Category: 2nd PUC

Students can Download Basic Maths Exercise 18.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.3 One or Two Marks Questions and Answers

Question 1.
3x² + 4y² = 10
Answer:
Given 3x² + 4y² = 10
Diff w.r.t x
6x + 8y \(\frac{d y}{d x}\) = 0
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-8 y}{6 x}=\frac{-4 y}{3 x}\)

Question 2.
\(\sqrt{x}+\sqrt{y}=3\)
Answer:

Question 3.
y² = 4ax.
Answer:
Given y² = 4ax.
Differentiate with respect to x
2y \(\frac{d y}{d x}\) = 4a.1 ⇒ \(\frac{d y}{d x}\) = \(\frac{4 a}{2 y}=\frac{2 a}{y}\)

Question 4.
\(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\)
Answer:

Question 5.
x² = 4ay
Answer:
Given x² = 4ay
Differentiate with respect to x, 2x = 4a \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{4 a}=\frac{x}{2 a}\)

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:

Question 7.
x³ + y³ = 3axy
Answer:
Given x³ + y³ = 3axy
3x² + 3y² \(\frac{d y}{d x}\) = 3a \(\left(x \cdot \frac{d y}{d x}+y \cdot 1\right)\)
\(\frac{d y}{d x}\) (3y² – 3ax) = 3ay – 3x² = \(\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}\)

Question 8.
x – y = 0
Answer:
Given x – y = 0
Differentiate with respect to x,
1 – \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = 1

Question 9.
x² – y² = a²
Answer:
Given x² – y² = a²
Differentiate with respect to x we get,
2x – 2y. \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{2 y}=\frac{x}{y}\)

Question 10.
x + \(\sqrt{x y}\) = x².
Answer:

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.3 Three marks Questions and Answers

Question 1.
log(xy) = x² + y²
Answer:
Given log(xy) = x² + y²
log x + log y = x² + y² differentiate w.r.t x

Question 2.
2^x + 2^y = 2^x+y
Answer:
Given 2^x + 2^y = 2^x+y
Differentiate w.r.t. x we get
2^x log 2 + 2^y log 2 \(\frac{d y}{d x}\)

Question 3.
x^y = y^x.
Answer:
Given x^y = y^x., taking log^m both sides
y log x = x log y differentiate
Both sides w.r.t x

Question 4.
sin xy = cos(x + y).
Answer:
Given sin xy = cos(x + y), diff w.r.t x.
cos(xy) \(\left[\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right]\)
\(\frac{d y}{d x}\) [sin(x + y) + xcos (xy)]
= -sin(x + y) – y cos (xy)
\(\frac{d y}{d x}=\frac{-[\sin (x+y)+\cos x y]}{(\sin (x+y)+x \cos x y)}\)

Question 5.
y = 4^x+y
Answer:
Given y = 4^x+y, diff. w r.t. x
\(\frac{d y}{d x}\) = 4^x+y log 4(1 + \(\frac{d y}{d x}\)) = 4^x+y
\(\frac{d y}{d x}\)(1 – 4^x+y log 4) = 4^x+y log 4
∴ \(\frac{d y}{d x}=\frac{4^{x+y} \cdot \log 4}{1-4^{x+y} \cdot \log 4}\)

Part-C

2nd PUC Basic Maths Differential Calculus Ex 18.3 Five Marks Questions and Answers.

Question 1.
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = a, Prove that x . \(\frac{d y}{d x}\) = y.
Answer:

Question 2.
If x^y = e^{y – x}, show that \(\frac{d y}{d x}\) = \(\frac{2-\log x}{(1-\log x)^{2}}\)
Answer:
Given x^y = e^{y – x} . Taking log both sides
y log x = (y – x)log e^e
x = y (1 – log x) ∵ log e^e = 1

Question 3.
If cos y = x cos(a + y). show that \(\frac{d y}{d x}\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\)
Answer:

Question 4.
If e^x = y^x show that \(\frac{d y}{d x}\) = \(\frac{(\log y)^{2}}{\log y-1}\)
Answer:
Given e^x = y^x Taking log^m both sides
y log e^e = x log y
y = x log y differentiate w.r.t x

Question 5.
If e^x+y = xy show that \(\frac{d y}{d x}\) = \(\frac{y(1-x)}{x(y-1)}\)
Answer:
Given ye^x+y = xy
Taking log m both sides
(x + y) loge = log(xy)
x+y = log x + log y diff w.r.t x

Question 6.
If y^x = x^y show that \(\frac{d y}{d x}\) = \(\frac{y(y=x \log y)}{x(x-y \log x)}\)
Answer:
Given y^x = x^y, Taking log^m both sides
x log y = y log x, diff w.r.t x

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.5

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.5

Question 1.
Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x – 1)² + 3
Answer:
f (x) = (2x – 1)² + 3
For all values of x, f (x) > 3 (2x – 1)² + 3 > 3
∴ minimum value is 3 when 2x -1 = 0 is x = \(\frac{1}{2}\) however the function has no maximum values as f (x) → ∞ as |x| ∞

(ii) f (x) = 9x² + 12x + 2
Answer:
f (x) = 9x² + 12x + 2

(iii) f (x) = – (x – 1)² + 10
Answer:
f (x) = – (x – 1)² + 10
f(x)= 10 – (x – 1)²
10 – (x – 1)² < 10
f(x) has a maximum value when x – 1 = 0,x= 1.
how ever f (x) has no minimum value.

(iv) g(x) = x³ + 1
Answer:
g(x) = x³ + 1 as x → ∞ g (x) → ∞ and a
s -+ – ∞, g (x) ⇒ – ∞
∴ g (x) has neither minimum value nor maximum value.

Question 2.
Find the maximum and minimum values, if any, of the following functions given by

(i) f (x) = |x + 2| – 1
Answer:
f (x) = |x + 2| – 1
f (x) = |x + 2| – 1 ≥ -1
minimum value is – 1 when x + 2 = 0, x = -2
however it has no maximum value.

(ii) g(x) = – | x + 1| + 3
Answer:
g(x) = – | x + 1| + 3 = 3,-1 x + 1|
g (x) < 3 v x + 1 = 0
∴ max. value is 3
when x = – 1
how ever no minimum value.

(iii) h(x) = sin (2x) + 5
Answer:
h(x) = sin (2x) + 5
maximum value of sin 2x = 1 and minimum value is -1
f(x) = 1 + 5 is 1 + 5
∴ max. h (x) = 6 and
min. h (x) = 4.

(iv) f (x) = |sin 4x + 3|
Answer:
f (x) = |sin 4x + 3|
-1 < sin 4x < 1 ⇒ 3 -1 < sin 4x + 3 < + 1 + 3 + 2′< sin 4 x + 3 < 4
2 < | sin 4x + 3 | < 4
f (x) > 2 and f (x) < 4
min. f (x) = 2 when sin 4x + 3 = 0
max f (x) = 4 when sin 4x + 3 = 0
∴ minimum value is 2 at sin 4x = -1
maximum value is 4 at sin 4x = 1

(v) h(x) = x + 1, x ∈ (- 1, 1)
Ans:
h(x) = x + 1, x ∈ (-1,1)
given that x ∈ (-1, 1)
ie. -1 < x < 1
-1 + 1< x + 1 < 1 + 10 < x + 1< 2
∴ x + 1 > 0 or x + 1 < 2
x + 1 > 0 so no minimum value
x + 1 < 2, so no maximum value.

Question 3.
Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x²
Answer:
f (x) = x²
f’ (x) = 2x, f’ (x) = 0
⇒ 2x = 0, x = 0
f'(x)= 2
f'(x)> 0, hence f (x) has minimum value at x = 0
and the minimum value is (0)²= 0.

(ii) g(x) = x³ – 3x
Answer:

(iii) h(x) = sin x + cos x, o < x< \(\frac{\pi}{2}\)
Answer:

(iv) f (x) = sin x – cos x, 0 < x < 2π
Answer:

(v) f (x) = x³ – 6x² + 9x + 15
Answer:

(vi) \(g(x)=\frac{x}{2}+\frac{2}{x}, x>0\)
Answer:

(vii) \(g(x)=\frac{1}{x^{2}+2}\)
Answer:

(viii) \(f(x)=x \sqrt{1-x}, x>0\)
Answer:

Question 4.
Prove that the following functions do not have maxima or minima :

(i) f (x) = e^x
Answer:
f (x) = e^x
f’ (x) = e^x
f'(x) > 0 ∀ x ∈ R
hence function has no critical point There is no point at which the function is maximum or minimum.

(ii) g(x) = log x, x > 0
Answer:
g (x) = log x, x > 0
g'(x) = \(\frac{1}{x}\), where x > 0
hence the function has no critical point
∴ There is no point at which the function is maximum or minimum.

(iii) h (x) = x³ + x¹ + x +1
Answer:
h (x) = x³+ x²+ x +1
h’ (x) = 3x² + 2x + 1
h’ (x) = 0 ⇒ 3x² + 2x + 1 = 0,
x has no real value, hence there is no critical point.
∴ For no point the function has max. or min. value.

Question 5.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) = x³,x ∈ [-2,2]
Answer:
f (x) = x³,x ∈ [- 2,2]
f’ (x) = 0 ⇒ 3x² = 0 ⇒ x = 0 for finding the absolute maximum and absolute minimum, we have to evaluate f (0), f (2), f (-2)
F (0) = (0)³ = 0, F (2) = (2)³ = 8,
F (-2) = (-2)³ = – 8
Absolute maximum = 8 and
Absolute minimum = -8
∴ maximum at 2 is 8 and minimum at -2 is – 8.

(ii) f (x) = sin x + cos x , x ∈ [0, π]
Answer:

(iii) \(f(x)=4 x-\frac{1}{2} x^{2}, x \in[-2,9 / 2]\)
Answer:

(iv) f (x) = (x – 1)² + 3, x ∈ [-3,1]
Answer:
f (x) = (x – 1)² + 3, x ∈ [-3,1]
f'(x) = 2 (x – 1)
f’ (x) = 0 ⇒ (x – 1) = 0
⇒ x = 1 we will evaluate f (-3) and f (1)
f(-3) = (-3 – 1)² + 3= 16 + 3 = 19
f (1) = (0)² + 3 = 3
Absolute maximum at x = -3 is 19 and
Absolute minimum x = 1 is 3.

