2nd PUC

Students can Download Basic Maths Exercise 18.4 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 18.4

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.4 Three Marks Questions and Answers

Question 1.
If y = \(y=\sqrt{x+\sqrt{x+\sqrt{x \ldots \ldots \infty}}}\) then prove that \(\frac{d y}{d x}=\frac{1}{2 y-1}\)
Answer:
Given \(y=\sqrt{x+\sqrt{x+\sqrt{x \ldots \ldots \infty}}}\)
⇒ y = \(\sqrt{x+y}\) S.R.S
y² = x + y diff w.r.t x
\(2 y \frac{d y}{d x}=1+\frac{d y}{d x} ; \quad \frac{d y}{d x}(2 y-1)=1\)
⇒ \(\frac{d y}{d x}=\frac{1}{2 y-1}\)

Question 2.
If y = \(\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots \ldots \infty}}\) then prove that \(\frac{d y}{d x}=\frac{\sec ^{2} x}{2 y-1}\)
Answer:
Given y = \(\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots \ldots \infty}}\)
y = \(\sqrt{\tan x+y}\) S.B.S
y² = tan x + y Diff w.r.t x
2y \(\frac{d y}{d x}\) = sec²x; \(\frac{d y}{d x}\) (2y – 1) = sec²x
\(\frac{d y}{d x}=\frac{\sec ^{2} x}{2 y-1}\)

Question 3.
If \(y=(\sin x)^{(\sin x)} \cdot \cdot^{\cdot \infty}\) , show that \(\frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log (\sin x)}\)
Answer:
Given \(y=(\sin x)^{(\sin x)} \cdot \cdot^{\cdot \infty}\)
y = (sin x)^y taking log^m both sides
log y = y log (sin x) diff w.r.t x

Question 4.
If y = (tan x)^(tanx)(tanx), show that \(\frac{d y}{d x}=\frac{2 y^{2} \cdot \csc x}{1-y \log (\tan x)}\)
Answer:
Given y = (tan x)^(tanx)(tanx)
⇒ y = (tan x)^y, taking log on both sides
log y = y log (tan x) diff w.r.t, x

Question 5.
If y = (e^x)^{(ex) …… ∞} , show that \(\frac{d y}{d x}=\frac{y^{2}}{1-x y}\)
Answer:
Given y = (e^x)^{(ex) …… ∞}
⇒ y = (e^x)^y taking log^m
⇒ log y = xy log e, diff w.r. t x

Students can Download Maths Chapter 7 Integrals Ex 7.3 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 7 Integrals Ex 7.3

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.3

Find the integrals of the functions in Exer­cises 1 to 22:

Question 1.
sin² (2x + 5)
Answer:

Question 2.
sin 3x cos 4x
Answer:

Question 3.
cos 2x cos 4x cos 6x
Answer:

Question 4.
Sin³ (2x + 1)
Answer:

Question 5.
sin³ x cos³ x
Answer:

Question 6.
sin x sin 2x sin 3x
Answer:

Question 7.
sin x sin 2x sin 3x
Answer:

Question 8.
\(\frac{1-\cos x}{1+\cos x}\)
Answer:

Question 9.
\(\frac{cos x}{1+\cos x}\)
Answer:

Question 10.
sin⁴ x
Answer:

Question 11.
cos⁴ 2x
Answer:

Question 12.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Answer:

Question 13.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Answer:

Question 14.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Answer:

Question 15.
tan³ 2x sec 2x
Answer:

Question 16.
tan⁴x
Answer:

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

Question 18.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Answer:

Question 19.
\(\frac{1}{\sin x \cos ^{3} x}\)
Answer:

Question 20.
\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\)
Answer:

Question 21.
sin^-1 (cos x)
Answer:

Question 22.
\(\frac{1}{\cos (x-a) \cos (x-b)}\)
Answer:

Question 23.
\(\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x \text { is equal to }\)
(A) tan x + cot x + C
(B) tan x + cosec + C
(C) – tan x + cot x + C
(D) tan x + sec x + C
Answer:

