Class 12

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

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

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

Question 1.
The order and degree of the differential equation \(\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}\) are respectively……..
(a) 2, 1
(b) 1, 2
(c) 3, 2
(d) 2, 3
Answer:
(d) 2, 3

Question 2.
The differential equation of y = c² + \(\frac{c}{x}\) is…….
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)
(b) \(\frac{d y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
(c) \(x^{3}\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}=y\)
(d) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
Answer:
(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

Question 3.
x² + y² = a² is a solution of ………
(a) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)
(b) \(y=x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}+a^{2} y\)
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)
(d) \(\frac{d^{2} y}{d x^{2}}=(x+1) \frac{d y}{d x}\)
Answer:
(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)

Question 4.
The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is
(a) \(y^{2}\left(1+\frac{d y}{d x}\right)=25\)
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(c) \((y-5)^{2}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)
(d) \((y-5)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=25\)
Answer:
(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

Question 5.
The differential equation y \(\frac{d y}{d x}\) + x = 0 represents family of ………
(a) circles
(b) parabolas
(c) ellipses
(d) hyperbolas
Answer:
(a) circles

Hint:
y \(\frac{d y}{d x}\) + x = 0
∴ ∫y dy + ∫x dx = c
∴ \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=c\)
∴ x² + y² = 2c which is a circle.

Question 6.
The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……
(a) \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)
(b) x tan^-1x + c = 0
(c) x – tan^-1x = c
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)
Answer:
(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)

Question 7.
The solution of (x + y)² \(\frac{d y}{d x}\) = 1 is…….
(a) x = tan^-1(x + y) + c
(b) y tan^-1(\(\frac{x}{y}\)) = c
(c) y = tan^-1(x + y) + c
(d) y + tan^-1(x + y) = c
Answer:
(c) y = tan^-1(x + y) + c

Question 8.
The Solution of \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}-y^{2}}}{2}\) is……
(a) sin^-1(\(\frac{y}{x}\)) = 2 log |x| + c
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c
(c) sin(\(\frac{x}{y}\)) = log |x| + c
(d) sin(\(\frac{y}{x}\)) = log |y| + c
Answer:
(b) sin^-1(\(\frac{y}{x}\)) = log |x| + c

Question 9.
The solution of \(\frac{d y}{d x}\) + y = cos x – sin x is……
(a) y e^x = cos x + c
(b) y e^x + e^x cos x = c
(c) y e^x = e^x cos x + c
(d) y² e^x = e^x cos x + c
Answer:
(c) y e^x = e^x cos x + c
Hint:
\(\frac{d y}{d x}\) + y = cos x – sin x
I.F. = \(e^{\int 1 d x}=e^{x}\)
∴ the solution is y . e^x = ∫(cos x – sin x) e^x + c
∴ y . e^x = e^x cos x + c

Question 10.
The integrating factor of linear differential equation x \(\frac{d y}{d x}\) + 2y = x² log x is……..
(a) \(\frac{1}{x}\)
(b) k
(c) \(\frac{1}{n^{2}}\)
(d) x²
Answer:
(d) x²
Hint:
I.F. = \(e^{\int \frac{2}{x} d x}\)
= e^{2 log x}
= x²

Question 11.
The solution of the differential equation \(\frac{d y}{d x}\) = sec x – y tan x is…….
(a) y sec x + tan x = c
(b) y sec x = tan x + c
(c) sec x + y tan x = c
(d) sec x = y tan x + c
Answer:
(b) y sec x = tan x + c

Hint:
\(\frac{d y}{d x}\) = sec x – y tan x
∴ \(\frac{d y}{d x}\) + y tan x = sec x
I.F. = \(e^{\int \tan x d x}=e^{\log \sec x}\) = sec x
∴ the solution is
y . sec x = ∫sec x . sec x dx + c
∴ y sec x = tan x + c

Question 12.
The particular solution of \(\frac{d y}{d x}=x e^{y-x}\), when x = y = 0 is……
(a) e^x-y = x + 1
(b) e^x+y = x + 1
(c) e^x + e^y = x + 1
(d) e^y-x = x – 1
Answer:
(a) e^x-y = x + 1

Question 13.
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is a solution of……..
(a) \(\frac{d^{2} y}{d x^{2}}+y x+\left(\frac{d y}{d x}\right)^{2}=0\)
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)
(c) \(y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+y=0\)
(d) \(x y \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}=0\)
Answer:
(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)

Question 14.
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……
(a) 5\(\frac{2}{3}\) grams
(b) 5\(\frac{1}{3}\) grams
(c) 5.1 grams
(d) 5 grams
Answer:
(b) 5\(\frac{1}{3}\) grams

Question 15.
If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..
(a) 51.7°C
(b) 54.7°C
(c) 52.7°C
(d) 50.7°C
Answer:
(b) 54.7°C

(II) Solve the following:

Question 1.
Determine the order and degree of the following differential equations:
(i) \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)
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.

(ii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2 \times 5}=1+\frac{d y}{d x}\)
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{10}=1+\frac{d y}{d x}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 10.
∴ the given D.E. is of order 3 and degree 10.

(iii) \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
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.

(iv) \(\frac{d y}{d x}=3 y+\sqrt[4]{1+5\left(\frac{d y}{d x}\right)^{2}}\)
Solution:
The given D.E. is

This D.E. has the highest order derivative \(\frac{d y}{d x}\) with power 4.
∴ the given D.E. is of order 1 and degree 4.

(v) \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\).
∴ order = 4
Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

Question 2.
In each of the following examples verify that the given function is a solution of the differential equation.
(i) \(x^{2}+y^{2}=r^{2} ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)
Solution:
x² + y² = r² ……. (1)
Differentiating both sides w.r.t. x, we get

Hence, x² + y² = r² is a solution of the D.E.
\(x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

(ii) y = e^ax sin bx; \(\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0\)
Solution:

(iii) y = 3 cos(log x) + 4 sin(log x); \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)
Solution:
y = 3 cos(log x) + 4 sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(iv) xy = ae^x + be^-x + x²; \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+x^{2}=x y+2\)
Solution:

(v) x² = 2y² log y, x² + y² = xy \(\frac{d x}{d y}\)
Solution:
x² = 2y² log y ……(1)
Differentiating both sides w.r.t. y, we get

∴ x² + y² = xy \(\frac{d x}{d y}\)
Hence, x² = 2y² log y is a solution of the D.E.
x² + y² = xy \(\frac{d x}{d y}\)

Question 3.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) y² = a(b – x)(b + x)
Solution:
y² = a(b – x)(b + x) = a(b² – x²)
Differentiating both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = a(0 – 2x) = -2ax
∴ y \(\frac{d y}{d x}\) = -ax …….(1)
Differentiating again w.r.t. x, we get

This is the required D.E.

(ii) y = a sin(x + b)
Solution:
y = a sin(x + b)

This is the required D.E.

(iii) (y – a)² = b(x + 4)
Solution:
(y – a)² = b(x + 4) …….(1)
Differentiating both sides w.r.t. x, we get
\(2(y-a) \cdot \frac{d}{d x}(y-a)=b \frac{d}{d x}(x+4)\)

(iv) y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
Solution:
y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)
∴ y² = a cos (log x) + b sin (log x) …….(1)
Differentiating both sides w.r.t. x, we get

(v) y = Ae^3x+1 + Be^-3x+1
Solution:
y = Ae^3x+1 + Be^-3x+1 …… (1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

Question 4.
Form the differential equation of:
(i) all circles which pass through the origin and whose centres lie on X-axis.
Solution:

Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.
∴ equation of the circle is (x – h)² + (y – 0)² = h²
∴ x² – 2hx + h² + y² = h²
∴ x² + y² = 2hx ……..(1)
Differentiating both sides w.r.t. x, we get
2x + 2y \(\frac{d y}{d x}\) = 2h
Substituting the value of 2h in equation (1), we get
x² + y² = (2x + 2y \(\frac{d y}{d x}\)) x
∴ x² + y² = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) + x² – y² = 0
This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.
Solution:
Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.
Then equation of the parabola is
(x – h)² = 4b(y – k) ……. (1)
where h and k are arbitrary constants.

Differentiating both sides of (1) w.r.t. x, we get
2(x – h). \(\frac{d}{d x}\)(x – h) = 4b . \(\frac{d}{d x}\)(y – k)
∴ 2(x – h) x (1 – 0) = 4b(\(\frac{d y}{d x}\) – 0)
∴ (x – h) = 2b \(\frac{d y}{d x}\)
Differentiating again w.r.t. x, we get
1 – 0 = 2b \(\frac{d^{2} y}{d x^{2}}\)
∴ 2b \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0
This is the required D.E.

(iii) an ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ 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.