Question 6.
Find the maximum profit that a company can make, if the profit function is given by
p(x) = 41+24x – 18x²
Answer:

Question 7.
Find both the maximum value and the minimum value of
3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].
Answer:
f (x) = 3x⁴ – 8x³ + 12x² – 48 x + 25
f’ (x) = 12x³ – 24x² + 24x – 48
_f‘(x) = 0 ⇒ 12 (x³ – 2x² + 2x -4) = 0
⇒ 12 (x² (x – 2) + 2 (x – 2))
⇒ 12 ( (x – 2) (x² + 2) ) = 0 (x – 2)(x² + 2) = 0
⇒ x = 2 but x² + 2 ≠ 0
The points are f (0), f (2), f (3)
f (x) = 3x⁴ – 8x³ + 12x² – 48x + 25
f (0) = 25
f (2) = 3 (16) – 8 (8) + 12 (4) – 48 (2) + 25 = 48 – 64 + 48 – 96 + 25 = -39
f(3) = 3 (3⁴) – 8 (3³) + 12 (3²) – 48 (3) + 25 = 243 – 216 + 108 – 144 + 25
376 – 360 = 16
∴ maximum of f (x) at x = 0 is 25
minimum of f (x) at x = 2 is – 39.

Question 8.
At what points in the interval [0,2π] does the function sin 2x attain its maximum value?
Answer:

Question 9.
What is the maximum value of the function sin x + cos x on \(\left[0, \frac{2}{\pi}\right]\)
Answer:

Question 10.
Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1]
Answer:
f (x) = 2x³ – 24x +107
f’ (x) = 6x² – 24 ⇒ x² = 4, x = ± 2 + 2 s [1,3]
∴ we have to evaluate
f (1). f (2), f (3)
f (1) = 2 (1)³ – 24 x 1 + 107 = 85
f (2) = 2 (2)³– 24 x 2+ 107 = 75
f(3) = 2(3)³-24 x 3 + 107 = 89
maximum at x = 3 and miximum value is 89 minimum at x = 2, is 75
Now the interval is [-3, -1]
∴ points are -3, -2, -1
f (-3) = 2 (-3)³ – 24 (-3) + 107 = 125
f (-2) = 2 (-2)³ – 24 (-2) + 107
= 2 (-8+ 48 + 107 = 139 = 139
f (-1) = 2 (-1) 3-24 (-1) + 107 = 129
maximum at -2 is 139
minimum at (-3) is 125.

Question 11.
It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0,2]. Find the value of a.
Answer:
f (x) = x⁴ – 62x² + ax + 9
f'(x) = 4x³ – 124x + a
at x = 1, the function has maximum value
∴ f'(1) = 0 ⇒ 4 (1)² -124 (1) + a = 0, a = 120
∴ when a = 120, the function attains maximum value.

Question 12.
Find the maximum and minimum values of x + sin 2x on [0,2π]
Answer:

Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Answer:

Question 14.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Answer:
x + y = 60 ⇒ y = 60 – x

Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x² y⁵ is a maximum.
Answer:

Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Answer:

Question 17.
A square piece of tin of side 18 cm ¡s to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
Answer:


x = 9 is not possible, it so side = 0
volume is maximum when the sides are 12,12,3
volume =12 x 12 x 3 = 432 cm³
The volume is maximum when the side of the square to be cut off is 3cm.

Question 18.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?
Answer:
The sides of the box are x, 45 – 24, 45 – 2x

Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Answer:
Let the sides of rectangle be x and y and radius of the circle is r

hence rectangle becomes square.
∴ The area is maximum when the rectangle is a square.

Question 20.
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Answer:

Question 21.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Answer:

Question 22.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Answer:

Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius
R is \(\frac{8}{27} \)of the volume of the sphere.
Answer:

Question 24.
Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) time the radius of the base.
Answer:

Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is \(\tan ^{-1} \sqrt{2}\).
Answer:

Question 26.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is \(\sin ^{-1}\left(\frac{1}{3}\right)\)
Answer:

Choose the correct answer in the Exercises 27 and 29.

Question 27.
The point on the curve x² = 2y which is nearest to the point (0, 5) is
(A) (2,\(\sqrt{2}\),4)
(B) (2, \(\sqrt{2}\), 0)
(C) (0,0)
(D) (2,2)
Answer:

Question 28.
For all real values of x, the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
(A) 0
(B) 1
(C) 3
(D) 1/3
Answer:

Question 29.
The maximum value of [x(x – 1) + 1]^1/3< x < 1 is 0 ≤ x ≤ 1 is
Answer:

Students can Download Basic Maths Exercise 18.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.2

Part-A

2nd PUC Basic Maths Differential Calculus Ex 18.2 One or Two Marks Questions and Answers

Question 1.
(a² – x²)¹⁰
Answer:
Let y(a² – x²)¹⁰
\(\frac{d y}{d x}\) = 10(a² – x²)^{10 – 1}. \(\frac{d y}{d x}\)(a² – x²)
= 10(a² – x²)⁹ (-2x) = -20x (a² – x²)⁹

Question 2.
log[log(log x)]
Answer:
Let y = log x (log log(x))
\(\frac{d y}{d x}=\frac{1}{\log (\log x)} \cdot \frac{1}{\log x} \cdot \frac{1}{x}\)

Question 3.
cos x³
Answer:
Let y = cosx³
\(\frac{d y}{d x}\) = -sinx³ .3x²

Question 4.
sin³\(\sqrt{x}\)
Answer:
Let y = sin³(\(\sqrt{x}\)) = (sin \(\sqrt{x}\)))³
\(\frac{d y}{d x}\) = 3.sin² \(\sqrt{x}\) . cos \(\sqrt{x}\) . \(\frac{1}{2 \sqrt{x}}\)

Question 5.
[log(cos x)]²
Answer:
Let y = [log(cos x)]²
\(\frac{d y}{d x}\) = 2log(cos x). \(\frac{1}{\cos x}\)(-sin x)
= -2 tan x log(cos x)

Question 6.
\(\sec \left(x+\frac{1}{x}\right)\)
Answer:

Question 7.
\(7^{\sin \sqrt{x}}\)
Answer:
Let y = \(7^{\sin \sqrt{x}}\)
\(\frac{d y}{d x}\) = \(7^{\sin \sqrt{x}}\) . log 7 . cos \(\sqrt{x}\) . \(\frac{1}{2 \sqrt{x}}\)

Question 8.
\(\sqrt{\cot \sqrt{x}}\)
Answer:

Question 9.
log(sin \(\sqrt{x}\))
Answer:

Question 10.
log[log (tan x)]
Answer:
Let y = log(log (tan x))
\(\frac{d y}{d x}\) = \(\frac{1}{\log (\tan x)} \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\)

Question 11.
cos 3x . sin 5x
Answer:
Let y = cos 3x . sin 5x (Trans using formula)
sy = 2[sin 8x – sin(-2x)] = 2 [sin 8x] + 2sin 2x
\(\frac{d y}{d x}\) = 16 cos 8x + 4 cos 2x
OR
Let y = cos 3x . sin 5x
\(\frac{d y}{d x}\) = cos 3x(5 cos 5x) + sin 5x (-3 sin 3x) = 5 cos 3x sin5x – 3 sin 5x.sin 3x.

Question 12.
sin x . sin 2x
Answer:
Let y = sin x . sin 2x
\(\frac{d y}{d x}\) = sin x(2 cos2x) + sin 2x cos x = 2 sin x cos 2x + cos x sin 2x

Question 13.
elog_e(x + \(\sqrt{x^{2}+a^{2}}\)).
Answer:

Question 14.
e^2x . sin 3x.
Answer:
Let y = e^2x . sin 3x
\(\frac{d y}{d x}\) = e^2x(3 cos 3x) + sin 3x(2e^2x)

Question 15.
cos⁵x . cos(x⁵).
Answer:
Let y = cos⁵x . cos(x⁵)
\(\frac{d y}{d x}\) = cos⁵x(-sin(x⁵)5x⁴) + cos(x⁵).5cos⁴x.(-sin x)
= – cos⁵x sin(x⁵) 5x⁴ – 5 cos(x⁵) . cos⁴x . sin x

Question 16.
3^x2 .log x.
Answer:
Let y = 3^x2 .log x.
\(\frac{d y}{d x}\) = 3^x2 . \(\frac { 1 }{ x }\) + logx . 3^x2 . log_e 3 . 2x .

Question 17.
\(\frac{x}{\sqrt{x^{2}-1}}\)
Answer:

Question 18.
\(\frac{x}{\sqrt{2 x-1}}\)
Answer:

Question 19.
\(\frac{\mathrm{e}^{\sin \mathrm{x}}}{\sqrt{\log \mathrm{x}}}\)
Answer:

Question 20.
\(\log \left(\frac{1+\sin x}{1-\sin x}\right)\)
Answer:

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.2 Three Marks Questions and Answers

Question 1.
If y = \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\) , show that \(\frac{d y}{d x}\) = sec² \(\left(x+\frac{\pi}{4}\right)\)
Answer:

Question 2.
If y = log \(\left[\frac{1-\cos x}{1+\cos x}\right]\) , prove that \(\frac{d y}{d x}\) = 2 cosec x.
Answer:

Question 3.
Differentiate e^2x w.r.t x from first principles
Answer:

Question 4.
Differentiate sin 2x w.r.t x from first principles.
Answer:

Question 5.
Differentiate tan ax w.r.t x froom the principles.
Answers:

Students can Download Basic Maths Exercise 7.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.3

Part – A

2nd PUC Basic Maths Ratios and Proportions Ex 7.3 Three Marks Questions and Answers ( 3 × 4 = 12)

Question 1.
If ₹ 150 maintains a family of 4 persons for 30 days. How long 7600 maintain a family of 6 persons?
Answer:

Question 2.
300 workers can finish a work in 8 days. How many workers will finish the same work in 5 days.
Answer:

Workers and days are in inverse proportion 300 : x = 5 : 8
x = \(\frac{300 \times 8}{5}\) = 480 workers

Question 3.
5 carpenters can earn ₹540 in 6 days working 9 hours a day. How much will 8 carpenters can earn in 12 days working 6 hours a day?
Answer:

Days and amount are in direct proportion, hours and amount are in direct proportion

Question 4.
A mixture contains milk and water in the ratio 6:1 on adding 5 litres of water, the ratio of milk and water becomes 7 : 2, find the quantity of milk in the original mixture.
Answer:
Quantity of milk is 6x and water is 1x, 5 liters of water is added, the new ratio is 7:2
\(\frac{6 x}{x+5}=\frac{7}{2}\)
12x = 7x + 35 ⇒ 5x = 35
x = 7
The quantity of milk is 6(x) = 6 (7) = 42

Part – B

2nd PUC Basic Maths Ratios and Proportions Ex 7.3 Five Mark Questions and Answers 

Question 1.
A jar contains two liquids X and Y in the ration 7:5. When 6 litres of the mixture is drawn and the jar in filled with the same quantity of Y, the ratio of Xand Y becomes 7:9. Find the quantity X in the jar initially,
Answer:
Let quantity of liquid X is 7x and Y is 5x 6 liters of the mixture is drawn.
i.e.,\(\frac{7 \times 6}{12}=\frac{7}{2}\) litres of X is removed
\(\frac{5 \times 6}{12}=\frac{5}{2}\) litres of Y is removed
∴ The remaining quantity of X and Y is 7x – \(\frac{7}{2}\) and 5x –\(\frac{5}{2}\) respectively.
6 litres of Y is added to get ratio 7:9.