Question 24.
\(\int \frac{\mathbf{e}^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x \text { is equals to }\)
(A) – cot (ex^x) + C
(B) tan (xe^x) + C
(C) tan (e^x) + C
(D) cot (e^x) + C
Answer:
= ∫ sec² (xe^x). e^x (1 + x) dx = tan (xe^x) + C
Answer (B)

Students can Download Maths Chapter 7 Integrals Ex 7.2 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 7 Integrals Ex 7.2

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.2

Integrate the function in Exercises 1 to 37:

Question 1.
\(\cfrac{2 x}{1+x^{2}}\)
Answer:

Question 2.
\(\cfrac{(\log x)^{2}}{x}\)
Answer:

Question 3.
\(\cfrac{1}{x+x \log x}\)
Answer:

Question 4.
sin x sin (cos x)
Answer:
∫ sinx sin (cosx) dx
cosx = t ⇒ – sinx . dx = dt
∫- sin t. dt = cos t + C = cos (cos x) + C

Question 5.
sin (ax + b) cos (ax + b)
Answer:

Question 6.
\(\sqrt{a x+b}\)
Answer:

Question 7.
\(x \sqrt{x+2}\)
Answer:

Question 8.
\(x \sqrt{1+2 x^{2}}\)
Answer:

Question 9.
\((4 x+2) \sqrt{x^{2}+x+1}\)
Answer:

Question 10.
\(\frac{1}{x-\sqrt{x}}\)
Answer:

Question 11.
\(\frac{\mathbf{x}}{\sqrt{\mathbf{x}+4}}, \mathbf{x}>0\)
Answer:

Question 12.
\(\frac{\mathbf{x}}{\sqrt{\mathbf{x}+4}}, \mathbf{x}>0\)
Answer:

Question 13.
\(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\)
Answer:

Question 14.
\(\frac{1}{\mathbf{x}(\log \mathbf{x})^{\mathrm{m}}} \mathbf{x}>0\)
Answer:

Question 15.
\(\frac{x}{9-4 x^{2}}\)
Answer:

Question 16.
e^{2x + 3}
Answer:

Question 17.
\(\frac{\mathbf{x}}{\mathbf{c}^{x^{2}}}\)
Answer:

Question 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
Answer:

Question 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Answer:

Question 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Answer:

Question 21.
tan² (2x – 3)
Answer:

Question 22.
sec² (7 – 4x)
Answer:

Question 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Answer:

Question 24.
\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Answer:

Question 25.
\(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\)
Answer:

 

Question 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Answer:

Question 27.
\(\sqrt{\sin 2 x} \cos 2 x\)
Answer:

Question 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
Answer:

Question 29.
cot x log sin x
Answer:

Question 30.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
Answer:

Question 31.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
Answer:

Question 32.
\(\frac{1}{1+\cot x}\)
Answer:

Question 33.
\(\frac{1}{1-\tan x}\)
Answer:

Question 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Answer:

Question 35.
\(\frac{(1+\log x)^{2}}{x}\)
Answer:

Question 36.
\(\frac{(x+1)(x+\log x)^{2}}{x}\)
Answer:

Question 37.
\(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\)
Answer:

Question 38.
\(\int \frac { 10x^{ 9 }+10^{ x }\log _{ x^{ 10 } } dx }{ x^{ 10 }+10^{ x } } { equals }\)
Answer:

Question 39.
\(\int \frac{d x}{\sin ^{2} x \cos ^{2} x} \text { equals }\)
Answer:

Students can Download Maths Chapter 7 Integrals Ex 7.1 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 7 Integrals Ex 7.1

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.1

Find an anti derivative (of integral) of the following functions by the method of inspection.

Question 1.
sin 2x
Answer:

Question 2.
cos 3x
Answer:

Question 3.
e^2x
Answer:

Question 4.
(ax + b)²
Answer:

Question 5.
sin 2x – 4 e^3x

Find the following integrals in Exercises 6 to 20:

Question 6.
(4e^3x + 1)dx
Answer:

Question 7.
\(\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x\)
Answer:

Question 8.
∫(ax² + bx+c)dx
Answer:

Question 9.
∫(2x²+e^x)dx
Answer:

Question 10.
\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x\)
Answer:

Question 11.
\(\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x\)
Answer:

Question 12.
\(\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x\)
Answer:

Question 13.
\(\int \frac{x^{3}-x^{2}+x-1}{x-1} d x\)
Answer:

Question 14.
\(\int(1 – x) \sqrt{x} d x\)
Answer:

Question 15.
\(\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x\)
Answer:

Question 16.
∫(2x – 3cosx + e^x)dx
Answer:

Question 17.
\(\int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x\)
Answer:

Question 18.
∫sec x(sec x + tan x)dx
Answer:
∫sec x (sec x + tan x) dx
= ∫(sec²x+sec x tan x) dx
= tan x + sec x + C

Question 19.
\(\int \frac { \sec ^{ 2 } x }{ { cosec }^{ 2 }x } dx\)
Answer:

Question 20.
\(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\)
Answer:
∫(2sec² x – 3sec x tan x)dx
= 2 tan x – 3 sec x + C

Choose the correct answer in Exercises 21 and 22.

Question 21.
The anti derivative of \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) equals
(A) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)
(B) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+C\)
(C) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
(D) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Answer:

Question 22.
\(\frac{d}{d x} f(x)=4 x^{3} \cdot \frac{3}{x^{4}}\) such that f(2) Then f(x) is ………
(A)\(x^{4}+\frac{1}{x^{3}}-\frac{129}{8}\)
(B)\(x^{3}+\frac{1}{x^{4}}+\frac{129}{8}\)
(C)\(x^{4}+\frac{1}{x^{3}}+\frac{129}{8}\)
(D)\(x^{3}+\frac{1}{x^{4}}-\frac{129}{8}\)
Answer:

Students can Download Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 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 Miscellaneous Exercise

Question 1.
Using differentials, find the approximate value of each of the following

(a) \(\left(\frac{17}{81}\right)^{1 / 4}\)
Answer:

(b) (33)^-1/5
Answer:

Question 2.
Show that the function given by log has maximum at \(f(x)=\frac{\log x}{x}\) has maximum at x = e
Answer:

Question 3.
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?
Answer:

Question 4.
Find the equation of the normal to curve y² = 4x which passes through the point (1, 2).
Answer:

Question 5.
Show that the normal at any point 6, to the curve x = a cos θ +a θ sin θ ,y = a sin θ –
a θ cos θ is at a constant distance from the origin.
Answer:

∴ equation of normal is
y – (a sin θ – a θ cos θ)
= – cot θ (x – a cos θ – a θ sin θ)
y + x cot θ= a cot θ cos θ+a θ cot θ sin θ – a θ cos θ
\(y+\frac{x \cos \theta}{\sin \theta}=\frac{a \cos ^{2} \theta}{\sin \theta}+\frac{a \sin ^{2} \theta}{\sin \theta}\)
x cos θ a cos²θ, a sin²θ
normal form is x cos θ+ y sin θ = p when p is the normal from the normal to the given curve is at a consistent distance ‘a’ from the origin.

Question 6.
Find the intervals in which the function f given by
\(f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\)
(i) increasing
(ii) decreasing
Answer:

Question 7.
Find the intervals in which the function f given by
x³ + -1/x³, x ≠ 0
(i) increasing
(ii) decreasing
Answer:

Question 8.
Find the maximum area of an isosceles triangle inscribed in the ellipse
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with its vertex at one end of the major axis.
Answer:

Question 9.
‘A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq metres for the base and ₹ 45 per square metre for sides. What is the cost of least expensive tank?
Answer:

Question 10.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Answer:

Question 11.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Answer:

Question 12.
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is
\(\left(a^{\frac{3}{2}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\)
Answer:

Question 13.
Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Answer:
f (x) = (x – 2)⁴ (x + 1)³
f’ (x) = (x – 2)⁴ 3 (x + 1)² + (x + 1)³ 4(x – 2)³
= (x – 2)³ (x + 1) [3 (x -2) + 4 (x + 1)]
= (x – 2)³ (x + 1)² [3x – 6 + 4x + 4]
= (x – 2)³ (x + 1)² [7x – 2]
f’ (x) = 0 ⇒ x = 2, x = -1, x = 2/7
f’ (x)=(x – 2)³ (x+1)² x 7 + (x – 2)³ (7x – 2) 2 (x +1) + (x + 1)² (7x – 2)3 (x – 2)²
here second demivalive test fails as f” (x) = 0 then we apply first demivalive test
At x = 2 f'(x) changes from negative to positive.
∴ function has minimum value and the local minimum value is f (2) = 0
At x = -1
f’ (x) does not change (positive to negative)
∴ f^x has neither maximum value nor minimum value.
∴ x = -1 is the point of inflection at x = 2/7
f’ (x) change from positive to negative
hence function has been maximum at x = 2/7 and local maximum value is