(iv) all the lines which are normal to the line 3x + 2y + 7 = 0.
Solution:
Slope of the line 3x – 2y + 7 = 0 is \(\frac{-3}{-2}=\frac{3}{2}\).
∴ slope of normal to this line is \(-\frac{2}{3}\)
Then the equation of the normal is
y = \(-\frac{2}{3}\)x + k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{2}{3} \times 1+0\)
∴ 3\(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

(v) the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\).
Solution:
The equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\)
i.e., \(\frac{x^{2}}{16 k}-\frac{y^{2}}{36 k}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get
a² = 16k, b² = 36k
∴ a = 4√k, b = 6√k
∴ l(transverse axis) = 2a = 8√k
and l(conjugate axis) = 2b = 12√k
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = 4√k and 2B = b = 6√k
∴ A = 2√k and B = 3√k
∴ equation of the required hyperbola is
\(\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1\)
i.e., \(\frac{x^{2}}{4 k}-\frac{y^{2}}{9 k}=1\)
∴ 9x² – 4y² = 36k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
9 × 2x – 4 × 2y \(\frac{d y}{d x}\) = 0
∴ 9x – 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Question 5.
Solve the following differential equations:
(i) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:

(ii) \(\frac{d y}{d x}\) = x²y + y
Solution:

(iii) \(\frac{d y}{d x}=\frac{2 y-x}{2 y+x}\)
Solution:

(iv) x dy = (x + y + 1) dx
Solution:

(v) \(\frac{d y}{d x}\) + y cot x = x² cot x + 2x
Solution:
\(\frac{d y}{d x}\) + y cot x = x cot x + 2x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = cot x and Q = x² cot x + 2x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \cot x d x}\)
= \(e^{\log (\sin x)}\)
= sin x
∴ the solution of (1) is given by
y(I.F.) = ∫Q . (I.F.) dx + c
∴ y sin x = ∫(x² cot x + 2x) sin x dx + c
∴ y sinx = ∫(x² cot x . sin x + 2x sin x) dx + c
∴ y sinx = ∫x² cos x dx + 2∫x sin x dx + c
∴ y sinx = x² ∫cos x dx – ∫[\(\frac{d}{d x}\left(x^{2}\right)\) ∫cos x dx] dx + 2∫x sin x dx + c
∴ y sin x = x² (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c
∴ y sin x = x² sin x – 2∫x sin x dx + 2∫x sin x dx + c
∴ y sin x = x² sin x + c
∴ y = x² + c cosec x
This is the general solution.

(vi) y log y = (log y² – x) \(\frac{d y}{d x}\)
Solution:

(vii) 4 \(\frac{d x}{d y}\) + 8x = 5e^-3y
Solution:

Question 6.
Find the particular solution of the following differential equations:
(i) y(1 + log x) = (log xx) \(\frac{d y}{d x}\), when y(e) = e²
Solution:

(ii) (x + 2y²) \(\frac{d y}{d x}\) = y, when x = 2, y = 1
Solution:


This is the general solution.
When x = 2, y = 1, we have
2 = 2(1)² + c(1)
∴ c = 0
∴ the particular solution is x = 2y².

(iii) \(\frac{d y}{d x}\) – 3y cot x = sin 2x, when y(\(\frac{\pi}{2}\)) = 2
Solution:
\(\frac{d y}{d x}\) – 3y cot x = sin 2x
\(\frac{d y}{d x}\) = (3 cot x) y = sin 2x ……..(1)
This is the linear differential equation of the form

(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y
Solution:

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

Question 7.
Show that the general solution of defferential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) is given by (x + y + 1) = c(1 – x – y – 2xy).
Solution:

Question 8.
The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Solution:
Let P(x, y) be a point on the curve y = f(x).
Then slope of the normal to the curve is \(-\frac{1}{\left(\frac{d y}{d x}\right)}\)
∴ equation of the normal is

This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
2² + 3² = 4(2) + c
∴ c = 5
∴ equation of the required curve is x² + y² = 4x + 5.

Question 9.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.
Solution:
Let r be the radius and V be the volume of the spherical balloon at any time t.
Then the rate of change in volume of the spherical balloon is \(\frac{d V}{d t}\) which is a constant.


Hence, the radius of the spherical balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\) units.

Question 10.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in 2\(\frac{2}{9}\) years from the start.
Solution:
Let x be the assets of the presort at time t years.
Then the rate of reduction is \(\frac{d x}{d t}\) which is proportional to √x.
∴ \(\frac{d x}{d t}\) ∝ √x
∴ \(\frac{d x}{d t}\) = -k√x, where k > 0
∴ \(\frac{d x}{\sqrt{x}}\) = -k dt
Integrating both sides, we get
\(\int x^{-\frac{1}{2}} d x\) = -k∫dt
∴ \(\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\) = -kt + c
∴ 2√x = -kt + c
At the beginning, i.e. at t = 0, x = 10,00,000
2√10,00,000 = -k(0) + c
∴ c = 2000
∴ 2√x = -kt + 2000 ……..(1)
Also, when t = 2, x = 10,000
∴ 2√10000 = -k × 2 + 2000
∴ 2k = 1800
∴ k = 900
∴ (1) becomes,
∴ 2√x = -900t + 2000
When the person will be bankrupt, x = 0
∴ 0 = -900t + 2000
∴ 900t = 2000
∴ t = \(\frac{20}{9}=2 \frac{2}{9}\)
Hence, the person will be bankrupt in \(2 \frac{2}{9}\) years.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.6 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.6

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x₀
log x₀ = k × 0 + c
∴ c = log x₀
∴ log x = kt + log x₀
∴ log x – log x₀ = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ………(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀

∴ the number of bacteria will be 8 times the original number in 12 hours.

Question 2.
If the population of a country doubles in 60 years; in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
[Given log 2 = 0.6912, log 3 = 1.0986]
Solution:
Let P be the population at time t years.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{d P}{P}\) = k∫dt + c
∴ log P = kt + c
Initially i.e. when t = 0, let P = P₀
∴ log P₀ = k x 0 + c
∴ c = log P₀
∴ log P = kt + log P₀
∴ log P – log P₀ = kt
∴ log(\(\frac{P}{P_{0}}\)) = kt ……(1)
Since, the population doubles in 60 years, i.e. when t = 60, P = 2P₀

∴ the population becomes triple in 95.4 years (approximately).

Question 3.
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
Solution:
Let θ°C be the temperature of the body at time t minutes.
The room temperature is given to be 25°C.
Then by Newton’s law of cooling, \(\frac{d \theta}{d t}\), the rate of change of temperature, is proportional to (θ – 25).


∴ the temperature of the body will be 36.36°C after 1 hour.

Question 4.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 2½ hours. [Take √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
\(\int \frac{d x}{x}\) = k∫dt + c
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt ……(1)
Now, when t = 1, x = 2 × 1000 = 2000
∴ log(\(\frac{2000}{1000}\)) = k
∴ k = log 2

∴ the number of bacteria after 2½ hours = 5656.

Question 5.
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
Solution:
Let m be the mass of the radioactive element at time t.
Then the rate of disintegration is \(\frac{d m}{d t}\) which is proportional to m.


∴ log(3)^-1 = -kt
∴ -log 3 = -kt
∴ t = \(\frac{1}{k}\) log 3
∴ the original mass will disintegrate to 0.5 gm when t = \(\frac{1}{k}\) log 3

Question 6.
The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
Solution:
Let x gm be the amount of the substance left at time t.
Then the rate of decay is \(\frac{d x}{d t}\), which is proportional to x.


∴ \(\frac{x}{25}=\frac{27}{125}\)
∴ x = \(\frac{27}{5}\)
∴ the amount left after 3 hours \(\frac{27}{5}\) gm.

Question 7.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
\(\int \frac{1}{P} d P\) = k∫dt + c
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k × 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt
∴ log(\(\frac{P}{30000}\)) = kt …….(1)
Now, when t = 40, P = 40000

∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Question 8.
A body cools according to Newton’s law from 100°C to 60°C in 20 minutes. The temperature of the surroundings is 20°C. How long will it take to cool down to 30°C?
Solution:
Let θ°C be the temperature of the body at time t.
The temperature of the surrounding is given to be 20°C.
According to Newton’s law of cooling


∴ the body will cool down to 30°C in 60 minutes, i.e. in 1 hour.

Question 9.
A right circular cone has a height of 9 cm and a radius of the base of 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec, where A is the area of the water surface
at that instant, show that the vessel will be full in 75 seconds.
Solution:

Let r be the radius of the water surface and h be the height of the water at time t.
∴ area of the water surface A = πr² sq cm.
Since height of the right circular cone is 9 cm and radius of the base is 5 cm.
\(\frac{r}{h}=\frac{5}{9}\)
∴ r = \(\frac{5}{9} h\)
∴ area of water surface, i.e. A = \(\pi\left(\frac{5}{9} h\right)^{2}\)
∴ A = \(\frac{25 \pi h^{2}}{81}\) ……..(1)
The water level, i.e. the rate of change of h is \(\frac{d h}{d t}\) rises at the rate of \(\left(\frac{\pi}{A}\right)\) cm/sec.

∴ t = \(\frac{81 \times 9 \times 25}{3 \times 81}\) = 75
Hence, the vessel will be full in 75 seconds.

Question 10.
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
Solution:
Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then V = \(\frac{4}{3}\)πr³ and S = 4πr²
The rate at which the raindrop evaporates is \(\frac{d V}{d t}\) which is proportional to the surface area.
∴ \(\frac{d V}{d t}\) ∝ S
∴ \(\frac{d V}{d t}\) = -kS, where k > 0 ………(1)

On integrating, we get
∫dr = -k∫dt + c
∴ r = -kt + c
Initially, i.e. when t = 0, r = 3
∴ 3 = -k × 0 + c
∴ c = 3
∴ r = -kt + 3
When t = 1, r = 2
∴ 2 = -k × 1 + 3
∴ k = 1
∴ r = -t + 3
∴ r = 3 – t, where 0 ≤ t ≤ 3.
This is the required expression for the radius of the raindrop at any time t.

