2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.7

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Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.7

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.7 One Mark Questions and Answers

Question 1.
Find \(\frac{d^{2} y}{d x^{2}}\).
1. y = 3x³ + 4x² + 7
2. y = \(\sqrt{2 x+3}\)
3. y = e^{3x + 2}
4. y = x³ . logx
5. y = log x + a^x
6. y = e^-x sin 2x
7. y = log(log x)
8. y = cos 4x cos 2x
9. y = sin 3x sin 2x
10. y = cos mx sin nx
Answer:
1. Given y = 3x³ + 4x² + 7
\(\frac{d y}{d x}\) = 9x² + 8x
\(\frac{d^{2} y}{d x^{2}}\) = 18x + 8

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

3. y = e^{3x + 2}
\(\frac{d y}{d x}\) = 3e^{3x + 2}
\(\frac{d^{2} y}{d x^{2}}\) = 9e^{3x + 2} = 9y

4. y = x³ logx
\(\frac{d y}{d x}\) = x³ . \(\frac { 1 }{ x }\) + log x . 3x² = x² + 3x² log x
\(\frac{d^{2} y}{d x^{2}}\) = 2x + 3x² \(\frac { 1 }{ x }\) + log x . 6x = 5x + 6x log x

5. y = log x + a^x
\(\frac{d y}{d x}\) = \(\frac { 1 }{ x }\) + a^x . log a
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{-1}{x^{2}}\) + log a . a^xlog a
= \(\frac{-1}{x^{2}}\) + a^x(log a)²

6. Given y = e^-x sin 2x
\(\frac{d y}{d x}\) = e^-x(2 cos2x) + sin 2x .(-e^-x)
= e^-x( 2 cos2x – sin2x)
\(\frac{d^{2} y}{d x^{2}}\) = e^-x(-4 sin 2x – 2 cos 2x) + (2 cos 2x – sin 2x) (-e^-x)
= e^-x [-4 sin 2x – 2 cos 2x – 2 cos 2x + sin 2x]
= e^-x (-3 sin2x – 4 cos 2x)

7. y = log(log x)

8. y = cos 4x cpos 2x
y = \(\frac { 1 }{ 2 }\) (cos 6x + cos 2x)
\(\frac{d y}{d x}\) = \(\frac { 1 }{ 2 }\) (-6 sin6x – 2 sin2x) = -3 sin 6x – sin2x
\(\frac{d^{2} y}{d x^{2}}\) = -18 cos 6x – 2 cos 2x

9. Given y = sin 3x sin 2x
y = \(\frac { 1 }{ 2 }\) (cos 5x – cos 7x);
\(\frac{d y}{d x}\) = \(\frac { 1 }{ 2 }\) (-5 sin 5x + 7 sin 7x)
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac { 1 }{ 2 }\) (- 25 cos 5x + 49 cos 7x)

10. y = cosmx.sinnx
y = \(\frac { 1 }{ 2 }\)[sin (m + n)x – sin(m – n)x]
\(\frac{d y}{d x}\) = \(\frac { 1 }{ 2 }\)[(m + n)cos(m – n)x – (m – n)cos (m – n)x]
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac { 1 }{ 2 }\)[-sin(m + n)x .(m + n)² + (m – n)² . sin (m – n)x]
= \(\frac { 1 }{ 2 }\)[-(m + n)² . sin (m + n)x + (m – n)² sin(m – n)x].

Part – B

2nd PUC Basic Maths Differential Calculus Ex 18.7 Two and Three Marks Questions and Answers

Question 1.
Find \(\frac{d^{2} y}{d x^{2}}\) if.
1. x = a cos θ, y = a sin θ
2. x = a(θ + sinθ), y = a(1 – cos θ)
3. x = a cos³t, y = -a sin³t
Answer:
1. Given x = a cos θ, y = a sin θ
Differentiate both W.r.t. θ

2. Given x = a(θ + sinθ), y = a(1 – cos θ)
Differentiate both w.r.t θ

3. Given x = a cos³t, y = -a sin³t
Differentiate bpth w.r.t t

Question 2.
If y = sin mx, show that \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Answer:
Given y = sin mx
\(\frac{d y}{d x}\) = -m cos mx
\(\frac{d^{2} y}{d x^{2}}\) = m² sin mx = -m²y
∴ \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)