Question 14.
Find the absolute maximum and minimum values of the function f given by
f (x) = cos² x + sin x, x ∈ [o,π]
Answer:

Question 15.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\)
Answer:
Let the radius of sphere be ‘r’

Question 16.
Let f be a function defined on [a, b] such that f’ (x) > 0, for all xe (a, b). Then prove that f is an increasing function on (a, b).
Answer:
Let x₁, x₂ be any two real number [a, b] such that x₁ <  x₂, then f (x) satisfies the conditions of L.M.V theorem. ∋ C ∈  (a, b)

Question 17.
Show that the height of the cylinder of maximum volume that can be inscribed in sphere of radius R is \(\frac{2 \mathbf{R}}{\sqrt{3}}\). Also find the maximum volume.
Answer:
Let r be the radius of the cylinder and height 2x

Question 18.
Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and α the greatest volume of cylinder is \(\frac{4}{27} \pi \mathrm{h}^{3}\)
Answer:
Height of cone – h
Radius of cone – r
Height of cylinder – y
Radius of cylinder – x



Choose the correct answer in the Exercises from 19 to 24.

Question 19.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m³/h
(B) 0.1 m³h
(C) 1.1 m³/h
(D) 0.5 m³/h
Answer:

Question 20.
The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2,-1) is
(A) \(\frac{22}{7}\)
(B) \(\frac{6}{7}\)
(C) \(\frac{7}{6}\)
(D)\(\frac{-6}{7}\)
Answer:

Question 21.
The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is  ………..
(A) 1
(B) 2
(C) 3
(D) \(\frac{1}{2}\)
Answer:

Question 22.
The normal at the point (1,1) on the curve 2y + x² = 3 is ……………….
(A) x + y = 0
(B) x – y = 0
(C) x + y = 1
(D) x – y = 1
Answer:

Question 23.
The normal to the curve x² = 4y passing (1,2) is
(A) x + y = 3
(B) x – y = 3
(C) x + y = 1
(D) x – y = 1
Answer:

Question 24.
The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are ………………
(A) \(\left(4, \pm \frac{8}{3}\right)\)
(B) \(\left(4, \frac{-8}{3}\right)\)
(C) \(\left(4,+\frac{3}{8}\right)\)
(D) \(\left(\pm 4, \frac{8}{3}\right)\)
Answer:

2nd PUC Maths Application of Derivatives Miscellaneous Exercise Additional Questions and Answers

Question 1.
If the length of the three sides of a trapezium other than the base is 10cm, find the area of the trapezium when it is maximum (CBSE 2010)
Answer:

Question 2.
Find the intervals in which the function sinx + cos x x ∈ [0,π] is (1) strictly increasing (2) strictly decreasing. (CBSE 2010)
Answer:

Question 3.
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, find the apps error in. Calculating its surface area, (CBSE 2011)
Answer:

Question 4.
Find the relationship between a and b so that the function of defined by
\(f(x)=\left\{\begin{array}{l}{a x+1, \text { If } x \leq 3 \text { is continous at } x=3} \\{b x+3 \text { if } x>3} \end{array}\right.\)
Answer:

Question 5.
An open box with a square box is to be made out of a given quantity of sheet of area a². Show that the maximum volume of the box is \(a^{3} / 6 \sqrt{3}\) (PUC 2011)
Answer:

Question 6.
Show that the rectangle of maximum perimeter which can be insenibed in a circle of radius a is a square of side \(\sqrt{2}\) a. (CBSE 2008)
Answer:

Question 7.
Discuss the continuity of
\(f(x)=\left\{\begin{array}{cl}{\frac{1-\cos x}{x^{2}}} & {x \neq 0} \\{1} & {x=0} \end{array}\right.\) (KPUC 2008)
Answer:

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 6.3 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.