Question 11.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Solution:
Let P be the population of the city at time t.
Then the rate of growth of population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k = 0.04
∴ \(\frac{d P}{d t}\) = (0.04)P
∴ \(\frac{1}{P}\) dP = (0.04)dt
On integrating, we get
\(\int \frac{1}{P} d P\) = (0.04) ∫dt + c
∴ log P = (0.04)t + c
Initially, i.e., when t = 0, P = 10000
∴ log 10000 = (0.04) × 0 + c
∴ c = log 10000
∴ log P = (0.04)t + log 10000
∴ log P – log 10000 = (0.04)t
∴ log(\(\frac{P}{10000}\)) = (0.04)t
When t = 25, then
∴ log(\(\frac{P}{10000}\)) = 0.04 × 25 = 1
∴ log(\(\frac{P}{10000}\)) = log e ……[∵ log e = 1]
∴ \(\frac{P}{10000}\) = e = 2.7182
∴ P = 2.7182 × 10000 = 27182
∴ the population of the city after 25 years will be 27,182.

Question 12.
Radium decomposes at a rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Solution:
Let x be the amount of the radium at time t.
Then the rate of decomposition is \(\frac{d x}{d t}\) which is proportional to x.


Hence, \(\left(10-\frac{p}{10}\right)^{2} \%\) of the amount will be left after 2 years.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\)
Solution:
\(\frac{d y}{d x}+\frac{y}{x}=x^{3}-3\) …….(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = \(\frac{1}{x}\) and Q = x³ – 3

This is the general solution.

(ii) cos²x . \(\frac{d y}{d x}\) + y = tan x
Solution:

(iii) (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:

(iv) \(\frac{d y}{d x}\) + y . sec x = tan x
Solution:
\(\frac{d y}{d x}\) + y sec x = tan x
∴ \(\frac{d y}{d x}\) + (sec x) . y = tan x ……..(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = sec x and Q = tan x
∴ I.F. = \(e^{\int P d x}\)
= \(e^{\int \sec x d x}\)
= \(e^{\log (\sec x+\tan x)}\)
= sec x + tan x
∴ the solution of (1) is given by
y (I.F.) = ∫Q . (I.F.) dx + c
∴ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + tan²x) dx + c
∴ (sec x + tan x) . y = ∫(sec x tan x + sec²x – 1) dx + c
∴ (sec x + tan x) . y = sec x + tan x – x + c
∴ y(sec x + tan x) = sec x + tan x – x + c
This is the general solution.

(v) x \(\frac{d y}{d x}\) + 2y = x² . log x
Solution:

(vi) (x + y) \(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……….(1)
This is the linear differential equation of the form

This is the general solution.

(vii) (x + a) \(\frac{d y}{d x}\) – 3y = (x + a)⁵
Solution:

(viii) dr + (2r cot θ + sin 2θ) dθ = 0
Solution:
dr + (2r cot θ + sin 2θ) dθ = 0
∴ \(\frac{d r}{d \theta}\) + (2r cot θ + sin 2θ) = 0
∴ \(\frac{d r}{d \theta}\) + (2 cot θ)r = -sin 2θ ………(1)
This is the linear differential equation of the form dr
\(\frac{d r}{d \theta}\) + P . r = Q, where P = 2 cot θ and Q = -sin 2θ

This is the general solution.

(ix) y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ………(1)
This is the linear differential equation of the form

This is the general solution.

(x) \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x\left(1-x^{2}\right)^{\frac{1}{2}}\)
Solution:

(xi) \(\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}\)
Solution:

Question 2.
Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition,
\(\frac{d y}{d x}\) = x + 3y – 1
∴ \(\frac{d y}{d x}\) – 3y = x – 1 ………(1)
This is the linear differential equation of the form

This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y = 0 in (2), we get
0 = 2 + c
∴ c = -2
∴ from (2), the equation of the required curve is 3(x + 3y) = 2 – 2e^3x i.e. 3(x + 3y) = 2 (1 – e^3x).

Question 3.
Find the equation of the curve passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\) having slope of the tangent to the curve at any point (x, y) is \(-\frac{4 x}{9 y}\).
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}=-\frac{4 x}{9 y}\)
∴ y dy = \(-\frac{4}{9}\) x dx
Integrating both sides, we get
∫y dy= \(-\frac{4}{9}\) ∫x dx
∴ \(\frac{y^{2}}{2}=-\frac{4}{9} \cdot \frac{x^{2}}{2}+c_{1}\)
∴ 9y² = -4x² + 18c₁
∴ 4x² + 9y² = c where c = 18c₁
This is the general equation of the curve.
But the required curve is passing through the point \(\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)\).
∴ by putting x = \(\frac{3}{\sqrt{2}}\) and y = √2 in (1), we get
\(4\left(\frac{3}{\sqrt{2}}\right)^{2}+9(\sqrt{2})^{2}=c\)
∴ 18 + 18 = c
∴ c = 36
∴ from (1), the equation of the required curve is 4x² + 9y² = 36.

Question 4.
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
Solution:
Let A(x, y) be any point on the curve.
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
x + y = \(\frac{d y}{d x}\) + 5
∴ \(\frac{d y}{d x}\) – y = x – 5 ………(1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + P . y = Q, where P = -1 and Q = x – 5
∴ I.F. = \(e^{\int P d x}=e^{\int-1 d x}=e^{-x}\)
∴ the solution of (1) is given by

This is the general equation of the curve.
But the required curve is passing through the point (0, 2).
∴ by putting x = 0, y = 2 in (2), we get
2 = 4 – 0 + c
∴ c = -2
∴ from (2), the equation of the required curve is y = 4 – x – 2e^x.

Question 5.
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
Solution:
Let A(x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at the point A is \(\frac{d y}{d x}\).
According to the given condition
\(\frac{d y}{d x}\) = x + xy
∴ \(\frac{d y}{d x}\) – xy = x ……….. (1)
This is the linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q, where P = -x and Q = x
∴ I.F. = \(e^{\int P d x}=e^{\int-x d x}=e^{-\frac{x^{2}}{2}}\)
∴ the solution of (1) is given by
y . (I.F.) = ∫Q . (I.F.) dx + c

This is the general equation of the curve.
But the required curve is passing through the point (0, 1).
∴ by putting x = 0 and y = 1 in (2), we get
1 + 1 = c
∴ c = 2
∴ from (2), the equation of the required curve is 1 + y = \(2 e^{\frac{x^{2}}{2}}\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.4

I. Solve the following differential equations:

Question 1.
\(x \sin \left(\frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\)
Solution:

Question 2.
(x² + y²) dx – 2xy . dy = 0
Solution:
(x² + y²) dx – 2xy dy = 0
∴ 2xy dy = (x² + y²) dx
∴ \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}\) ………(1)

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

Question 4.
y² dx + (xy + x²) dy = 0
Solution:
y² dx + (xy + x²) dy = 0
∴ (xy + x²) dy = -y² dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ……..(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get

Question 5.
(x² – y²) dx + 2xy dy = 0
Solution:

Question 6.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:

Question 7.
\(x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0\)
Solution:

Question 8.
\(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{X}{y}\right) d y=0\)
Solution:

Question 9.
\(y^{2}-x^{2} \frac{d y}{d x}=x y \frac{d y}{d x}\)
Solution:

Question 10.
xy \(\frac{d y}{d x}\) = x² + 2y², y(1) = 0
Solution:

Question 11.
x dy + 2y · dx = 0, when x = 2, y = 1
Solution:
∴ x dy + 2y · dx = 0
∴ x dy = -2y dx
∴ \(\frac{1}{y} d y=\frac{-2}{x} d x\)
Integrating, we get

This is the general solution.
When x = 2, y = 1, we get
4(1) = c
∴ c = 4
∴ the particular solution is x²y = 4.

Question 12.
x² \(\frac{d y}{d x}\) = x² + xy + y²
Solution:
x² \(\frac{d y}{d x}\) = x² + xy + y²
∴ \(\frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Question 13.
(9x + 5y) dy + (15x + 11y) dx = 0
Solution:
(9x + 5y) dy + (15x + 11y) dx = 0
∴ (9x + 5y) dy = -(15x + 11y) dx
∴ \(\frac{d y}{d x}=\frac{-(15 x+11 y)}{9 x+5 y}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Question 14.
(x² + 3xy + y²) dx – x² dy = 0
Solution:
(x² + 3xy + y²) dx – x² dy = 0
∴ x² dy = (x² + 3xy + y²) dx
∴ \(\frac{d y}{d x}=\frac{x^{2}+3 x y+y^{2}}{x^{2}}\) ………(1)

Question 15.
(x² + y²) dx – 2xy dy = 0.
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

I. Evaluate the following:

Question 1.
∫x² log x dx
Solution:

Question 2.
∫x² sin 3x dx
Solution:

Question 3.
∫x tan^-1 x dx
Solution:

Question 4.
∫x² tan^-1 x dx
Solution:

Question 5.
∫x³ tan^-1 x dx
Solution:
Let I = ∫x³ tan^-1 x dx
= ∫(tan^-1 x) . x³ dx

Question 6.
∫(log x)² dx
Solution:
Let I = ∫(log x)² dx
Put log x = t

Question 7.
∫sec³ x dx
Solution:
Let I = ∫sec³ x dx
= ∫sec x sec² x dx
= sec x ∫sec² x dx – ∫[\(\frac{d}{d x}\)(sec x) ∫sec² x dx] dx
= sec x tan x – ∫(sec x tan x)(tan x) dx
= sec x tan x – ∫sec x tan² x dx
= sec x tan x – ∫sec x (sec² x – 1) dx
= sec x tan x – ∫sec³ x dx + ∫sec x dx
∴ I = sec x tan x – I + log|sec x + tan x|
∴ 2I = sec x tan x + log|sec x + tan x|
∴ I = \(\frac{1}{2}\) [sec x tan x + log|sec x + tan x|] + c.