Question 2.
Two taps fill a cistern separately in 20 minutes and 40 minutes respectively and a drain pipe can drain off 30 litres per minute. If all the three pipes are opened, the cistern fills in 72 minutes what is the capacity of the cistern?
Answer:
Time taken by tap A is 20 min
∴ \(\frac{1}{20}\)b of the Cistern is filled by tap A
Similarly \(\frac{1}{40}\) of the Cistern is filled by tap B
Both the taps can fill \(\frac{1}{20}+\frac{1}{40}=\frac{3}{40}\)
“204040 A drain tap can drain 30 liters per minute the cistern is filled in 60 minute.

The drain tap can drain in 17 minutes in 1minute it drains 30 litres
In 17 minutes it drains 30 × 17litres
∴ The capacity of the cistern = 510 litres

Question 3.
If ten persons can do a job in 60 days. In how many days can twenty persons do the same job?
Answer:

Persons & days are in inverse proportion
∴ 10:20 = x : 60
X= \(\frac{60 \times 10}{20}\) = 30days 20

Question 4.
A can do a piece of work in 20 days, B in 30 days and C in 60 days. All of them began to work together. However, A left the job after 6 days and B quit work 6 days before the completion of work. How many days
did the work last?
Answer:
In 1 day the work done by A, B & C is

In 6 days the work done is \(\frac{6}{10}=\frac{3}{5}\)
Remainin work is \(1-\frac{3}{5}=\frac{2}{5}\) of the work
Let the number of days to complete the work be x.C does \(\frac{x}{60}\) of the work & B does \(\frac{x-6}{30}\) of the work in x days

Question 5.
8 men and 16 women can finish a job in 6 days | but 12 men & 24 women can finish it in 8 days. How many days will 26 men and 20 women take to finish the job?
Answer:
8 men & 16 women can finish a Job in 6 days
∴ In 1 day the work done is of 48men and 96 women.
12 men & 24 women can finish a job in 8 days. In 1 day the work done is of 96 men & 192 women.
∴ 48 men + 96 women = 96 men + 192 women
26M + 20W = 52W + 20W = 72W
Let the required number of days be x.
192 : 72 = x : 5
X = \(\frac{192 \times 5}{72}=\frac{45}{3}\) = 15 days

Question 6.
4 men and 12 boys can do a piece of work in 5 … days by working 8 hours per day. In how many days 2 men & 4 boys can do the same piece of work working 12 hours a day.
Answer:
Given 4 m = 12 B
1 m = 3 B
2men & 4 boys = 6 boys + 4 boys = 10 boys

Boys & days are in inverse proposition days & hours are in inverse proportion.
∴ 10 : 12 :: 5 : x 12 : 8
x = \(\frac{12 \times 8 \times 5}{10 \times 12}=\) = 4days.

Question 7.
A railway train 100 metres long is running at the speed of 30 kmph. In what time will it pass (i) a man standing near the line (ii) a bridge 100 metres long?
Answer:
d = 100m Speed = 30km ph
Speed = \(\frac{\mathrm{d}}{\mathrm{t}}\)

Length of the bridge is 100m
d = 100 + 100 = 200
t = \(\frac{200 \times 18}{30 \times 5}\) = = 24 Sec.

Question 8.
The driver of car is traveling at a speed of 36 kmph and spots a bus 80 metres ahead of him. After 1 hour the bus is 120 metres behind the car. What is the speed of the bus?
Answer:
Speed of the car = 36 kmph
Speed of the bus = x kmph
Relative speed = (36 – x) kmph
Distance = 200m
Bus by 120m
∴ t = \(\frac{200}{36-x}\)
1 = \(\frac{200}{(36-x) 1000}\)
36 – x = 0.2
36 -0.2 = x
x = 35.8
∴ Speed of the bus is 35.8 kmph.

Students can Download Basic Maths Exercise 19.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3

Part- A

2nd PUC Basic Maths Application of Derivatives Ex 19.3 Two or Three Marks Questions with Answers.

Question 1.
Find the maximum and minimum value of the following function.
(i) f(x) = x³ – 3x
(ii) f(x) = x³ – 6x² + 9x + 15(0 ≤ x ≤ 6)
(iii) f(x) = x⁴ – 62x² + 120x + 9
(iv) f(x) = 2x³ – 3x² – 12x + 12
(v) f(x) = 2x³ – 3x² – 36x + 10
(vi) f(x) = 9x² + 12x + 2
(vii) f(x) = 2x³ – 15x² + 36x + 10
(viii) f(x) = 2x³ – 21x² + 36x – 20
(ix) f(x) = 2x³ – 15x² + 36x + 10
(x) f(x) = 12x⁵ – 45x⁴ + 40x³ + 6
Answer:
(i) Given f(x) = x³ – 3x …..(1)
f'(x) = 3x² – 3 = 3(x² – 1) = 3(x – 1) (x + 1) …..(2)
f'(x) = 0 ⇒ x = ±1
f”(x) = 6x – (3)
At x = 1, f”(1) = 6 > 0, f(x) is minimum at x = 1
& minimum value is f(1) = 1 – 3 = -2
At x = -1, f”(-1) = -6 < 0 f(x) is maximum at x = -1
Maximum value is f(-1) = -1 + 3 = 2

(ii) f(x) = x³ – 6x² + 9x + 15 (0 ≤ x ≤ 6) – (1)
f'(x) = 3x² – 12x + 9
= 3(x² – 4x + 3) = 3 (x – 3) (x + 1] = 0
f “(x) = 6x – 12
f'(x) = 0 ⇒ x = 1 or 3
At x = 1 f”(1) = 6 – 12 = -6 < 0 the function is maximum at x = 1.
And maximum value is f (1) = 1 – 6 + 9 + 15 = 19
At x = 3, f”(3) = 6 × 3 – 12 = 18 -12 = 6 > 0, f is minimum at x = 3
And minimum value is f(3) = 3³ – 6.3² + 9.3 + 15 = 27 – 54 + 27 + 15 = 15.

(iii) Given f(x) = x⁴ – 62x² + 120x + 9 …..(1)
f”(x) = 4x³ – 124x + 120 ….(2)
f”(x) = 12x² – 124 …..(3)
for a function to be maximum or minimum of f'(x) = 0.
⇒ x³ – 31x + 30 = 0 here x = 1 is a root

⇒ x² + x – 30 = 0 ⇒ (x + 6)(x – 5) = 0 ⇒ x = 5 – 6
Put x = 1 in (3] we get f”(x) = (12 – 124) < 0
f(x) att-ains maximum at = 1 & maximum at x = 1 & max value is
f(1) = 1 – 62 + 120 + 9 = 68.
At x = 5, f”(5) = 12(5)² 124 = 300 – 124 > 0 f(x) attains
minimum at x = 5, & minimum value is f(5) = 625 – 1550 + 600 + 9 = – 316
At x = – 6, f “(-6) = 12 (-6)² – 124 = 432 – 124 > 0
f(x) attains minimum at x = -6 & minimum value is
f (-6) = (-6)⁴ – 62(-6)² + 120 (-6) + 9
= 1296 – 2232 – 720 + 9 = -1647.

(iv) f(x) = 2x³ – 3x² – 12x + 12 ….(1)
f'(x) = 6x² – 6x – 12 = 6 (x² – x – 2) …(2)
f'(x) = 12x – 6 ….(3)
for a function to be maximum or minimum f ‘(x) = 0 ⇒ (x – 2) (x + 1) = 0
⇒ x = 2 or – 1
Put x = (-1) in (3) we get f “(-1) = -12 – 6 = -18 < 0
f(x) att-ains maximum at x = -1 & maximum value is
f(-1) = 2 (-1)3 – 3 (-1)2 – 12(-1) + 12 = 19
Put x = 2 in(3) f”(2) = 24 – 6 = 18 > 0
f(x) attains minimum at x = 2 & minimum value is
f(2) = 2 (2)³ – 3(2)² – 12(2) + 12 = 6- ^-24+ >^=8

(v) Given f(x) = 2x³ – 3x² – 36x + 10 ….(1)
f ‘(x) = 6x² – 6x – 36 = 6 (x² – x – 6) …..(2)
f “(x) = 12x – 6 ……(3)
For a function to be maximum of minimum f ‘(x) = 0
⇒ (x² – x – 6) = 0 => (x – 3) (x + 2) = 0
⇒ x = 3 or – 2
Put x = 3 in equation (3) we get
f “(3) = 36 – 6 = 30 >0
⇒ f(x) attains minimum at x = 3
Minimum value is f(3) = 2(3)³ – 3 (3²) – 36 (3) + 10
f(3) = 54 – 27 – 108 + 10 = -71
Put x = -2 in equation (3) we get
f”(-2) = -24 – 6 = – 30 < 0 ⇒ f (x) attains maximum at x = -2
Maximum value is f(-2) = 2(-2)³ -3(-2)² – 36 (-2] + 10
f(-2) -16-12 + 72 + 10 = 54

(vi) Given f(x) = 9x² + 12x + 2 ….(1)
f'(x) = 18x + 12 …(2)
f”(x) = 18 > 0 ……(3)
⇒ f(x) attains minimum
f'(x) = 0 ⇒ 18x+ 12 = 0 ⇒ x = \(-\frac{2}{3}\)
& f” \(\left(-\frac{2}{3}\right)\) 18 > 0 ⇒ f(x) is minimum & the minimum value is
f\(\left(-\frac{2}{3}\right)\) = 9\(\left(\frac{4}{9}\right)\) + 12 \(\left(-\frac{2}{3}\right)\) + 2
= 4 – 8 + 2 = -2

(vii) f(x) = 2x³ – 15x² + 36x + 10 …….(1)
f ‘(x) = 6x² – 30x + 36 = 6 (x² – 5x + 6) ……..(2)
f”(x) = 12x – 30 …..(3)
f'(x) = 0 ⇒ x² -5x + 6 = 0 ⇒ (x – 3)(x – 2) = 0 ⇒ x = 3 or 2
when x = 3 f “(x) = 12x – 30
f “(3) = 36 – 30 = 6 > 0 ⇒ f(x) has minimum
Minimum value is f(3) = 2(3)³ – 15(3)² + 36(3) + 10
f(3) = 54 – 135 + 108 + 10 = 37
when x = 2, f”(2) = 24 – 30 = -6 < 0 ⇒ f(x) has maximum
maximum value is f(2) = 2(2)³ – 15(2) + 36(2) + 10
f(2) = 16 – 60 + 72 + 10 = 38.