Question 3.
If y = 500e^7x + 600e^-7x, show that \(\frac{d^{2} y}{d x^{2}}\) = 49y.
Answer:
y = 500e^7x + 600e^-7x
\(\frac{d y}{d x}\) = 500.7e^7x + 600(-7. e^-7x)
\(\frac{d^{2} y}{d x^{2}}\) = 500(49e^7x) + 600(+49 .e^-7x)
= 49 (500^7x + 600e^-7x) = 49y
∴ \(\frac{d^{2} y}{d x^{2}}\) = 49y

Question 4.
If y = e^ax + e^-ax, Show that y₂ – a²y = 0
Answer:
y = e^ax + e^-ax
y₁ = ae^ax – ae^-ax
Y₂ = a²e^ax + a²e^-ax
= a²(e^ax + e^-ax) = a²y
∴ y₂ – a²y = 0.

Question 5.
If y = 2 + log x, show that xy₂ + y₁ = 0.
Answer:
y_i = \(\frac { 1 }{ x }\) ⇒ xy₁ = 1
∴ = xy₂ + y₁ = 0

Part – C

2nd PUC Basic Maths Differential Calculus Ex 18.7 Five Marks Questions and Answers

Question 1.
If x² – xy + y² = a², show that \(\frac{d^{2} y}{d x^{3}}=\frac{6 a^{2}}{(x-2 y)^{3}}\)
Answer:

Question 2.
If x² + 2xy + 3y² = 1, show that y² = \(\frac{-2}{(x+3 y)^{3}}\)
Answer:
x² + 2xy + 3y² = 1
Differentiate w.r.t x we get
2x + 2(x.y₁ + y.1) + 6y.y₁ = 0
y₁(2x + 6y) = -2x -2y

Hence proved.

Question 3.
If y = a cos mx + b sin mx, show that \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)
Answer:
Given y = a cos mx + b sin mx
\(\frac{d y}{d x}\) = -am sin mx + bm cos mx
\(\frac{d^{2} y}{d x^{2}}\) = -am² cos mx – bm² sin mx
= m²(a cos mx + b sin mx) = -m²y
⇒ \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)

Question 4.
If y = a cos (log x) + b sin (log x), show that x²y₂ + xy₁ + y = 0
Answer:
y = a cos(log x) + b sin (log x)
\(y_{1}=\frac{-a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\)
xy₁ = -a sin (log x) + b cos (log x) Again diff w.r.t x
xy₂ + y₁1 = \(\frac{-a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\)
x²y₂ + xy₁ = -(a cos logx + b sin (log x)) = =y
⇒ x²y₂ + xy₁ + y = 0

Question 5.
If y = log (x – \(\sqrt{x^{2}+1}\)), show that (x² + 1)y₂ + xy₁ = 0
Answer:
If y = log (x – \(\sqrt{x^{2}+1}\))

2y₁ . y₂(1 + x²) + y₁²(2x) = 0 ÷ 2y₁
⇒ (1 + x²)y₂ + xy₁ = 0.

Question 6.
If y = x + \(\sqrt{x^{2} – 1}\) , show that (x² – 1)y₂ + xy₁ – y = 0
Answer:
y = x + \(\sqrt{x^{2} – 1}\) differentiate w.r.t x

∴ (x² – 1) y₁² = y² Differentiate agin w.r.t x
(x² – 1)2y₁y₂ + y₁² (2x) = 2y .y₁ (÷ 2y₁ we get)
(x² – 1)y₂ + xy₁ – y = 0 Hence proved

Question 7.
If y = \((x+\sqrt{x^{2}+1})^{m}\) , show that (x² + 1)y₂ + xy₁ – m²y = 0
Answer:
y = \((x+\sqrt{x^{2}+1})^{m}\)

(x²)y₁² = m²y² ; (x² + 1) . 2y₁y₂ + y²₁(2x) = m².2yy₁ ÷ by 2y₁
(x² + 1)y₂ + xy₁ – m² = 0

Question 8.
If y = sin(log x), show that x²y₂ + xy₁ + y = 0/.
Answers:
y = sin (log x)
y₁ = \(\frac{\cos (\log x)}{x}\)
xy₁ = cos (log x).
Again differentiate w.r.t. x
xy₂ + y₁1 = \(\frac{-\sin (\log x)}{x}\)
x²y₂ + xy₁ = -y
⇒ x²y₂ + xy₁ + y = 0.