Question 8.
∫x . sin² x dx
Solution:

Question 9.
∫x³ log x dx
Solution:

Question 10.
∫e^2x cos 3x dx
Solution:

Question 11.
∫x sin^-1 x dx
Solution:

Question 12.
∫x² cos^-1 x dx
Solution:

Question 13.
\(\int \frac{\log (\log x)}{x} d x\)
Solution:

= t(log t – 1) + c
= (log x) . [log(log x) – 1] + c.

Question 14.
\(\int \frac{t \cdot \sin ^{-1} t}{\sqrt{1-t^{2}}} d t\)
Solution:

Question 15.
∫cos√x dx
Solution:
Let I = ∫cos√x dx
Put √x = t
∴ x = t²
∴ dx = 2t dt
∴ I = ∫(cos t) 2t dt
= ∫2t cos t dt
= 2t ∫cos t dt – ∫[\(\frac{d}{d t}\)(2t) ∫cos t dt]dt
= 2t sin t – ∫2 sin t dt
= 2t sin t + 2 cos t + c
= 2[√x sin√x + cos√x] + c.

Question 16.
∫sin θ . log(cos θ) dθ
Solution:
Let I = ∫sin θ . log (cos θ) dθ
= ∫log(cos θ) . sin θ dθ
Put cos θ = t
∴ -sin θ dθ = dt
∴ sin θ dθ = -dt

= -t log t + t + c
= -cos θ . log(cos θ) + cos θ + c
= -cos θ [log(cos θ) – 1] + c.

Question 17.
∫x cos³ x dx
Solution:
cos 3x = 4 cos³ x – 3 cos x
∴ cos³ x + 3 cos x = 4cos3x
∴ cos³ x = \(\frac{1}{4}\) cos 3x + \(\frac{3}{4}\) cos x

Question 18.
\(\int \frac{\sin (\log x)^{2}}{x} \cdot \log x d x\)
Solution:

Question 19.
\(\int \frac{\log x}{x} d x\)
Solution:
Let I = \(\int \frac{\log x}{x} d x\)
Put log x = t
\(\frac{1}{x}\) dx = dt
∴ I = ∫t dt
= \(\frac{1}{2}\) t² + c
= \(\frac{1}{2}\) (log x)² + c

Question 20.
∫x sin 2x cos 5x dx.
Solution:
Let I = ∫x sin 2x cos 5x dx
sin 2x cos 5x = \(\frac{1}{2}\)[2 sin 2x cos 5x]
= \(\frac{1}{2}\) [sin(2x + 5x) + sin(2x – 5x)]
= \(\frac{1}{2}\) [sin 7x – sin 3x]
∴ ∫sin 2x cos 5x dx = \(\frac{1}{2}\) [∫sin 7x dx – ∫sin 3x dx]

Question 21.
\(\int \cos (\sqrt[3]{x}) d x\)
Solution:
Let I = \(\int \cos (\sqrt[3]{x}) d x\)

II. Integrate the following functions w.r.t. x:

Question 1.
e^2x sin 3x
Solution:

Question 2.
e^-x cos 2x
Solution:

Question 3.
sin(log x)
Solution:

Question 4.
\(\sqrt{5 x^{2}+3}\)
Solution:
Let I = \(\sqrt{5 x^{2}+3}\) dx

Question 5.
\(x^{2} \sqrt{a^{2}-x^{6}}\)
Solution:

Question 6.
\(\sqrt{(x-3)(7-x)}\)
Solution:

Question 7.
\(\sqrt{4^{x}\left(4^{x}+4\right)}\)
Solution:

Question 8.
(x + 1) \(\sqrt{2 x^{2}+3}\)
Solution:
Let I = ∫(x + 1) \(\sqrt{2 x^{2}+3}\) dx
Let x + 1 = A[\(\frac{d}{d x}\)(2x² + 3)] + B
= A(4x) + B
= 4Ax + B
Comparing the coefficients of x and constant term on both the sides, we get
4A = 1, B = 1
∴ A = \(\frac{1}{4}\), B = 1

Question 9.
\(x \sqrt{5-4 x-x^{2}}\)
Solution:
Let I = ∫\(x \sqrt{5-4 x-x^{2}}\) dx
Let x = A[\(\frac{d}{d x}\)(5 – 4x – x²)] + B
= A[-4 – 2x] + B
= -2Ax + (B – 4A)
Comparing the coefficients of x and the constant term on both sides, we get
-2A = 1, B – 4A = 0

Question 10.
\(\sec ^{2} x \sqrt{\tan ^{2} x+\tan x-7}\)
Solution:

Question 11.
\(\sqrt{x^{2}+2 x+5}\)
Solution:

Question 12.
\(\sqrt{2 x^{2}+3 x+4}\)
Solution:

III. Integrate the following functions w.r.t. x:

Question 1.
[2 + cot x – cosec² x] e^x
Solution:
Let I = ∫e^x [2 + cot x – cosec² x] dx
Put f(x) = 2 + cot x
∴ f'(x) = \(\frac{d}{d x}\)(2 + cot x)
= \(\frac{d}{d x}\)(2) + \(\frac{d}{d x}\)(cot x)
= 0 – cosec² x
= -cosec² x
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x (2 + cot x) + c.

Question 2.
\(\left(\frac{1+\sin x}{1+\cos x}\right) e^{x}\)
Solution:

Question 3.
\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Solution:
Let I = ∫\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Let f(x) = \(\frac{1}{x}\), f'(x) = \(-\frac{1}{x^{2}}\)
∴ I = ∫e^x [f(x) + f'(x)] dx
= e^x f(x) + c
= e^x . \(\frac{1}{x}\) + c

Question 4.
\(\left[\frac{x}{(x+1)^{2}}\right] e^{x}\)
Solution:

Question 5.
\(\frac{e^{x}}{x}\) . [x(log x)² + 2 log x]
Solution:

Question 6.
\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)
Solution:
Let I = ∫\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)

Question 7.
\(e^{\sin ^{-1} x}\left[\frac{x+\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right]\)
Solution:

Question 8.
log(1 + x)^(1+x)
Solution :
Let I = ∫log(1 + x)^(1+x) dx

Question 9.
cosec (log x)[1 – cot(log x)]
Solution:
Let I = ∫cosec (log x)[1 – cot(log x)] dx
Put log x = t
x = e^t
dx = e^t dt
I = ∫cosec t (1 – cot t). e^t dt
= ∫e^t [cosec t – cosec t cot t] dt
= ∫e^t [cosec t + \(\frac{d}{d t}\) (cosec t)] dt
= e^t cosec t + c ….. [∵ e^t [f(t) +f'(t) ] dt = e^t f(t) + c ]
= x . cosec(log x) + c.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(C) Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C)

I. Evaluate:

Question 1.
\(\int \frac{3 x+4}{x^{2}+6 x+5} d x\)
Solution:
Let I = \(\int \frac{3 x+4}{x^{2}+6 x+5} d x\)
Let 3x + 4 = A[\(\frac{d}{d x}\)(x² + 6x + 5)] + B
= A(2x + B) + B
∴ 3x + 4 = 2Ax + (6A + B)
Comparing the coefficient of x and constant on both sides, we get
2A = 3 and 6A + B = 4
∴ A = \(\frac{3}{2}\) and 6(\(\frac{3}{2}\)) + B = 4
∴ B = -5
3x + 4 = \(\frac{3}{2}\) (2x + 6) – 5

Question 2.
\(\int \frac{2 x+1}{x^{2}+4 x-5} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{x^{2}+4 x-5} d x\)
Let 2x + 1 = A[\(\frac{d}{d x}\)(x² + 4x – 5)] + B
2x + 1 = A(2x + 1) + B
∴ 2x + 1 = 2Ax + (4A + B)
Comparing the coefficient of x and constant on both sides, we get
4A = 2 and 4A + B = 4
∴ A = \(\frac{3}{2}\) and 6(\(\frac{3}{2}\)) + B = 4
∴ B = -5
∴ 2x + 1 = \(\frac{3}{2}\)(2x + 1) – 5

Question 3.
\(\int \frac{2 x+3}{2 x^{2}+3 x-1} d x\)
Solution:
Let I = \(\int \frac{2 x+3}{2 x^{2}+3 x-1} d x\)
Let 2x+ 3 = A[\(\frac{d}{d x}\)(2x² + 3x – 1)] + B
2x + 1 = A(4x + 3) + B
∴ 2x + 1 = 4Ax + (3A + B)
Comparing the coefficient of x and constant on both sides, we get
4A = 2 and 3A + B = 3
∴ A = \(\frac{1}{2}\) and 3(\(\frac{1}{2}\)) + B = 3
∴ B = \(\frac{3}{2}\)
∴ 2x + 3 = \(\frac{1}{2}\)(4x + 3) + \(\frac{3}{2}\)