(viii) Given f(x) = 2x³ – 21x² + 36x – 20 ….. (1)
f'(x) = 6x² – 42x + 36 …… (2)
= 6(x² – 7x + 6)
f'(x) = 6 (x – 1) (x – 6) = 0 ⇒ x = 1 or 6
f”(x) = 12x – 42 … (3)
when x = l,f “(1) = 12 – 42 = -30 < 0 ⇒ f(x) is maximum
maximum value is f(1) = 2 – 21 + 36 – 20 = -3
when x = 6, f “(6) = 72 – 42 = 30 > 0 ⇒ f(x) is minimum
minimum value is f(6) = 2(6)3³ – 21 (6)² + 36(6) – 20
f(6) = 432 – 756 + 216 – 20 = -128

(ix) Given f(x) = 12x⁵ – 45x⁴ + 40 x³ + 6 ….(1)
f'(x) = 60x⁴ – 180x³ + 120x² ….(2)
= 60x² (x² – 3x + 2)
= 60x² (x – 1) (x – 2)
f ‘(x) = 0 ⇒ 60x² (x – 1) (x – 2) = 0 ⇒ x = 0, 1, 2
f “(x) = 60 (4x³ – 6x + 4x) ……. (3)
when x = 0, f”(x) = 0 ⇒ f(x) has neither maximum nor minimum
when x = 1, f”(x) = -1 < 0 ⇒ f(x) has a maximum & maximum value is
f(1) = 12 – 45 + 40 + 6 = 13
When x =2 f”(x) = 4 > 0 ∴ f(x) has a minimum
minimum value is f(2) = 12(32) – 45(16) + 40(8) + 6
f(2) = 384 – 720 + 320 + 6 = -10.

Question 2.
The sum of two natural numbers is 48. Find the numbers when their product is maximum.
Answer:
Let the two numbers be x & y.
Given x + y = 48 & product: = xy where y 48 – x.
Let p = xy = x (48 – x) = 48x – x².
\(\frac{d p}{d x}\) = 48 – 2x
\(\frac{d p}{d x}\) = 0 ⇒ 48 – 2x = 0 x = 24
\(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{d} \mathrm{x}^{2}}\) = -2 < 0 ⇒ product is maximum
x = 24 ⇒ y = 48 – 24 = 24
⇒ the two numbers are 24, 24.

Question 3.
Find two positive numbers whose sum is 14 and the sum of whose square is minimum.
Answer:
Let the two numbers be x and y
Given x + y = 14 & S = x² + y² where y = 14 – x
∴ S = x² + (14 – x)² = x² + 14² + x² – 28x = 2x² – 28x + 14²
\(\frac{d s}{d x}\) = 4x – 28 → (1) \(\frac{d s}{d x}\) = 0 ⇒ 4x – 28 = 0 ⇒ x = 7
\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 4 > 0, sum is minimum.
∴ y = 14 – x = 14 – 7 = 7
∴ the two positive number are 7 & 7.

Question 4.
Find two positive numbers whose sum is 30 and the sum of their cubes is minimum.
Answer:
Let the two numbers be x & y.
Given x + y = 30 & S = x³ + y³ where y = 30 – x
S = x³ + (30 – x)³ = x³ + (30)³ – x³ – 2700x + 90x²
\(\frac{d s}{d x}\) = – 2700 + 180x
\(\frac{d s}{d x}\) = 0 ⇒ x = \(\frac{2700}{180}\) = 15
\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 180 > 0 ⇒ sum of cubes is minimum & y = 30 – 15 = 15
∴ two positive number are 15 & 15.

Question 5.
The product of two natural numbers is 64. Find the numbers is their sum is minimum
Answer:
Let the two numbers be x & y
Given xy = 64 ⇒ y = \(\frac{64}{x}\)
Let s = x + y = x + \(\frac{64}{x}\).
\(\frac{d s}{d x}\) = 1 – \(\frac{64}{x^{2}}, \frac{d s}{d x}\) = 0 ⇒ x² = 64 ⇒ x = ±8
\(\frac{d^{2} s}{d x^{2}}=+\frac{128}{x^{3}}\)
When x = 8, \(\frac{d^{2} s}{d x^{2}}=\frac{128}{8^{3}}\) > 0 ⇒ s is minimum
When x = -8, \(\frac{d^{2} s}{d x^{2}}=\frac{128}{8^{3}}\) < 0 ⇒ s is maximum
The two numbers are 8 & 8.

Students can Download Basic Maths Exercise 6.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.1

Part – A

2nd PUC Basic Maths Mathematical Logic Ex 6.1 One Mark Questions and Answers

Question 1.
Symbolise the following propositions:
(1) 3x = 9 and x<7
(ii) 33 + 11 ≠ 3 or 8 – 6 = 2
(iii) If two numbers and equal then their squares are not equal.
(iv) If oxygen is a gas then gold is a compound
(v) y + 4 ≠ 4 ore is not a vowel
Answer:
(i) Let p:3x = 9, q = x < 7
Given in symbols is p ∧ q

(ii) Let p:33 = 11 = 3, q= 8 – 6 = 2
Given is ~p ∨ q

(iii) Let p: Two numbers are equal, q: Squares are equal then given is p → ~q

(iv) Let p: Oxygen is a gas
q: Gold is a compound
Then given is p → q

(v) P: y + 4 = 4, q: e is a vowel given proposition is ~p v ~q

Part – B

2nd PUC Basic Maths Mathematical Logic Ex 6.1 Two or Three Marks Questions and Answers

Question 1.
if p, q and r are propositions with truth values F, T and F respectively, then find the truth values of the following compound propositions:
(i) (~p → q) ∨ r
(ii) (p ∧ ~q) → r
(iii) p → (q → r)
(iv) ~(p → q) ∨~(p ↔ q)
(v) (p ∧ q) ∨ ~ r
(vi) ~(p ∨ r) → ~q
Answers:
(i) (~p → q)∨ r
(~F →T) ∨ r
(T →T) ∨ F
T ∨ F
= T

(ii) (p ∧~q) → r
(F∧ ~T) → F
(F ∧F) → F
F →F
= T

(iii) p → (q → r)
F →(T →F)
F →(T →F)
= T

(iv) ~(p → q) ∨~(p↔ q)
~(F →T) ∨ ~(F↔T)
~T ∨ ~F
F ∨ T = T

(v) (p ∧ q) ∨ ~ r
(F ∧ T) ∨ ~ r
F ∨ T
= T

(vi) ~(P ∨ r) → ~ q
~(F ∨ F) ∨ ~ T
~F → ~ T
T → F
= F

Question 2.
Answer:
(1) If the compound proposition “(p → q) ∧ (p ∧ r)” is false, then find the truth values of p, q and r.
(ii) If the compound proposition p→ (q ∨ r) is false, then find the truth values of p, q and r.
(iii) If the compound proposition p → (~q ∨ r) is false, then find the truth values of p, q and r.
(iv) If the truth value of the propositions (p ∧ q) → (r ∨ ~s) is false, then find the truth values of p, q, rand s.

Answers:
(i) Given (p → q) ∧ (p ∧ r) is false
(a) Case 1: p → q is true & p ∧ r is false
p is T q is T & p is T &ris F
p = T, q = T, r = F
Case 2(a): p = F, q = T p ∧ r= F → p = F, r = F
p = F, q = T, r = f

(b): (p →q) is F & par is true
T → F = F
T ∧ T is T
P=T, q = F, r=T

Case 3: (p → q) is F & (p ∧ r) is false
T → F = F
F ∧ F = F
F ∧ T = F
F ∧ F = F .
∴ p = T, q = F, r= F.

(ii) Given p → (q ∨ r) is false
T → F = F
∴ p = T & q ∨ r is false =
F ∨ F= F
∴ p = T, q = F & r= F

(iii) Given p → (q ∨ r) is false
Then T → F= F
∴ P = T, ~ q ∨ r= F
F ∨ F = F
∴ q = T, q = T, r = F.

(iv) Given (p ^ q) → (r ∨ ~s) is false
We know that T → F = F
∴ p∧q = T and r ∨ ~s = F is false
T∧ T = T
F ∨ F= F is false
∴ p = T, q = T, r = F, S = T

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.3

Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.
Answer:

Question 2.
Find the slope of the tangent to the curve
\(y=\frac{x-1}{x-2}, x \neq 2 \text { at } x=10 \)
Answer:

Question 3.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x- coordinate is 2.
Answer:

Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x-coordinate is 3.
Answer:
\(\frac{d y}{d x}\) = 3x² – 3 dx
slope at x = 3 is 3 (9) – 3 = 24.

Question 5.
Find the slope of the normal to the curve
\(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta \text { at } \theta=\frac{\pi}{4}\)
Answer:

Question 6.
Find the slope of the normal to the curve
\(x=1-a \sin \theta, y=b \cos ^{2} \theta \text { at } \theta=\frac{\pi}{2}\)
Answer:

Question 7.
Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to
the x-axis.
Answer:
\(\frac{d y}{d x}\) = 3x² – 6x – 9, slope of the tangent
since the tangent is parallel to x-axis \(\frac{d y}{d x}\) = 0
3x² – 6x – 9 = 0
3 (x + 1) (x – 3) = 0, x = 3, x = -1
when x = 3, y = 27 – 27 – 27 4- 7 = -20
when x = -1, y = -1 -3 + 9 + 7 = 12
The points at which the tangent parallel to x – axis are (3, -20) and (-1, 12).

Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Answer:

Question 9.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.
Answer:
Slope of the line y = x – 11 is 1
slope of the curve \(\frac{d y}{d x}\) = 3x² – 11
∴ 3x² – 11 = 1                   ‘
3x² = 12 ⇒ x² = 4, x = +2
when x = 2, y = (2)³ – 11 (2) + 5 = -9
when x = -2, y = -8 + 22 + 5 = 19
points are (2, -9) and (-2, 19)
equation of tangent at (2, -9) and slope is 1
y + 9 = 1 (x – 2)
y = x- 11 equation of tangent at (-2, 19)
y – 19 = 1 (x + 2) ⇒ y = x + 211
∴ (2, -9) is the only point at which the tangent is y = x – 11.10.

Question 10.
Find the equation of all lines having slope – 1 that are tangents to the curve
\(y=\frac{1}{x-1}, x \neq 3\)
Answer:

x = 2, x = 0
when x = 2, y = 1
when x = 0, y = -1
The points are (2, 1) and (0,-1)
equation of line though (2,1) having slope = -1
is y – 1 = -1 x (x – 2) ⇒ y + x = 3
or x + y – 3 = 0 and equation of line through (0,- 1)y + 1 = -1 (x – 0)
∴  y + x + 1 = 0.

Question 11.
Find the equation of all lines having slope 2 which are tangents to the curve
\(y=\frac{1}{x-3}, x \neq 3\)
Answer:
slope of the line = 2
slope of the curve = \(\frac{-1}{(x-3)^{2}}\)
\(\frac{-1}{(x-3)^{2}}\) =2
⇒ 2 (x -3)² = -1
⇒ 2 (x² – 6x + 9) = -1
⇒ 2x² – 12x + 19 = 0
which has no real roots b² – 4ac < 0
hence there is no point on the curve.

Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve
\(y=\frac{1}{x^{2}-2 x+3}\)
Answer:

Question 13.
Find the points of curve \(\frac{x^{2}}{9}+\frac{y^{2}}{16}=1\) which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis.
Answer:
16x² + 9y² = 144
16 x 2 x + 9 x 2y x \(\frac{d y}{d x}\) = 0

(0,4) and (0, -4) are the points on the curve at which tangent is parallel to x – axis.

(ii) For tangents parallel to y – axis \(\frac{d y}{d x}\) = \(\frac{1}{0}\)
\(\frac{-16 x}{9 y}=\frac{1}{0} \Rightarrow y=0\)
when y = 0, x² = 9x = ± 3 point is (3,0) and (-3,0)
(3, 0) and (-3, 0) are the points at which the curve is parallel to y – axis.

Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
Answer:
y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) dy
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10 dx
slope at (0,5) = -10
∴ Equation of tangent at (0, 5) is
y – 5 = -10 (x – 0)
y – 5 = -10 x 10 x + y – 5 = 0
slope of the normal at (0,5)
\((0,5)=\frac{-1}{-10}=\frac{1}{10}\)
∴ equation of normal is
y – 5 =\(\frac{1}{10}\) (x – 0) = 10y – 50 = x
x – 10y + 50 = 0

(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1, 3)
Answer:
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10 dx
slope of the tangent at x = 1

(iii) y = x³ at (1,1)
Answer:
\(\frac{d y}{d x}=3 x^{2}\) 
∴ slope of the tangent at x = 1 = 3
∴ Equation of tangent is y – 1 = 3(x – 1)
3x – y – 2 = 0
∴ slope of normal = \(\frac{-1}{3} \)
∴ equation of normal is y – 1 = \(\frac{-1}{3} \)(x – 1)
3y – 3 = – (x – 1)
x + 3y – 4 = 0.

(iv) y = x² at (0, 0)
Answer:
\(\frac{d y}{d x}=2 x\)
∴ slope at x = 0 =0
∴ Equation of tangent is y – 0 = 0 (x – 0) ⇒ y = 0
slope of the normal = -1/(0)
∴ equation of normal \(y – 0=\frac{-1}{0}(x-0) \Rightarrow x=0\)

(v) x = cos t, y = sin t at \(\frac{\pi}{4}\)
Answer:

Question 15.
Find the equation of the tangent line to the curve y = x² – 2x +7 which is
(a) parallel to the line 2x – y + 9 = 0
Answer:
\(\frac{d y}{d x}\) = 2x – 2 also slope = 2
2 x – 2 = 2 ⇒2x = 4 ⇒ x = 2
when x = 2, y = (2)² – 2 (2) + 7 = 7
∴ equation of tangent is y – 7 = 2 (x – 2)
2x – y + 3 = 0 ⇒ 2x – y + 3 = 0.

Students can Download Basic Maths Exercise 5.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.2

Part – A

2nd PUC Basic Maths Partial Fractions Ex 5.2 Five Marks Questions and Answers

I. Resolve the following into partial fractions; (3 and 5 marks)

Question 1.
\(\frac{x}{(x+1)(x-4)}\)
Answer:

Put
x = 4, 4 = A(0) +B (4 + 1) ⇒ 4 = 5B ⇒ B = \(\frac{4}{5}\)
Put x = -1, -1 = A(-1-4) +B(0) ⇒ -1 = -5A = A=\(\frac{1}{5}\)
Substituting both A & B in equation (1) we get

Question 2.
\(\frac{x+1}{(x+2)(x-3)}\)
Answer:
\(\frac{x+1}{(x+2)(x-3)} =\frac{A}{x+2}+\frac{B}{x-3}\)………………….. (1)
x = 3, 3 + 1 = A(0) +B(3 + 2)
4 = 5B = B = \(\frac{4}{5}\)
Put
x = -2, -2 + 1 = A(-2 – 3) + B(0)
-1 = A(-5)
A = 1/5;
3 = -A ⇒ A = -3

Question 3.
\(\frac{7 x-1}{(1-2 x)(1-3 x)}\)
Answer:
Let
\(\frac{7 x-1}{(1-2 x)(1-3 x)}=\frac{A}{1-2 x}+\frac{B}{1-3 x}\)
⇒ (7x – 1) = A(1 – 3x) + B(1 – 2x)
Put x = 0, -1 = A + B …………..1
Put x=1 6 = -2A-B …………………….2
Solving 1 & 2 (adding) -A = 5 ⇒ A = -5
B = -1-A = -1 + 5 = 4.

Question 4.
\(\frac{3-7 x^{2}}{(1-3 x)(1+2 x)(1+x)}\)
Answer:
Let
\(\frac{3-7 x^{2}}{(1-3 x)(1+2 x)(1+x)}=\frac{A}{(1-3 x)}+\frac{B}{1+2 x}+\frac{C}{1+x}\) …………… (1)
∴ 3 – 7x² = A(1 + 2x) (1+x) +B(1 – 3x) (1 + x) +C (1 – 3x) (1 + 2x)
Put x = -1, 3 – 7 = A(0) + B(0) +C(1 +3) (1-2)
-4 = -4C ⇒ C = 1
Put x = 0, 3 = A + B + C = A + B = 2
Put x = 1, 3 – 7 = A(1 + 2) (1 + 1) + B(1 – 3) (1 + 1) +C(1 – 3) (1 + 2)
-4 = 6A – 4B – 6C ∵ C= 1
-4 + 6 = 6A – 4B = 2
3A – 2B = 1 2A + 2B = 4
⇒ A + B=2

Question 5.
\(\frac{x^{2}}{(x+1)(x+2)(x+3)}\)
Answer:
Let \(\frac{x^{2}}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}\)
x² = A(x +2) (x + 3) +B(x + 1) (x + 3) + C(x + 1) (x + 2)
Put x=-1, (-1)² = A(-1 +2) (-1 + 3) + 0 + 0
1 = 2A ⇒ A =\(\frac{1}{2}\)
Put x = -2, (-2)² = A(0) + B(-2 + 1) (-2 + 3) + C(0)
4 = -B ⇒ b = -4
Put x = -3, (-3)² = A(0) + B(0) +C(-3 + 1) (-3 + 2)
9 = -2C:(-1) ⇒ 9 = 2c ⇒ c = \(\frac{9}{2}\)
Equation 1 be becomes

Question 6.
Answer:
\(\frac{2 x^{2}+10 x-3}{(x+1)(x-3)(x+3)}\)
Let

2x² +10x – 3 = A[x – 3) (x + 3) + B(x + 1) (x + 3) + C(x +1) (x – 3)
Put x=-1, 2 – 10 – 3 = A(-4) (2) + B(0) + C(0)
-11 = -8A = A = \(\frac{11}{8}\)
Put x = 3, 18 + 30 – 3 = A(0) + B(4)(6) + C(0)
+45 = 24B ⇒ B = \(\frac{45}{24}=\frac{15}{8}\)
Put x = -3, 18 – 30 – 3 = C(-3 + 1) (-3 -3) = C(-2) (-6)-15 = + 12C
C = \(\frac{-15}{12}=\frac{-5}{4}\)

Question 7.
\(\frac{3 x+1}{x^{2}-6 x+8}\)
Answer:

Question 8.
\(\frac{x^{2}-10 x+13}{(x+1)\left(x^{2}-5 x+6\right)}\)
Answer:
Let

x² – 10x + 13 = A(x-2)(x – 3) + B(x +1) (x – 3) + C(x + 1) (x – 2)
Put x=-1, 1 + 10 + 13 = A (-3) (-4) + 0 + 0
24 = 12A ⇒ A = \(\frac{24}{12}\)=2
Put x = 2, 4 – 20 + 13 A(0) + B(3) (-1) + C(0)
-3 = -3B ⇒ B = 1
Put x = 3, 9 – 30 + 13 = A(0) + B(0) + C(4) (1)
-8 = 47 ⇒ C= -2

Question 9.
\(\frac{3 x+20}{x^{2}+4 x}\)
Answer:
Let \(\frac{3 x+20}{x^{2}+4 x}=\frac{3 x+20}{x(x+4)}=\frac{A}{x}+\frac{B}{x+4}\) ……..(1)
3x + 20 = A(x + 4) + B(x)
Put x=0, 20 = 4A ⇒ A = 5
Put x = -4, -12 + 20 = A(0) + B(-4)
8 = -4B ⇒ B = -2
\(\frac{3 x+20}{x^{2}+4 x}=\frac{5}{x}-\frac{2}{x+4}\)

Question 10.
\(\frac{x+3}{(x-1)\left(x^{2}-4\right)}\)
Answer:
Let \(\frac{x+3}{(x-1)\left(x^{2}-4\right)}=\frac{x+3}{(x-1)(x-2)(x+2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+2}\) ……(1)
(x + 3) = A[x -2) (x + 2) +B(x – 1) (x + 2) + C(x – 1) (x – 2)
Put x= 1, 4 = A(-1)(3) ⇒ A=\(\frac{-4}{3}\)
Put x = 2, 5 = A(0) + B(1) (4) + C(O)
Put 5 = 4B ⇒ B = \(\frac{-4}{3}\)
Put x = -2, -2 + 3 = A(0) + B(0) +C(-3) (-4)
1 = 12C ⇒ C = \(\frac{1}{12}\)

Question 11.
\(\frac{x+3}{x^{3}-x}\)
Answer:
Let \(\frac{x+3}{x^{3}-x}=\frac{x+3}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\) ….. (1)
x + 3 = A(x-1) (x + 1) + B(x) (x + 1) + C(x) (x – 1)
Put x = 0 3 = -A + 0 + 0 ⇒ A=-3
Put x = 1, 4 = A(0) + B(1) (2) + 0 = 4 = 2B ⇒ B = 2
Put x = -1, 2 = A(0) + B(0) + C(-1) (-2); 2 = 2c ⇒ c = 1
∴ \(\frac{x-3}{x^{3}-x}=\frac{-3}{x}+\frac{2}{x-1}+\frac{1}{x+1}\)

Question 12.
\(\frac{1+3 x+2 x^{2}}{(1-2 x)\left(1-x^{2}\right)}\)
Answer:
Let


1 + 3x + 2x² = A(1 – x)(1 + x) +B(1 – 2x)(1 + x) + C(1 – 2x) (1 – x)
Put x = 1, 6 = A(0) + B(-1)(2) +C(0);
6 = -2B ⇒ B = -3
Put x=-1, 0) = A(0) + B(0) +C(1 + 2)(2)
6C = 0 ⇒ C = 0
Put x = 0, 1 = A + B + C = A – 3 = 1 ⇒ A = 1 + 3 = 4

Part – B

1. Resolve the following into partial fractions; (5 Marks)

Question 1.
\(\frac{4}{(x-3)(x+1)^{2}}\)
Answer:
Let \(\frac{4}{(x-3)(x+1)^{2}}=\frac{A}{x-3}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}\)
∴4 = A(x + 1)² + B(x + 1)(x – 3) + C(x – 3)
Put x = 3, 4 = A(4)² + B(0) + C(0)
4 = 16A ⇒ A = \(\frac{1}{4}\)
Put x =-1, 4 = A(0) + B(0) + C(-1-3); Put 4 = -4C ⇒ C = -1
Put x = 0 4 = A- 3B – 3C
3B = A – 3C – 4

