+4 y^{2}}\right)=a^{2}\)
Solution:
\(\tan ^{-1}\left(\frac{3 x^{2}-4 y^{2}}{3 x^{2}+4 y^{2}}\right)=a^{2}\)

Differentiating both sides w.r.t. x, we get

(v) \(\cos ^{-1}\left(\frac{7 x^{4}+5 y^{4}}{7 x^{4}-5 y^{4}}\right)=\tan ^{-1} a\)
Solution:

(vi) \(\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20\)
Solution:
\(\log \left(\frac{x^{20}-y^{20}}{x^{20}+y^{20}}\right)=20\)

Differentiating both sides w.r.t. x, we get

(vii) \(e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a\)
Solution:
\(e^{\frac{x^{7}-y^{7}}{x^{7}+y^{7}}}=a\)

Differentiating both sides w.r.t. x, we get

(viii) \(\sin \left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=a^{3}\)
Solution:

Differentiating both sides w.r.t. x, we get

Question 5.
(i) If log (x + y) = log (xy) + p, where p is a constant, then prove that \(\frac{d y}{d x}=-\frac{y^{2}}{x^{2}}\).
Solution:
log (x + y) = log (xy) + p
∴ log (x + y) = log x + log y + p
Differentiating both sides w.r.t. x, we get

(ii) If \(\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2\), show that \(\frac{d y}{d x}=-\frac{99 x^{2}}{101 y^{2}}\)
Solution:
\(\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2\)

(iii) If \(\log _{5}\left(\frac{x^{4}+y^{4}}{x^{4}-y^{4}}\right)=2\), show that \(\frac{d y}{d x}=-\frac{12 x^{3}}{13 y^{3}}\)
Solution:

Differentiating both sides w.r.t. x, we get

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

(v) If \(\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}\), show that \(\frac{d y}{d x}=\frac{x^{4}}{3 y^{4}}\)
Solution:
\(\sin ^{-1}\left(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}\right)=\frac{\pi}{6}\)
\(\frac{x^{5}-y^{5}}{x^{5}+y^{5}}=\sin \frac{\pi}{6}=\frac{1}{2}\)
2x⁵ – 2y⁵ = x⁵ + y⁵
3y⁵ = x⁵
Differentiating both sides w.r.t. x, we get
\(3 \times 5 y^{4} \frac{d y}{d x}=5 x^{4}\)
∴ \(\frac{d y}{d x}=\frac{x^{4}}{3 y^{4}}\)

(vi) If x^y = e^x-y, then show that \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}}\)
Solution:
x^y = e^x-y
log x^y = log e^x-y
y log x = (x – y) log e
y log x = (x – y) ….. [∵ log e = 1]
y + y log x = x – y
y + y log x = x
y(1 + log x) = x
y = \(\frac{x}{1+\log x}\)

(vii) If \(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}\), then show that \(\frac{d y}{d x}=\frac{\sin x}{1-2 y}\)
Solution:
\(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}\)
y² = cos x + \(\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}\)
y² = cos x + y
Differentiating both sides w.r.t. x, we get

(viii) If \(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}\), then show that \(\frac{d y}{d x}=\frac{1}{x(2 y-1)}\)
Solution:
\(y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}\)

(ix) If \(y=x^{x^{x^{-\infty}}}\), then show that \(\frac{d y}{d x}=\frac{y^{2}}{x(1-\log y)}\)
Solution:
\(y=x^{x^{x^{-\infty}}}\)

(x) If e^y = y^x, then show that \(\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}\)
Solution:
e^y = y^x
log e^y = log y^x
y log e = x log y
y = x log y …… [∵log e = 1] ……….(1)
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=x \frac{d}{d x}(\log y)+(\log y) \cdot \frac{d}{d x}(x)\)