Question 4.
\(\int \frac{3 x+4}{\sqrt{2 x^{2}+2 x+1}} d x\)
Solution:
Let I = \(\int \frac{3 x+4}{\sqrt{2 x^{2}+2 x+1}} d x\)
Let 3x + 4 = A[\(\frac{d}{d x}\)(2x² + 2x + 1)] + B
∴ 3x + 4 = A (4x + 2) + B
∴ 3x + 4 = 4Ax + (2A + B)
Comparing the coefficient of x and the constant on both the sides, we get
4A = 3 and 2A + B = 4
∴ A = \(\frac{3}{4}\) and 2(\(\frac{3}{4}\)) + B = 4
∴ B = \(\frac{5}{2}\)
∴ 3x + 4 = \(\frac{3}{4}\) (4x + 2) + \(\frac{5}{2}\)

Question 5.
\(\int \frac{7 x+3}{\sqrt{3+2 x-x^{2}}} d x\)
Solution:
Let I = \(\int \frac{7 x+3}{\sqrt{3+2 x-x^{2}}} d x\)
Let 7x + 3 = A[\(\frac{d}{d x}\)(3 + 2x – x²)] + B
7x + 3 = A(2 – 2x) + B
∴ 7x + 3 = -2Ax + (2A + B)
Comparing the coefficient of x and constant on both the sides, we get
-2A = 7 and 2A + B = 3

Question 6.
\(\int \sqrt{\frac{x-7}{x-9}} d x\)
Solution:

Comparing the coefficients of x and constant term on both sides, we get

Question 7.
\(\int \sqrt{\frac{9-x}{x}} d x\)
Solution:

Question 8.
\(\int \frac{3 \cos x}{4 \sin ^{2} x+4 \sin x-1} d x\)
Solution:

Question 9.
\(\int \sqrt{\frac{e^{3 x}-e^{2 x}}{e^{x}+1}} d x\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(B) Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B)

I. Evaluate the following:

Question 1.
\(\int \frac{1}{4 x^{2}-3} \cdot d x\)
Solution:

Question 2.
\(\int \frac{1}{25-9 x^{2}} \cdot d x\)
Solution:

Question 3.
\(\int \frac{1}{7+2 x^{2}} \cdot d x\)
Solution:

Question 4.
\(\int \frac{1}{\sqrt{3 x^{2}+8}} \cdot d x\)
Solution:

Question 5.
\(\int \frac{1}{\sqrt{11-4 x^{2}}} \cdot d x\)
Solution:

Question 6.
\(\int \frac{1}{\sqrt{2 x^{2}-5}} \cdot d x\)
Solution:

Question 7.
\(\int \sqrt{\frac{9+x}{9-x}} \cdot d x\)
Solution:

Question 8.
\(\int \sqrt{\frac{2+x}{2-x}} \cdot d x\)
Solution:

Question 9.
\(\int \sqrt{\frac{10+x}{10-x}} \cdot d x\)
Solution:

Question 10.
\(\int \frac{1}{x^{2}+8 x+12} \cdot d x\)
Solution:

Question 11.
\(\int \frac{1}{1+x-x^{2}} \cdot d x\)
Solution:

Question 12.
\(\int \frac{1}{4 x^{2}-20 x+17} \cdot d x\)
Solution:

Question 13.
\(\int \frac{1}{5-4 x-3 x^{2}} \cdot d x\)
Solution:

Question 14.
\(\int \frac{1}{\sqrt{3 x^{2}+5 x+7}} \cdot d x\)
Solution:

Question 15.
\(\int \frac{1}{\sqrt{x^{2}+8 x-20}} \cdot d x\)
Solution:

Question 16.
\(\int \frac{1}{\sqrt{8-3 x+2 x^{2}}} \cdot d x\)
Solution:

Question 17.
\(\int \frac{1}{\sqrt{(x-3)(x+2)}} \cdot d x\)
Solution:

Question 18.
\(\int \frac{1}{4+3 \cos ^{2} x} \cdot d x\)
Solution:

Question 19.
\(\int \frac{1}{\cos 2 x+3 \sin ^{2} x} \cdot d x\)
Solution:

Question 20.
\(\int \frac{\sin x}{\sin 3 x} \cdot d x\)
Solution:

II. Integrate the following functions w. r. t. x:

Question 1.
\(\int \frac{1}{3+2 \sin x} \cdot d x\)
Solution:

Question 2.
\(\int \frac{1}{4-5 \cos x} \cdot d x\)
Solution:

Question 3.
\(\int \frac{1}{2+\cos x-\sin x} \cdot d x\)
Solution:

Question 4.
\(\int \frac{1}{3+2 \sin x-\cos x} \cdot d x\)
Solution:

Question 5.
\(\int \frac{1}{3-2 \cos 2 x} \cdot d x\)
Solution:

Question 6.
\(\int \frac{1}{2 \sin 2 x-3} \cdot d x\)
Solution:

Question 7.
\(\int \frac{1}{3+2 \sin 2 x+4 \cos 2 x} \cdot d x\)
Solution:

Question 8.
\(\int \frac{1}{\cos x-\sin x} \cdot d x\)
Solution:

Question 9.
\(\int \frac{1}{\cos x-\sqrt{3} \sin x} \cdot d x\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(A) Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

I. Integrate the following functions w.r.t. x:

Question 1.
\(\frac{(\log x)^{n}}{x}\)
Solution:

Question 2.
\(\frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}}\)
Solution:
Let I = \(\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x\)

Question 3.
\(\frac{1+x}{x \cdot \sin (x+\log x)}\)
Solution:

Question 4.
\(\frac{x \cdot \sec ^{2}\left(x^{2}\right)}{\sqrt{\tan ^{3}\left(x^{2}\right)}}\)
Solution:

Question 5.
\(\frac{e^{3 x}}{e^{3 x}+1}\)
Solution:

Question 6.
\(\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)} \cdot a^{x+\tan ^{-1} x}\)
Solution:

Question 7.
\(\frac{e^{x} \cdot \log \left(\sin e^{x}\right)}{\tan \left(e^{x}\right)}\)
Solution:

Question 8.
\(\frac{e^{2 x}+1}{e^{2 x}-1}\)
Solution:

Question 9.
sin⁴x . cos³x
Solution:

Question 10.
\(\frac{1}{4 x+5 x^{-11}}\)
Solution:

Question 11.
x⁹ . sec²(x¹⁰)
Solution:

Question 12.
\(e^{3 \log x} \cdot\left(x^{4}+1\right)^{-1}\)
Solution:

Question 13.
\(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Solution:
Let I = \(\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x\)
Dividing numerator and denominator by cos²x, we get

Question 14.
\(\frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}}\)
Solution:

Question 15.
\(\frac{2 \sin x \cos x}{3 \cos ^{2} x+4 \sin ^{2} x}\)
Solution:

Question 16.
\(\frac{1}{\sqrt{x}+\sqrt{x^{3}}}\)
Solution:

Question 17.
\(\frac{10 x^{9}+10^{x} \cdot \log 10}{10^{x}+x^{10}}\)
Solution:

Question 18.
\(\frac{x^{n-1}}{\sqrt{1+4 x^{n}}}\)
Solution:

Question 19.
(2x + 1) \(\sqrt{x+2}\)
Solution:

Question 20.
\(x^{5} \sqrt{a^{2}+x^{2}}\)
Solution:

Question 21.
\((5-3 x)(2-3 x)^{-\frac{1}{2}}\)
Solution:

Question 22.
\(\frac{7+4 x+5 x^{2}}{(2 x+3)^{\frac{3}{2}}}\)
Solution:

Question 23.
\(\frac{x^{2}}{\sqrt{9-x^{6}}}\)
Solution:

Question 24.
\(\frac{1}{x\left(x^{3}-1\right)}\)
Solution:

Question 25.
\(\frac{1}{x \cdot \log x \cdot \log (\log x)}\)
Solution:

II. Integrate the following functions w.r.t x:

Question 1.
\(\frac{\cos 3 x-\cos 4 x}{\sin 3 x+\sin 4 x}\)
Solution:

Question 2.
\(\frac{\cos x}{\sin (x-a)}\)
Solution:

Question 3.
\(\frac{\sin (x-a)}{\cos (x+b)}\)
Solution:

Question 4.
\(\frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x}\)
Solution:
Let I = \(\int \frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x} d x\)
Dividing numerator and denominator of cos²x, we get

Question 5.
\(\frac{\sin x+2 \cos x}{3 \sin x+4 \cos x}\)
Solution:
Let I = \(\int \frac{\sin x+2 \cos x}{3 \sin x+4 \cos x} d x\)
Put, Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ sin x+ 2 cos x = A(3 sin x + 4 cos x) + B [\(\frac{d}{d x}\) (3 sin x + 4 cos x)]
= A(3 sin x + 4 cos x) + B (3 cos x – 4 sin x)
∴ sin x + 2 cos x = (3A – 4B) sin x + (4A + 3B) cos x
Equating the coefficients of sin x and cos x on both the sides, we get
3A – 4B = 1 …… (1)
and 4A + 3B = 2 …… (2)
Multiplying equation (1) by 3 and equation (2) by 4, we get
9A – 12B = 3
16A + 12B = 8
On adding, we get

Question 6.
\(\frac{1}{2+3 \tan x}\)
Solution:
Let I = \(\int \frac{1}{2+3 \tan x} d x\)

Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ cos x = A(2 cos x + 3 sin x) + B [\(\frac{d}{d x}\) (2 cos x + 3 sin x)]
= A (2 cos x + 3 sin x) + B (-2 sin x + 3 cos x)
∴ cos x = (2A + 3B) cos x + (3A – 2B) sin x
Equating the coefficients of cosx and sinx on both the sides, we get
2A + 3B = 1 …… (1)
and 3A – 2B = 0 ……. (2)
Multiplying equation (1) by 2 and equation (2) by 3, we get
4A + 6B = 2
9A – 6B = 0
On adding, we get