Question 2.
\(\frac{9}{(x+1)(x+2)^{2}}\)
Answer:
Let \(\frac{9}{(x+1)(x+2)^{2}}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}\)
9 = A(x + 2)² + B(x + 1)(x + 2) +C(x + 1)
Put x = -1, 9 = A(-1 + 2)² + 0 + 0
A = 9
Put x= -2, 9 = A(0) + B(0) + CC – 10 = -9
Put x = 0, 9 = 4A + 2B + C
9 = 36 + 2B -9
9 = 27 + 2B \(\frac{-18}{2}\) = B B = -9
∴ \(\frac{9}{x-1}-\frac{9}{x+2}-\frac{9}{(x+1)^{2}}\)

Question 3.
\(\frac{3 x+4}{(x+1)^{2}(x-1)}\)
Answer:
Let \(\frac{3 x+4}{(x+1)^{2}(x-1)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x-1}\)
3x+4 = A(x + 1)(x – 1) + B(x – 1) + C (x + 1)²
Put x = -1, ⇒ 3 + 4 = A(0) + B(-2) + C(0)
1 = -2B ⇒ B = \(-\frac{1}{2}\)
Put x = 1, = 3 + 4 = A(0) +B(0) + C(2)²
7 = 4C ⇒ C = 5
Comparing the coefficients of x² on both sides
0 = A +C ⇒ A = -C = \(-\frac{7}{4}\)

Question 4.
\(\frac{3 x+2}{(x-2)(x+3)^{2}}\)
Answer:
Let \(\frac{3 x+2}{(x-2)(x+3)^{2}}=\frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^{2}}\)
(3x + 2) = A(x + 3)² + B(x – 2) (x + 3) + C(x – 2)
Put x = 2, 6 + 2 = A(2 + 3)² + B(0)+C (0)
8 = 25A ⇒ A = \(\frac{8}{25}\)
Put X=-3, -9 + 2 = A(0) + B(0) +C(-5)
-7 = -5C ⇒ C = \(\frac{7}{5}\)
Comparing the co-efficients of x2 on both sides
0 = A + B ⇒ B = -A = \(\frac{8}{25}\)

Question 5.
\(\frac{2 x^{2}-4 x+1}{(x+2)(x-3)^{2}}\)
Answer:
2x² – 4x + 1 = A[x – 3)² + B(x – 2) (x-3) + C (x – 2)
Put x = 2, 8-8 + 1 = A(-1)² = A = 1
Put x = 3, 18 – 12 + 1 = A(0) + B(0) + C(3 – 2) = 7 = C
Comparing coefficient of x² = 2 = A + B = B = 2-A = 2 -1 =1
∴ \(\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}=\frac{1}{x-2}+\frac{1}{x-3}+\frac{7}{(x-3)^{2}}\)

Question 6.
\(\frac{x^{2}}{(x+1)^{2}(x-5)}\)
Answer:
Let \(\frac{x^{2}}{(x+1)^{2}(x-5)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x-5}\)
x² = A (x + 1)(x – 5) + B(x – 5) + C(x + 1)²
Put x = -1, 1 = A(0) + B(-1-5) + C(0)
1 = -6B ⇒ B = \(-\frac{1}{6}\)
Put x = 5, 25 = A(0) + B(0) + C(6)²
C = \(\frac{25}{36}\)
Compare the coefficients of x2 on both sides we get

Question 7.
\(\frac{4-7 x}{(2+3 x)(1+x)^{2}}\)
Answer:
Let \(\frac{4-7 x}{(2+3 x)(1+x)^{2}}=\frac{A}{2+3 x}+\frac{B}{1+x}+\frac{C}{(1+x)^{2}}\)
⇒ 4 – 7x = A(1 + x)² +B(1 + x) (2 + 3x) + C(2 + 3x)
Put x=-1, 11 = A(0) + B(0) + C(-1)
C = -11
Put x = 0 4 = A + 2B + 2C ⇒ 26 = A + 2B …..(1)
Compare the coefficients off of x² on both sides
0 = A + 3B …….(2)
Equations 1 – 2 gives 26 =-B = 0 ⇒B = -26& A = 78
∴ \(\frac{4-7 x}{(2+3 x)(1+x)^{2}}=\frac{78}{2+3 x}-\frac{26}{1+x}-\frac{11}{(1+x)^{2}}\)

Question 8.
\(\frac{1+2 x}{(x+2)^{2}(x-1)}\)
Answer:
Let \(\frac{1+2 x}{(x+2)^{2}(x-1)}=\frac{A}{x+2}+\frac{B}{(x+2)^{2}}+\frac{C}{x-1}\)
1 + 2x = A (x + 2) (x – 1) + B(x – 1) +C(x + 2)²
3 = A(0) + B(0) + C(9) ⇒ C = \(\frac{1}{3}\)
Put x = -2, 1 – 4 = B(-3) ⇒ B = 1
Comparing the coefficient of x² on both sides we get

Question 9.
\(\frac{9 x-27}{(2+1)(x-2)^{2}}\)
Answer:
Let \(\frac{9 x-27}{(2+1)(x-2)^{2}}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}}\)
9x – 27 = A(x – 2)² + B(x + 1)(x – 2) + C(x + 1)
Put x = -1, -36 = A(-3)² + B(0) + C(0)
-36 = 9A ⇒ A=-4
Put x = 2, 18 – 27 = A(0) + B(0) +C(2+1)
-9 = 3C ⇒ C =-3
Comparing the coefficients of x? on both sides
0 = A + B = B = -A ⇒ B =-(-4) = 4

Question 10.
\(\frac{2 x+5}{(x+2)(x-1)^{2}}\)
Answer:
Let \(\frac{2 x+5}{(x+2)(x-1)^{2}}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{(x-1)^{2}}\)
2x + 5 = A(x – 1)² + B(x – 1)(x + 2) +C(x + 2)
Put x = -2, -4 + 5 = A(-3)² + 0 + 0
1 = 9A ⇒ A =\(\frac{1}{9}\)
Put x= 1, 2 + 5 = A(0) + B(0) + C(1 + 2)
7 = 3C ⇒ C = \(\frac{7}{3}\)
Comparing the coeff of x2 on both sides
0 = A+B ⇒ B = A = \(-\frac{1}{9}\)

Question 11.
\(\frac{x}{(1+2 x)^{2}(1-3 x)}\)
Answer:
Let \(\frac{x}{(1+2 x)^{2}(1-3 x)}=\frac{A}{1+2 x}+\frac{B}{(1+2 x)^{2}}+\frac{C}{1-3 x}\)
X = A(1 + 2x) (1 – 3x) + B(1 – 3x) + C(1 + 2x)2
Put x = 0 we get A + B + C = 0
Comparing co efficient of x2 both sides
0 = -6A + 40 ⇒ 6A = 4C ⇒3A = 20
We have A+B+C = 0
⇒ 3A + 3B + 3C = 0
2C + 3B + 3C = 0
5C = -3B
: 3A = 2C
Comparing the coefficient of x both sides

Part – C

III. Resolve into partial fractions: (5 Marks)

Question 1.
\(\frac{2 x^{2}-7 x+1}{x^{2}-3 x-4}\)
Answer:
\(\frac{2 x^{2}-7 x+1}{x^{2}-3 x-4}\)
This is an improper fraction
∴ Convert into proper fraction by actual division.

9- x = A(x + 1) + B(x – 4)
Put x = 4, 9-4 = A(4 + 1) + B(0)
5 = 5A ⇒ A = 1
Put x=-1, 9-(-1) = A(0) +B(-1-4)
10 = -5B ⇒ B = -2

Question 2.
\(\frac{4 x^{2}-3 x+5}{(2-x)(1+x)}\)
Answer:

x + 13 = A(1 + x) + B(2 – x)
Put x = 2, 15 = A[1 + 2) + B(0)
15 = 3A ⇒ A = \(\frac{15}{3}\) = 5
Put x=-1, -1 + 13 = A(0) +B(2-(-1))
12 = 3B ⇒ B = 4

Question 3.
\(\frac{x^{3}+7 x^{2}+17 x+11}{x^{2}+5 x+6}\)
Answer:

(x – 1) = A(x + 3) + B(x + 2)
Put x = -2, -3 = A(-2 + 3) + B(0)
-3 = A ⇒ A=-3
Put X=-3, -4 = A(0) + B(-3 + 2)
-4 = -B ⇒ B = 4

Question 4.
\(\frac{2 x^{2}+3 x+2}{x^{2}-x-2}\)
Answer:

∴ 5x + 6 = A (x + 1) +B(x – 2)
Put x = 2, 16 = A(3) ⇒ A = \(\frac{16}{3}\)
Put x = -1, 1 = A(0) + B(-3)=B = \(\frac{1}{-3}\)
1 2x+3x+2-21 16 1 x – x – 2 =2+3(x-2) 3(x+1)

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

∴ 2x – 3 = A[x -1) (2x + 3) +B(x) (2x + 3) + C(x) (x – 1)
Put x = 0 -3 = A(-1) (3) + B(0) + C(0)
-3 = -3A ⇒ A = 1
Put x = 1, -1 = A(0) + B(1) (2 + 3) + C(0)
-1 = 5B ⇒ B = \(-\frac{1}{5}\)
Compare the co-efficient of x² both sides

Students can Download Basic Maths Exercise 19.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.2

Part-A

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Two of Three Marks Questions and Answers.

Question 1.
Find whether the following functions are increasing or decreasing or neither.
(i) f(x) = x⁴ – 8x³ + 22x² – 24x + 5 at x = 0, -2
(ii) f(x) = 4x³ – 15x² + 12x – 2 at x = 1,-1
(iii) f(x) = (x – 1)(x – 2)² at x = 1,3.
Answer:
(i) Given
f(x) = x⁴ – 8x³ + 22x² – 24x + 5 at x = 0, -2
f'(x) = 4x³ – 24x² + 44x – 24
At x = 0, f'(0) = -24 < 0 ∴ f(x) is decreasing at x = 0
At x = -2, f'(-2) = 4(-2)³ – 24(-2)² + 44 (-2) – 24 < 0
= 32 – 96 – 88 – 24 <0 (negative)
∴ f(x) is decreasing at x = -2

(ii) Given
f(x) = 4x³ – 15x² + 12x – 2 at x = 1,-1
f'(x) = 12x² – 30x + 12
At x = 1, f'(x) = 12 – 30 + 12 = -6 < 0
∴ f(x) is increasing at x = 1
At x = -1, f'(-1) = 12(-1)² – 30 (-1) + 12 = 54 > 0
∴ f(x) is increasing at x = -1

(iii) Given
f(x] = (x – 1) (x – 2)² at x = 1, 3
f'(x) = (x – 1) (2 (x – 2} + (x – 2)^2>)
At x = 1, f'(1) = 0 + (-1)² = 1 > 0
∴ f(x) is increasing at x = 1
At x = 3, f'(3) = 2 × 2(1] + 1 = 5 > 0
∴ f(x) is increasing at x = 3.