Question 7.
\(\frac{4 e^{x}-25}{2 e^{x}-5}\)
Solution:
Let I = \(\int \frac{4 e^{x}-25}{2 e^{x}-5} d x\)
Put, Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ 4ex – 25 = A(2ex – 5) + B[\(\frac{d}{d x}\) (2ex – 5)]
= A(2ex – 5) + B(2ex – 0)
∴ 4ex – 25 = (2A + 2B) ex – 5A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 4 …….(1)
and 5A = 25
∴ A = 5
from (1), 2(5) + 2B = 4
∴ 2B = -6
∴ B = -3

Question 8.
\(\frac{20+12 e^{x}}{3 e^{x}+4}\)
Solution:

Question 9.
\(\frac{3 e^{2 x}+5}{4 e^{2 x}-5}\)
Solution:
Let I = \(\int \frac{3 e^{2 x}+5}{4 e^{2 x}-5} d x\)
Put, Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ 3e2x + 5 = A(4e2x – 5) + B [\(\frac{d}{d x}\) (4e2x – 5)]
= A(4e2x – 5) + B(4 . e2x × 2 – 0)
∴ 3e2x + 5 = (4A + 8B) e2x – 5A
Equating the coefficient of e2x and constant on both sides, we get
4A + 8B = 3 …….. (1)
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = \(\frac{7}{8}\)

Question 10.
cos⁸ x . cot x
Solution:

Question 11.
tan⁵x
Solution:
Let I = ∫ tan⁵x dx

Question 12.
cos⁷x
Solution:

Question 13.
tan 3x tan 2x tan x
Solution:
Let I = ∫ tan 3x tan 2x tan x dx
Consider tan 3x = tan (2x + x) = \(\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}\)
tan 3x (1 – tan 2x tan x) = tan 2x + tan x
tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
I = ∫(tan 3x – tan 2x – tan x) dx
= ∫tan3x dx – ∫tan 2x dx – ∫tan x dx
= \(\frac{1}{3}\) log | sec 3x| – \(\frac{1}{2}\) log |sec 2x| – log |sec x| + c.

Question 14.
sin⁵x cos⁸x
Solution:

Question 15.
\(3^{\cos ^{2} x \cdot} \sin 2 x\)
Solution:

Question 16.
\(\frac{\sin 6 x}{\sin 10 x \sin 4 x}\)
Solution:

Question 17.
\(\frac{\sin x \cos ^{3} x}{1+\cos ^{2} x}\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

I. Integrate the following functions w.r.t. x:

(i) x³ + x² – x + 1
Solution:

(ii) \(x^{2}\left(1-\frac{2}{x}\right)^{2}\)
Solution:

(iii) \(3 \sec ^{2} x-\frac{4}{x}+\frac{1}{x \sqrt{x}}-7\)
Solution:

(iv) \(2 x^{3}-5 x+\frac{3}{x}+\frac{4}{x^{5}}\)
Solution:

(v) \(\frac{3 x^{3}-2 x+5}{x \sqrt{x}}\)
Solution:

II. Evaluate:

(i) ∫tan² x . dx
Solution:

(ii) \(\int \frac{\sin 2 x}{\cos x} \cdot d x\)
Solution:

(iii) \(\int \frac{\sin x}{\cos ^{2} x} \cdot d x\)
Solution:

(iv) \(\int \frac{\cos 2 x}{\sin ^{2} x} \cdot d x\)
Solution:

(v) \(\int \frac{\cos 2 x}{\sin ^{2} x \cdot \cos ^{2} x} \cdot d x\)
Solution:

= -cot x – tan x + c

(vi) \(\int \frac{\sin x}{1+\sin x} \cdot d x\)
Solution:

(vii) \(\int \frac{\tan x}{\sec x+\tan x} \cdot d x\)
Solution:

(viii) \(\int \sqrt{1+\sin 2 x} \cdot d x\)
Solution:

(ix) \(\int \sqrt{1-\cos 2 x} \cdot d x\)
Solution:

(x) ∫sin 4x cos 3x dx
Solution:

III. Evaluate:

(i) \(\int \frac{x}{x+2} \cdot d x\)
Solution:

(ii) \(\int \frac{4 x+3}{2 x+1} \cdot d x\)
Solution:

(iii) \(\int \frac{5 x+2}{3 x-4} \cdot d x\)
Solution:

(iv) \(\int \frac{x-2}{\sqrt{x+5}} \cdot d x\)
Solution:

(v) \(\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x\)
Solution:

(vi) \(\int \frac{\sin 4 x}{\cos 2 x} \cdot d x\)
Solution:

(vii) \(\int \sqrt{1+\sin 5 x} \cdot d x\)
Solution:

(viii) ∫cos² x . dx
Solution:

(ix) \(\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \cdot d x\)
Solution:

(x) \(\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \cdot d x\)
Solution:

IV.

Question 1.
If f'(x) = x – \(\frac{3}{x^{3}}\), f(1) = \(\frac{11}{2}\), find f(x).
Solution:
By the definition of integral,

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

I. Choose the correct option from the given alternatives:

Question 1.
If the function f(x) = ax³ + bx² + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and f'(2 + \(\frac{1}{\sqrt{3}}\)) = 0, then values of a and b are respectively.
(a) 1, -6
(b) -2, 1
(c) -1, -6
(d) -1, 6
Answer:
(a) 1, -6

Hint: f(x) = ax³ + bx² + 11x – 6 satisfies the conditions of Rolle’s theorem in [1, 3]
∴ f(1) = f(3)
a(1)³ + b(1)² + 11(1) – 6 = a(3)³ + b(3)² + 11(3) – 6
a + b + 11 = 27a + 9b + 33
26a + 8b = -22
13a + 4b = -11
Only a = 1, b = -6 satisfy this equation.

Question 2.
If f(x) = \(\frac{x^{2}-1}{x^{2}+1}\), for every real x, then the minimum value of f is
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
(c) -1

Question 3.
A ladder 5 m in length is resting against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the higher point of the ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of
(a) 1
(b) 2
(c) 2.5
(d) 3
Answer:
(b) 2

Question 4.
Let f(x) and g(x) be differentiable for 0 < x < 1 such that f(0) = 0, g(0) = 0, f(1) = 6. Let there exist a real number c in (0, 1) such that f'(c) = 2g'(c), then the value of g(1) must be
(a) 1
(b) 3
(c) 2.5
(d) -1
Answer:
(b) 3

Hint: f(x) and g(x) both satisfies the conditions of LMVT in (0, 1).
∴ f'(c) = \(\frac{f(1)-f(0)}{1-0}=\frac{6-0}{1}=6\)
and g'(c) = \(\frac{g(1)-g(0)}{1-0}=\frac{g(1)-0}{1}\) = g(1)
But f'(c) = 2g'(c)
6 = 2g(1)
∴ g(1) = 3

Question 5.
Let f(x) = x³ – 6x² + 9x + 18, then f(x) is strictly decreasing in
(a) (-∞, 1)
(b) [3, ∞)
(c) (-∞, 1] ∪ [3, ∞)
(d) (1, 3)
Answer:
(d) (1, 3)

Question 6.
If x = -1 and x = 2 are the extreme points of y = α log x + βx² + x, then
(a) α = -6, β = \(\frac{1}{2}\)
(b) α = -6, β = \(\frac{-1}{2}\)
(c) α = 2, β = \(\frac{-1}{2}\)
(d) α = 2, β = \(\frac{1}{2}\)
Answer:
(c) α = 2, β = \(\frac{-1}{2}\)

Hint: y = α log x + βx² + x
∴ \(\frac{d y}{d x}=\frac{\alpha}{x}+\beta \times 2 x+1=\frac{\alpha}{x}+2 \beta x+1\)
f(x) has extreme values at x = -1 and x = 2
∴ f'(-1) = 0 and f(2) = 0
α + 2β = 1
and \(\frac{\alpha}{2}\) + 4β = -1
By solving these two equations, we get
α = 2, β = \(\frac{-1}{2}\)

Question 7.
The normal to the curve x² + 2xy – 3y² = 0 at (1, 1)
(a) meets the curve again in the second quadrant
(b) does not meet the curve again
(c) meets the curve again in the third quadrant
(d) meets the curve again in the fourth quadrant
Answer:
(d) meets the curve again in fourth quadrant

Hint: x² + 2xy – 3y² = 0

= slope of the tangent at (1, 1)
∴ equation of the tangent at (1, 1) is -1
∴ equation of the normal is
y – 1= -1 (x – 1) = -x + 1
∴ x + y = 2
∴ y = 2 – x
Substituting y = 2 – x in x² + 2xy – 3y² = 0, we get
x² + 2x(2 – x) – 3 (2 – x)² = 0
⇒ x² + 4x – 2x² – 3(4 – 4x + x²) = 0
⇒ x² – 4x + 3 = 0
⇒ (x – 1)(x – 3) = 0
⇒ x = 1, x = 3
When x = 1, y = 2 – 1 = 1
When x = 3, y = 2 – 3 = -1
∴ the normal at (1, 1) meets the curve at (3, -1) which is in the fourth quadrant.