Question 2.
Find the value of x (Interval) for which the function is increasing or decreasing.
(i) f(x) = 2x³ – 15x² – 84x + 7
(ii) f(x) = x⁴ – 2x³ + 1
(iii) f(x) = x³ – 3x² + 3x – 100
(iv) f(x) = 2x² – 96x + 5
(v) f(x) = 10 – 6x – 2x²
(vi) f(x) = 2x³ + 9x² + 12x + 20
Answer:
(i) Given
f(x) = 2x³ – 15x² – 84x + 7
f'(x) = 6x² – 30x – 84
= 6(x² – 5x – 14) = 6(x – 7] (x + 2)
f(x) is increasing if f'(x) > 0
(x – 7) (x + 2) > 0
Case – 1
x – 7 > 0 & x + 2 > 0
x > 7 & x > -2
x > 7 ⇒ (7, ∞)

Case – 2:
x – 7 < 0 & x + 2 < 0
x < 7 & x < -2
x < -2 ⇒ (-∞ , -2) U (7,∞)
∴ Interval is (-∞ , -2) U (7,∞)
f(x) is decreasing if f'(x) < 0
(x – 7) (x + 2)
x – 7 < 0 & x + 2 > 0

Case -1:
x < 7 & x > -2
∴ -2 < x < 7

Case – 2 x – 7 > 0 & x + 2 < 0
x > 7 & x < -2 (not possible)

(ii) Given f(x) = x⁴ – 2x³ + 1
f'(x) = 4x³ – 6x²
= 2x² (2x – 3) Here 2x² > 0
f(x) is increasing if f'(x) > 0
⇒ 2x – 3 > 0
x > \(\frac { 3 }{ 2 }\) ⇒ (\(\frac { 3 }{ 2 }\), ∞) is the interval
f(x) is decreasing if f ‘(x) < 0
2x – 3 < 0
x < \(\frac { 3 }{ 2 }\) (-∞, \(\frac { 3 }{ 2 }\)) is the interval`

(iii)
Given f(x) = x³ – 3x² + 3x – 100
f'(x) = 3x² – 6x + 3
= 3(x² -2x + 1)
= 3(x – 1)²
f(x) is increasing
f ‘(x) > o ⇒ (x – 1)² > 0
increasing for all
∵ f'(x) > 0 it will not be decreasing

(iv) Given f(x) = 2x² – 96x + 5
f ‘(x) = 4x – 96 = 4 (x – 24)
f(x) is increasing if f’ (x) > 0
x – 24 > 0
x > 24
f(x) is decreasing if f'(x) < 0
x – 24 < 0
x < 24

(v) f(x) = 10 – 6x – 2x²
f'(x) = -6 -4x = -(6 + 4x) = -(6 + 4x)
f(x) is increasing if -(6 + 4x) > 0
⇒ 6 + 4x < 0
⇒ 4x < -6
⇒ x > –\(\frac { 3 }{ 2 }\)
f(x) is decreasing : if -(6 + 4x) < 0 ⇒ 6 + 4x > 0 ⇒ 4x > -6 ⇒ x > –\(\frac { 3 }{ 2 }\)

(vi) Given f(x) = 2x³ + 9x² + 12x + 20
f'(x) = 6x² + 18x + 12
= 6(x² + 3x + 2)
= 6(x + 2) (x + 1)
f(x) is increasing if f ‘(x) > 0
(x + 1) (x + 2) > 0

Case 1:
x + 2 > 0 and x + 1 > 0
x > -2 and x > -1
⇒ x > -1 ⇒ (-1, ∞)

Case 2:
x + 2 < o & x + 1 < 0
x < -2 & x < -1
x < -2 ⇒ (-∞, -2)
f(x) is decreasing if f ‘(x) < 0
(x + 2) (x + 1) < 0

Case 1:
x + 2 < 0 and x + 1 > 0
x < -2 and x > -1
Not possible

Case – 2:
x + 2 > 0 and x + 1 < 0
x > -2 and x < -1
x > -2 and x < -1 ⇒ -2 < x < -1.

Students can Download Basic Maths Exercise 19.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1

Part-A

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Two Marks Questions and Answers

Question 1.
The displacement ‘s’ of a particle at time ‘t’ is given by S = 4t³ – 6t² + t – 7. Find the velocity and acceleration when t = 2 sec.
Answer:
Given S = 4t₃ – 6t + t – 7
Velocity = v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12t² -12t + 1
At t = 2secs, v = 12(2)² – 12(2) + 1 = 48 – 24 + 1 = 25 units/sec.
At t = 2 sec, acceleration = 24.2 – 12 = 48 – 12 = 36 units/sec².

Question 2.
If S = 5t² + 4t – 8. Find the initial velocity and acceleration, (s = displacement, t = time).
Answer:
Given s = 5t² + 4t – 8
V = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 10t + 4 dt
Initial velocity is velocity when t = 0
i.e., = 10.0 + 4 = 4 units/sec.
Acceleration = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 10 units /sec².

Question 3.
A stone thrown vertically upward rises ‘s’ ft. in ‘t’ sec. where s = 80t – 16t². What its velocity after 2 sec.? Find the acceleration?
Answer:
Given S = 80t -16t²
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 80 – 32t
At t=2 sec, v = 80 – 32 (2)
= 80 – 64 = 16 ft./sec.
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = -22 ft/sec² .

Question 4.
A body is thrown vertically upwards its distance S feet is’t’ sec. is given by S = 5 + 12t – t². Find the greatest highest by the body.
Answer:
Given s = 5 + 12t – t²
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12 – 2t dt
Maximum height ⇒ Kinetic energy = 0 ⇒ v = 0 ⇒ 12 – 2t = 0 ⇒ t = 6 sec
∴ Greatest height = s = 5 + 12.6 – 6²
= 5 + 72 – 36 = 77 – 36 = 41 feet.

Question 5.
If v = \(\sqrt{s^{2}+1}\) prove that acceleration is ‘S’ (V = velocity, S = displacement).
Answer:
Given v = \(\sqrt{s^{2}+1}\)
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = \(\frac{1}{2 \sqrt{s^{2}+1}} \cdot 2 s \frac{d s}{d t}=\frac{1}{2 v}\) . 2 . s . v . s units / sec².

Question 6.
If S = at³ + bt. Find a and b given that when t = 3 velocity is ‘O’ and the acceleration is 14 unit. (S = displacement, t = time).
Answer:
Given s = at³ + bt; v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3at²+b
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 6at
When v = 0 then 3at² + b = 0 ⇒ 27a + b = 0.
t = 3 sec
When acceleration = 14 then 14 = 6a.3, b = -27a = \(-\frac{27.7}{9}\) = -21
F = 3 sec. a = \(\frac{14}{18}=\frac{7}{9}\)
∴ a = \(\frac { 7 }{ 9 }\) and b = -21.

Question 7.
When the brakes are applied to moving car, the car travels a distance ‘s’ ft. in ‘t’ see given by s = 8t – 6t² when does the car stop?
Answer:
Given s = 8t – 6t²
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 8 – 12t car stops when v = 0
∴ 8 – 12t = 0 ⇒ t = \(\frac{8}{12}=\frac{2}{3}\) sec.

Part-B

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Three Marks Questions and Answers.

Question 1.
The radius of sphere is increasing at the rate of 0.5 mt/sec. Find the rate of increase of its surface area and volume after 3 sec.
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.5 , t = 3 sec, \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = ? \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = ?
surface area = s = 4πr^{2
\(\frac{d s}{d t}\) = 4π . 2r . \(\frac{d r}{d t}\)}
= 4π × 2 × 1.5 × 0.5
= 6π m²/sec
dr = 0.5 × dt
⇒ r = 0.5 t
= 0.5 × 3
= 1.5

Question 2.
The surface area of a spherical bubble is increasing at the rate of a 0.8cm² / sec. Find at what rate is its volume increasing when r = .25cm [r = radius of the sphere].
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.8 cm² / sec. r = 2.5 cm, \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = ?
s = 4πr²
\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π . 2r × \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
0.8 = 4π × 2 × 2.5. \(\frac{\mathrm{dr}}{\mathrm{dt}}\) ⇒ \(\frac{0.8}{8 \pi \times 2.5}=\frac{0.1}{2.5 \pi}\)
v = \(\frac { 4 }{ 3 }\)πr³
\(\frac{d v}{d t}=\frac{4}{3} \pi 3 r^{2} \cdot \frac{d r}{d t}=4 \pi \times(2.5)^{2} \times \frac{0.1}{2.5 \pi}=1 \mathrm{cc} / \mathrm{sec}\)

Question 3.
A spherical balloon is being inflated at the rate 35cc/sec. Find the rate at which the surface area of the balloon increases when its diameter is 14cm.
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 35cc /sec 2r = 14 ⇒ r = 7, \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = ?, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = ?
v = \(\frac { 4 }{ 3 }\)πr³ s = 4πr²
\(\frac{d V}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t} \quad \frac{d S}{d t}=4 \pi 2 r \frac{d r}{d t}\)
35 = 4π . 7² \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π × 2 × 7 × \(\frac{5}{28 \pi}\)
\(\frac{d r}{d t}=\frac{35}{196 \pi}=\frac{5}{28 \pi}=10 \mathrm{cm}^{2} / \mathrm{sec}\)

Question 4.
The radius of a circular plate is increasing at the rate of \(\frac{2}{3 \pi}\) cm/sec. Find the rate of change of its area when the radius is 6cm.
Answer:
Given \(\frac{d r}{d t}=\frac{2}{3 \pi} r=6 \mathrm{cm} \frac{d A}{d t}=?\)
A = πr²

Question 5.
A circular patch of oil spreads on water the area growing at the rate of 16cm²/min. How fast are radius and the circumference increasing when the diameter is 12cm.
Answer:
Given \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 16cm²/min, d = 2r = 12cm, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = ? \(\frac{\mathrm{dc}}{\mathrm{dt}}\) = ? r = 6cm
A = πr² c = 2πr
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π . 2r . \(\frac{\mathrm{dr}}{\mathrm{dt}}\) \(\frac{\mathrm{dc}}{\mathrm{dt}}\) = 2π \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
16 = 2π .6 . \(\frac{\mathrm{dr}}{\mathrm{dt}}\) 2π . \(\frac{4}{3 \pi}=\frac{8}{3}\)cm / min
⇒ \(\frac{d r}{d t}=\frac{16}{12 \pi}=\frac{4}{3 \pi} \mathrm{cm} / \mathrm{min}\)

Question 6.
A stone is dropped into a pond waved in the form of circles are generated and the radius of the outer most ripple increases at the rate 2 inches/sec. How fast is the area increasing when the (a) radius is 5 inches (b) after 5 sec.?
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2inch/sec, r = 5 inch, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
(a) A = πr²
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π2r. \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = π × 2 × 5 × 2 = 20π sq. inches / sec.