Question 8.
The equation of the tangent to the curve y = 1 – \(e^{\frac{x}{2}}\) at the point of intersection with Y-axis is
(a) x + 2y = 0
(b) 2x + y = 0
(c) x – y = 2
(d) x + y = 2
Answer:
(a) x + 2y = 0
Hint: The point of intersection of the curve with the Y-axis is the origin (0, 0).

Question 9.
If the tangent at (1, 1) on y² = x(2 – x)² meets the curve again at P, then P is
(a) (4, 4)
(b) (-1, 2)
(c) (3, 6)
(d) \(\left(\frac{9}{4}, \frac{3}{8}\right)\)
Answer:
(d) \(\left(\frac{9}{4}, \frac{3}{8}\right)\)
Hint: y² = x(2 – x)²
= x(4 – 4x + x²)
= x³ – 4x² + 4x

= slope of the tangent at (1, 1)
∴ equation of the tangent at (1, 1) is
y – 1 = –\(\frac{1}{2}\) (x – 1)
∴ 2y – 2 = -x + 1
∴ x + 2y = 3
Only the coordinates \(\left(\frac{9}{4}, \frac{3}{8}\right)\) satisfy both the equations y² = x(2 – x)² and x + 2y = 3
∴ P is \(\left(\frac{9}{4}, \frac{3}{8}\right)\)

Question 10.
The approximate value of tan (44° 30′), given that 1° = 0.0175, is
(a) 0.8952
(b) 0.9528
(c) 0.9285
(d) 0.9825
Answer:
(d) 0.9825

II. Solve the following:

Question 1.
If the curves ax² + by² = 1 and a’x² + b’y² = 1, intersect orthogonally, then prove that \(\frac{1}{a}-\frac{1}{b}=\frac{1}{a^{\prime}}-\frac{1}{b^{\prime}}\)
Solution:
Let P(x₁, y₁) be the point of intersection of the curves.

Question 2.
Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x² drawn at the point (1, 2) and the coordinate axes.
Solution:
y = 3 – x²

= slope of the tangent at (1, 2)
∴ equation of the tangent at (1, 2) is
y – 2= -2(x – 1)
⇒ y – 2= -2x + 2
⇒ 2x + y = 4
Let this tangent cuts the coordinate axes at A(a, 0) and B(0, b).
∴ 2a + 0 = 4 and 2(0) + b = 4
∴ a = 2 and b = 4
∴ area of required triangle = \(\frac{1}{2}\) × l(OA) × l(OB)
= \(\frac{1}{2}\) ab
= \(\frac{1}{2}\) (2)(4)
= 4 sq units.

Question 3.
Find the equation of the tangent and normal drawn to the curve y⁴ – 4x⁴ – 6xy = 0 at the point M (1, 2).
Solution:
y⁴ – 4x⁴ – 6xy = 0
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 2)
∴ the equation of normal at M (1, 2) is
y – 2 = \(\frac{14}{13}\) (x – 1)
∴ 13y – 26 = 14x – 14
∴ 14x – 13y + 12 = 0
The slope of normal at (1, 2)
\(=\frac{-1}{\left(\frac{d y}{d x}\right)_{\mathrm{at}(1,2)}}=\frac{-1}{\left(\frac{14}{13}\right)}=-\frac{13}{14}\)
∴ the equation of normal at M (1, 2) is
y – 2 = \(\frac{-13}{14}\) (x – 1)
14y – 28 = -13x + 13
13x + 14y – 41 = 0
Hence, the equations of tangent and normal are 14x – 13y + 12 = 0 and 13x + 14y – 41 = 0 respectively.

Question 4.
A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the water level when the depth is 6 feet.
Solution:

Let r be the radius of the base, h be the height and V be the volume of the water level at any time t.
Since, the height of the cone is 8 feet and the radius is 4 feet,
\(\frac{r}{h}=\frac{4}{8}=\frac{1}{2}\)
r = \(\frac{h}{2}\) ……..(1)

Hence, the rate of change of water level is \(\left(\frac{2}{9 \pi}\right)\) ft/sec.

Question 5.
Find all points on the ellipse 9x² + 16y² = 400, at which the y-coordinate is decreasing and the x-coordinate is increasing at the same rate.
Solution:
Let P(x₁, y₁) be the point on the ellipse 9x² + 16y² = 400 whose y-coordinate decreasing x-coordinate is increasing at the same rate.

Question 6.
Verify Rolle’s theorem for the function f(x) = \(\frac{2}{e^{x}+e^{-x}}\) on [-1, 1].
Solution:
The functions e^x, e^-x, and 2 are continuous and differentiable in their respective domains.
∴ f(x) = \(\frac{2}{e^{x}+e^{-x}}\) is continuous on [-1, 1] and differentiable on (-1, 1), because e^x + e^-x ≠ 0 for all x ∈ [-1, 1].
Now, f(-1) = \(\frac{2}{e^{-1}+e}=\frac{2}{e+e^{-1}}\) and f(1) = \(\frac{2}{e+e^{-1}}\)
∴ f(-1) = f(1)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.
∴ there exist c ∈ (-1, 1) such that f'(c) = 0
Now, f(x) = \(\frac{2}{e^{x}+e^{-x}}\)

Hence, Rolle’s theorem is verified.

Question 7.
The position of a particle is given by the function s(t) = 2t² + 3t – 4. Find the time t = c in the interval 0 ≤ f ≤ 4 when the instantaneous velocity of the particle is equal to its average velocity in this interval.
Solution:
s(t) = 2t² + 3t – 4
∴ s(0) = 2(0)² + 3(0) – 4 = -4
and s(4) = 2(4)² + 3(4) – 4 = 32 + 12 – 4 = 40
∴ average velocity = \(\frac{s(4)-s(0)}{4-0}\)
= \(\frac{40-(-4)}{4}\)
= 11
Also, instantaneous velocity = \(\frac{d s}{d t}\)
= \(\frac{d}{d t}\) (2t² + 3t – 4)
= 2 × 2t + 3 × 1 – 0
= 4t + 3
∴ instantaneous velocity at t = c is \(\left(\frac{d s}{d t}\right)_{t=c}\) = 4c + 3
When instantaneous velocity at t = c equal to its average velocity, we get
4c + 3 = 11
4c = 8
∴ c = 2 ∈ [0, 4]
Hence, t = c = 2.

Question 8.
Find the approximate value of the function f(x) = \(\sqrt{x^{2}+3 x}\) at x = 1.02.
Solution:
f(x) = \(\sqrt{x^{2}+3 x}\)

Question 9.
Find the approximate value of cos^-1(0.51), given π = 3.1416, \(\frac{2}{\sqrt{3}}\) = 1.1547.
Solution:
Let f(x) = cos^-1 x

The formula for approximation is f(a + h)= f(a) + h . f'(a)
∴ cos^-1 (0.51) = f(0.51)
= f(0.5 + 0.01)
= f(0.5) + (0.01) f'(0.5)
= \(\frac{\pi}{3}\) + 0.01 × (-1.1547)
= \(\frac{3.1416}{3}\) – 0.011547
= 1.0472 – 0.011547
= 1.035653
∴ cos^-1 (0.51) = 1.035653.

Question 10.
Find the intervals on which the function y = x^x, (x > 0) is increasing and decreasing.
Solution:
y = x^x
∴ log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

y is increasing if \(\frac{d y}{d x}\) ≥ 0
i.e. if x^x (1 + log x) ≥ 0
i.e. if 1 + log x ≥ 0 ……[∵ x > 0]
i.e. if log x ≥ -1
i.e. if log x ≥ -log e …….[∵ log e = 1]
i.e. if logx ≥ log \(\frac{1}{e}\)
i.e. if x ≥ \(\frac{1}{e}\)
∴ y is increasing in \(\left[\frac{1}{e^{\prime}}, \infty\right)\)
y is decreasing if \(\frac{d y}{d x}\) ≤ 0
i.e. if x^x (1 + log x) ≤ 0
i.e. if 1 + log x ≤ 0 ……[∵ x > 0]
i.e. if log x ≤ -1
i.e. if log x ≤ -log e
i.e. if log x ≤ log \(\frac{1}{e}\)
i.e. if x ≤ \(\frac{1}{e}\) where x > 0
∴ y is decreasing is \(\left(0, \frac{1}{e}\right]\)
Hence, the given function is increasing in \(\left[\frac{1}{e^{\prime}}, \infty\right)\) and decreasing in \(\left(0, \frac{1}{e}\right]\)

Question 11.
Find the intervals on which the function f(x) = \(\frac{x}{\log x}\) is increasing and decreasing.
Solution:
f(x) = \(\frac{x}{\log x}\)

f is increasing if f'(x) ≥ 0
i.e. if \(\frac{\log x-1}{(\log x)^{2}}\) ≥ 0
i.e. if log x – 1 ≥ 0 ……..[∵ (log x)² > 0]
i.e. if log x ≥ 1
i.e. if log x ≥ log e ………[∵ log e = 1]
i.e. if x ≥ e
∴ f is increasing on [e, ∞)
f is decreasing if f'(x) ≤ 0
i.e. if \(\frac{\log x-1}{(\log x)^{2}}\) ≤ 0
i.e. if log x – 1 ≤ 0 ……..[∵ (log x)² > 0]
i.e. if log x ≤ 1
i.e. if log x ≤ log e
i.e. if x ≤ e
Also, x > 0 and x ≠ 1 because f(x) = \(\frac{x}{\log x}\) is not defined at x = 1.
∴ f is decreasing in (0, e] – {1}
Hence, f is increasing in [e, ∞) and decreasing in (0, e] – {1}.

Question 12.
An open box with a square base is to be made out of the given quantity of sheet of area a². Show that the maximum volume of the box is \(\frac{a^{3}}{6 \sqrt{3}}\).
Solution:
Let x be the side of square base and h be the height of the box.
Then x² + 4xh = a²
∴ h = \(\frac{a^{2}-x^{2}}{4 x}\) …….(1)
Let V be the volume of the box.
Then V = x²h


Hence, the maximum volume of the box is \(\frac{a^{3}}{6 \sqrt{3}}\) cu units.

Question 13.
Show that of all rectangles inscribed in a given circle, the square has the maximum area.
Solution:

Let ABCD be a rectangle inscribed in a circle of radius r.
Let AB = x and BC = y.
Then x² + y² = 4r² …….(1)
Area of rectangle = xy
= \(x \sqrt{4 r^{2}-x^{2}}\) ……[By (1)]
Let f(x) = x²(4r² – x²)
= 4r²x² – x⁴
∴ f'(x) = \(\frac{d}{d x}\) (4r2x2 – x4)
= 4r² × 2x – 4x³
= 8r²x – 4x³
and f”(x) = \(\frac{d}{d x}\) (8r²x – 4x³)
= 8r² × 1 – 4 × 3x²
= 8r² – 12x²
For maximum area, f'(x) = 0
⇒ 8r²x – 4x³ = 0
⇒ 4x³ = 8r²x
⇒ x² = 2r² ……..[∵ x ≠ 0]
⇒ x = √2r …..[x > 0]
and f”(√2r) = 8r² – 12(√2r)² = -16r² < 0
∴ f(x) is maximum when x = √2r
If x = √2r, then from (1),
(√2r)² + y² = 4r²
⇒ y² = 4r² – 2r² = 2r²
⇒ y = √2r ……[∵ y > 0]
⇒ x = y
∴ rectangle is a square.
Hence, amongst all rectangles inscribed in a circle, the square has maximum area.

Question 14.
Show that a closed right circular cylinder of a given surface area has maximum volume if its height equals the diameter of its base.
Solution:
Let r be the radius of the base, h be the height and V be the volume of the closed right circular cylinder, whose surface area is a² sq units (which is given).
2πrh + 2πr² = a²
⇒ 2πr(h + r) = a²
⇒ h = \(\frac{a^{2}}{2 \pi r}\) – r ……(1)


Hence, the volume of the cylinder is maximum if its height is equal to the diameter of the base.

Question 15.
A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is 30 m, find the dimensions so that the greatest possible amount of light may be admitted.
Solution:

Let x be the length, y be the breadth of the rectangle and r be the radius of the semicircle.
Then perimeter of the window = x + 2y + πr, where x = 2r
This is given to be 30 m
⇒ 2r + 2y + πr = 30
⇒ 2y = 30 – (π + 2)r
⇒ y = 15 – \(\frac{(\pi+2) r}{2}\) ……..(1)
The greatest possible amount of light may be admitted if the area of the window is maximized.
Let A be the area of the window.


Hence, the required dimensions of the window are as follows:
Length of rectangle = \(\left(\frac{60}{\pi+4}\right)\) metres
breadth of rectangle = \(\left(\frac{30}{\pi+4}\right)\) metres
radius of the semicircle = \(\left(\frac{30}{\pi+4}\right)\) metres

Question 16.
Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.
Solution:

Given the right circular cone of fixed height h and semi-vertical angle a.
Let R be the radius of the base and H be the height of the right circular cylinder that can be inscribed in the right circular cone.
In the figure, ∠GAO = α, OG = r, OA = h, OE = R, CE = H.
We have, \(\frac{r}{h}\) = tan α
∴ r = h tan α ……(1)
Since ∆AOG and ∆CEG are similar.



Hence, the height of the right circular cylinder is one-third of that of the cone.

Question 17.
A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least if the radius of the circle is half of the side of the square.
Solution:
Let r be the radius of the circle and x be the length of the side of the square. Then
(circumference of the circle) + (perimeter of the square) = l
∴ 2πr + 4x = l
∴ r = \(\frac{l-4 x}{2 \pi}\)
A = (area of the circle) + (area of the square)
= πr² + x²

This shows that the sum of the areas of circle and square is least when the radius of the circle = (\(\frac{1}{2}\)) side of the square.

Question 18.
A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares of the equal area from all comers. If the total area of the removed squares is 100, the resulting box has maximum volume. Find the lengths of the rectangular sheet of paper.
Solution:

The sides of the rectangular sheet of paper are in the ratio 8 : 15.
Let the sides of the rectangular sheet of paper be 8k and 15k respectively.
Let x be the side of the square which is removed from the comers of the sheet of paper.
The total area of removed squares is 4x², which is given to be 100.
4x² = 100
⇒ x² = 25
⇒x = 5 ……[x > 0]
Now, the length, breadth, and height of the rectangular box are 15k – 2x, 8k – 2x, and x respectively.
Let V be the volume of the box.
Then V = (15k – 2x) (8k – 2x) . x
⇒ V = (120k² – 16kx – 30kx + 4x²) . x
⇒ V = 4x³ – 46kx² + 120k²x
\(\frac{d V}{d x}=\frac{d}{d x}\) (4x³ – 46kx² + 120k²x)
= 4 × 3x² – 46k × 2x + 120k² × 1
= 12x² – 92kx + 120k²
Since, volume is maximum when the square of side x = 5 is removed from the corners,
\(\left(\frac{d V}{d x}\right)_{\text {at } x=5}=0\)
⇒ 12(5)² – 92k(5) + 120k² = 0
⇒ 60 – 92k + 24k² = 0
⇒ 6k² – 23k + 15 = 0
⇒ 6k² – 18k – 5k + 15 = 0
⇒ 6k(k – 3) – 5 (k – 3) = 0
⇒ (k – 3)(6k – 5) = 0
⇒ k = 3 or k = \(\frac{5}{6}\)
If k = \(\frac{5}{6}\), then
8k – 2x = 8k – 10 < 0
∴ k ≠ \(\frac{5}{6}\)
∴ k = 3
∴ 8k = 8 × 3 = 24 and 15k = 15 × 3 = 45
Hence, the lengths of the rectangular sheet are 24 and 45.

Question 19.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).
Solution:

Let x be the radius of the base and h be the height of the cone which is inscribed in a sphere of radius r.
In the figure, AD = h and CD = x = BD
Since, ΔABD and ΔBDE are similar,
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{BD}}{\mathrm{DE}}\)
BD² = AD . DE = AD (AE – AD)
x² = h(2r – h) …… (1)
Let V be the volume of the cone.


∴ V is maximum when h = \(\frac{4 r}{3}\)
Hence, the altitude (i.e. height) of the right circular cone of maximum volume = \(\frac{4 r}{3}\).

Question 20.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \(\frac{2 R}{\sqrt{3}}\). Also, find the maximum Volume.
Solution:
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
Then from the figure,
\(R^{2}+\left(\frac{h}{2}\right)^{2}=r^{2}\)
∴ R² = r² – \(\frac{h^{2}}{4}\) ………(1)
Let V be the volume of the cylinder.
Then V = πR²h



Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = \(\frac{4 R^{3}}{3 \sqrt{3}}\) cu cm.

Question 21.
Find the maximum and minimum values of the function f(x) = cos²x + sin x.
Solution:
f(x) = cos²x + sin x
∴ f'(x) = \(\frac{d}{d x}\) (cos²x + sin x)
= 2 cos x . \(\frac{d}{d x}\) (cos x) + cos x
= 2 cos x(-sin x) + cos x
= -sin 2x + cos x
and f”(x) = \(\frac{d}{d x}\) (-sin 2x + cos x)
= -cos 2x . \(\frac{d}{d x}\) (2x) – sin x
= -cos 2x × 2 – sin x
= -2 cos 2x – sin x
For extreme values of f(x), f'(x) = 0
-sin 2x + cos x = 0
-2 sin x cos x + cos x = 0
cos x (-2 sin x + 1) = 0
cos x = 0 or -2 sin x + 1 = 0
cos x = cos \(\frac{\pi}{2}\) or sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ x = \(\frac{\pi}{2}\) or x = \(\frac{\pi}{6}\)

(i) f”(\(\frac{\pi}{2}\)) = -2 cos π – sin \(\frac{\pi}{2}\)
= -2(-1) – 1
= 1 > 0
∴ by the second derivative test, f is minimum at x = \(\frac{\pi}{2}\) and minimum value of f at x = \(\frac{\pi}{2}\)
= f(\(\frac{\pi}{2}\))
= \(\cos ^{2} \frac{\pi}{2}+\sin \frac{\pi}{2}\)
= 0 + 1
= 1

(ii) f”(\(\frac{\pi}{6}\)) = \(-2 \cos \frac{\pi}{3}-\sin \frac{\pi}{6}\)
= \(-2\left(\frac{1}{2}\right)-\frac{1}{2}\)
= \(-\frac{3}{2}\) < 0
∴ by the second derivative test, f is maximum at x = \(\frac{\pi}{6}\) and maximum value of f at x = \(\frac{\pi}{6}\)
= f(\(\frac{\pi}{6}\))
= \(\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}\)
= \(\frac{5}{4}\)
Hence, the maximum and minimum values of the function f(x) are \(\frac{5}{4}\) and 1 respectively.