(b) After 5 sec, dr = 2dt
r = 2t
⇒ when t = 5, r = 10 inches
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π . 2r \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = π × 2 × 20 × 2 = 40π square inches/sec.

Question 7.
The side of an equilateral triangle is increasing at the rate \(\sqrt{3}\) cm./sec. Find the rate at which its area is increasing when its side is 2 meters.
Answer:
Given \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\sqrt{3}\) cm/sec., x = 2 meters, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
Area of equilateral Δ^le = A = \(\frac{\sqrt{3}}{4}\) x²
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \frac{d x}{d t}\)
= \(\frac{\sqrt{3}}{4}\) . 2. 200. \(\sqrt{3}\) = 300cm² / sec.

Question 8.
Water is being poured at the rate of 30 mt³/min. into a cylindrical vessel whose base is a circle of radius 3 mt. Find the rate at which the level of water is rising?
Answer:
Given r = 3mts, \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 30 m³/min, \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = ?
V = πr²h, r = constant
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = π . (3)² . \(\frac{\mathrm{dh}}{\mathrm{dt}}\)
30 = 9π \(\frac{\mathrm{dh}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = \(\frac{10}{3 \pi}\) meter/min.

Question 9.
Sand is being dropped at the rate of 10 mt³/sec. into a conical pile. If the height of the pile twice the radius of the base, at what rate is the height to the pile is increasing when the sand in the pile is 8mt high.
Given
\(\frac{d v}{d t}\) = 10m³/sec, h = 2r, h = 8, \(\frac{d h}{d t}\) = ?
r = \(\frac{\mathrm{h}}{2}\) \(\mathrm{v}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{h}=\frac{1}{3} \pi \cdot\left(\frac{\mathrm{h}}{2}\right)^{2} \cdot \mathrm{h}=\frac{1}{3} \pi \frac{\mathrm{h}^{3}}{4}\)
\(\mathrm{v}=\frac{1}{12} \pi \mathrm{h}^{3} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\pi}{12} \cdot 3 \mathrm{h}^{2} \frac{\mathrm{dh}}{\mathrm{dt}} ; \quad 10=\frac{\pi}{4} \cdot 8^{2} \cdot \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(\frac{d h}{d t}=\frac{40}{64 \pi}=\frac{5}{8 \pi} \mathrm{m} / \mathrm{sec}\)

Question 10.
A ladder of 15ft. long leans against a smooth vertical wall. If the top slides downwards at the rate of 2ft sec. Find how fast the lower and is moving when the lower end is 12ft. from the wall.
Answer:

Given \(\frac{\mathrm{d} y}{\mathrm{dt}}\) = 2ft /sec, x = 12
From fig, x² + y² = 15²
12² + y² = 15²
y² = 15² – 12²
y = \(\sqrt{225-144}\)
y = \(\sqrt{81}\) = 9
x² + y² = 15² ⇒ 2x \(\frac{\mathrm{dx}}{\mathrm{dt}}\) + 2y\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 0; 2.12.\(\frac{\mathrm{dx}}{\mathrm{dt}}\) + 2 .9 . 2 = 0
⇒ \(\frac{d x}{d t}=\frac{-36}{24}=\frac{-3}{2} f t / \sec\)

Question 11.
An edge of a variable cube is increasing at the rate of 10cm/sec. How fast the volume and also its surface area is increasing when the edge is 5cm long.
Answer:
Given \(\frac{d x}{d t}\) = 10 cm/sec, x = 5 cm, \(\frac{d v}{d t}\) = ? \(\frac{d s}{d t}\) = ?
(i) V = x³
\(\frac{d v}{d t}\) = 3x² \(\frac{d x}{d t}\) = 3 × (5²) × 10 = 750 cm³/sec.

(ii) S = 6x² .
\(\frac{d s}{d t}\) = 12 × .\(\frac{d x}{d t}\) = 12.5 .10 = 600 cm²/sec.

Question 12.
A man 6ft. tall is moving directly away from a lamp post of height 10ft. above the ground. If he is moving at the rate 3ft./sec. Find the rate at which the length of his shadow is increasing and also the tip of his shadow is moving?
Answer:

Let the shadow be x & y be the distance the man walks
Given \(\frac{d y}{d t}\) = 3ft/sec From a similar Δ^les we have
\(\frac{6}{10}=\frac{x}{x+y}\)
6x + 6y = 10x
6y = 4x ⇒ 3y = 2x
⇒ \(3 \frac{d y}{d t}=2 \frac{d x}{d t} \Rightarrow 2 \frac{d x}{d t}=3.3=9\)
∴ the shadow is increasing at the rate.
∴ \(\frac{d x}{d t}=\frac{9}{2}\) ft/sec & the tip of the shadow moves is
\(\frac{d x}{d t}+\frac{d y}{d t}=\frac{9+6}{2}=\frac{15}{2} \mathrm{ft} / \mathrm{sec}\)

Question 13.
The height of circular cone is 30 cm. and it is constant. The radius of the base is increasing at the rate of 0.25cm/sec. Find the rate of increase of volume of the cone when the radius of base is 10cm.
Answer:
dr
Given h = 30 cm, \(\frac{d r}{d t}\) = 0.25cm/sec. r = 10cm. dt
V = \(\frac { 1 }{ 3 }\) πr²h
\(\frac{d v}{d t}\) = \(\frac{\pi}{3}\)h.2r. \(\frac{d r}{d t}\) = π. \(\frac { 30 }{ 3 }\) . 20.(0.25) = 50π cm² / sec

Question 14.
The volume of a spherical ball in increasing at the rate 4πcc/sec. Find the rate of increase of the radius of the ball when the volume is 288πCC.
Answer:
Given V = 288π C.C., \(\frac{d v}{d t}\) = 4πcc/ sec \(\frac{d r}{d t}\) = ?
V = \(\frac { 4 }{ 3 }\) πr³

r = 6cm
\(\frac{d v}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t} 4 \pi=\frac{4}{3} \pi \times 3 \times 36 \times \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{36} \mathrm{cm} / \mathrm{sec}\)

Question 15.
A drop of ink spreads over a blotting paper so that the circumferences of the blot is 4πcm and it changes 3cm/sec. Find the rate of increase of its radius and also find the rate of increase of its area?
Answer:
Given c = 4π, \(\frac{d C}{d t}\) = 3cm / sec \(\frac{d A}{d t}\) = ? \(\frac{d r}{d t}\) = ?
Circumference = c = 2πr
4π = 2πr ⇒ r = 2
Again
C = 2πr & A = πr²
\(\frac{d c}{d t}\) = 2π . \(\frac{d r}{d t}\) \(\frac{d A}{d t}\) = π . 2r. \(\frac{d r}{d t}\)
3 = 2π . \(\frac{d r}{d t}\) = π . 2. 2. \(\frac{3}{2 \pi}\)
⇒ \(\frac{d r}{d t}\) = \(\frac{3}{2 \pi}\) cm/ sec \(\frac{d A}{d t}\) = 6cm² / sec

Question 16.
A circular plate of metal is heated so that its radius increase at the rate of O.lmm/min. At what rate is the [plate’s area increasing when the radius is 25cm [1cm = 10mm].
Answer:
Given \(\frac{d r}{d t}\) = 0.1 mm/min, r = 25 cm, \(\frac{d A}{d t}\) A = πr²
\(\frac{d A}{d t}\) = π. 2r . \(\frac{d r}{d t}\) = π . 2. /250 (0.1) = 50πmm² /min.

Question 17.
The surface area of a spherical soap bubble increasing at the rate of 0.6cm²/sec. Find the rate at which its volume is increasing when its radius is 3cm.
Answer:
Given \(\frac{ds}{d t}\) = 0.6 cm² / sec, r = 3cm, \(\frac{d v}{d t}\) = ?
s = 4 πr² &
\(\frac{d s}{d t}\) = 4π. 2r. \(\frac{d r}{d t}\)
0.6 = 4π × 3 × 3 × \(\frac{d r}{d t}\)
∴ \(\frac{d r}{d t}=\frac{0.6}{6 \times 4 \pi}=\frac{0.1}{4 \pi} \mathrm{cm} / \mathrm{sec}\)
\(\frac { 1 }{ 40 }\) πcm/sec.
v = \(\frac { 4 }{ 3 }\) πr³
\(\frac{d v}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}\)
= 4πr² \(\frac{d r}{d t}\)
= 4π(3)² . \(\frac{0.1}{4 \pi}\)
= 0.9 cm³ / sec.

Question 18.
A rod 13 feet long slides with it end A and B as two straight lines at right angles which meet at ‘O’. If A is moved away from O with a uniform speed at 4ft./sec., find the speed of the end B move when A is 5 feet from O.
Answer:

From fig we have
x² + y² = 13²
y² = 13² – 15² = 144
y = \(\sqrt{144}\) = 12
Als0
x² + y² = 13² ⇒ 2x \(\frac{d x}{d t}\) + 2y\(\frac{d y}{d t}\) = 0
\(5 \times 4=-12 \frac{\mathrm{dy}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{20}{12}=\frac{-5}{3}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-5}{3} \mathrm{ft.} / \mathrm{sec}\)

Question 19.
A street lamp is hung 12 feet above a straight horizontal floor on which a man of 5 feet is walking how fast his shadow lengthening when he is walking away from the lamp post at the rate of 175ft./min.
Answer:

Let the shadow be y & the distance from the man walks is x.
Given \(\frac{d x}{d t}\) = 175 ft/min.
From figure we have
\(\frac{12}{5}=\frac{x+y}{y}\) ⇒ 12y = 5x + 5y ⇒ 7y = 5x
⇒ \(\frac{7 \mathrm{dy}}{\mathrm{dt}}=5 \frac{\mathrm{dx}}{\mathrm{dt}}\)
∴ the shadow is lengthening ⇒ \(\frac{d y}{d t}=\frac{5}{7} \times 175\) = 125 ft/ min.

Question 20.
Find a point on the parabola y² = 4x at which the ordinate increases at twice the rate of the abscissa [Ordinate = y, abscissa = x].
Answer:
Given y² = 4x diff. w.r.t. x
2y \(\frac{d y}{d x}\) = 4 \(\frac{d x}{d t}\)
Also given \(\frac{d y}{d x}\) = 2. \(\frac{d x}{d t}\) ⇒ 2y .2 \(\frac{d x}{d t}\) = 4. \(\frac{d x}{d t}\)
⇒ y = 1 ⇒ 1² = 4x
⇒ x = \(\frac { 1 }{ 4 }\)
∴ the point on the parabola is (\(\frac { 1 }{ 4 }\), 1 ).