Students can Download Basic Maths Exercise 18.7 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 18.7

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 = 3x^{3} + 4x^{2} + 7

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

3. y = e^{3x + 2}

4. y = x^{3} . 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 = 3x^{3} + 4x^{2} + 7

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

\(\frac{d y}{d x}\) = x^{3} . \(\frac { 1 }{ x }\) + log x . 3x^{2} = x^{2} + 3x^{2} log x

\(\frac{d^{2} y}{d x^{2}}\) = 2x + 3x^{2 }\(\frac { 1 }{ x }\) + 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^{x}log 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 cos^{3}t, y = -a sin^{3}t

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 cos^{3}t, y = -a sin^{3}t

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

\(\frac{d^{2} y}{d x^{2}}\) = m^{2} sin mx = -m^{2}y

∴ \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)

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 y_{2} – a^{2}y = 0

Answer:

y = e^{ax} + e^{-ax}

y_{1} = ae^{ax} – ae^{-ax}

Y_{2} = a^{2}e^{ax} + a^{2}e^{-ax}

= a^{2}(e^{ax} + e^{-ax}) = a^{2}y

∴ y_{2} – a^{2}y = 0.

Question 5.

If y = 2 + log x, show that xy_{2} + y_{1} = 0.

Answer:

y_{i} = \(\frac { 1 }{ x }\) ⇒ xy_{1} = 1

∴ = xy_{2} + y_{1} = 0

Part – C

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

Question 1.

If x^{2} – xy + y^{2} = a^{2}, show that \(\frac{d^{2} y}{d x^{3}}=\frac{6 a^{2}}{(x-2 y)^{3}}\)

Answer:

Question 2.

If x^{2} + 2xy + 3y^{2} = 1, show that y^{2} = \(\frac{-2}{(x+3 y)^{3}}\)

Answer:

x^{2} + 2xy + 3y^{2} = 1

Differentiate w.r.t x we get

2x + 2(x.y_{1} + y.1) + 6y.y_{1} = 0

y_{1}(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

\(\frac{d^{2} y}{d x^{2}}\) = -am^{2} cos mx – bm^{2} sin mx

= m^{2}(a cos mx + b sin mx) = -m^{2}y

⇒ \(\frac{d^{2} y}{d x^{2}}+m^{2} y=0\)

Question 4.

If y = a cos (log x) + b sin (log x), show that x^{2}y_{2} + xy_{1} + y = 0

Answer:

y = a cos(log x) + b sin (log x)

\(y_{1}=\frac{-a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\)

xy_{1} = -a sin (log x) + b cos (log x) Again diff w.r.t x

xy_{2} + y_{1}1 = \(\frac{-a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x}\)

x^{2}y_{2} + xy_{1} = -(a cos logx + b sin (log x)) = =y

⇒ x^{2}y_{2} + xy_{1} + y = 0

Question 5.

If y = log (x – \(\sqrt{x^{2}+1}\)), show that (x^{2} + 1)y_{2} + xy_{1} = 0

Answer:

If y = log (x – \(\sqrt{x^{2}+1}\))

2y_{1} . y_{2}(1 + x^{2}) + y_{1}^{2}(2x) = 0 ÷ 2y_{1}

⇒ (1 + x^{2})y_{2} + xy_{1} = 0.

Question 6.

If y = x + \(\sqrt{x^{2} – 1}\) , show that (x^{2} – 1)y_{2} + xy_{1} – y = 0

Answer:

y = x + \(\sqrt{x^{2} – 1}\) differentiate w.r.t x

∴ (x^{2} – 1) y_{1}^{2} = y^{2} Differentiate agin w.r.t x

(x^{2} – 1)2y_{1}y_{2} + y_{1}^{2} (2x) = 2y .y_{1} (÷ 2y_{1} we get)

(x^{2} – 1)y_{2} + xy_{1} – y = 0 Hence proved

Question 7.

If y = \((x+\sqrt{x^{2}+1})^{m}\) , show that (x^{2} + 1)y_{2} + xy_{1} – m^{2}y = 0

Answer:

y = \((x+\sqrt{x^{2}+1})^{m}\)

(x^{2})y_{1}^{2} = m^{2}y^{2} ; (x^{2} + 1) . 2y_{1}y_{2} + y^{2}_{1}(2x) = m^{2}.2yy_{1} ÷ by 2y_{1}

(x^{2} + 1)y_{2} + xy_{1} – m^{2} = 0

Question 8.

If y = sin(log x), show that x^{2}y_{2} + xy_{1} + y = 0/.

Answers:

y = sin (log x)

y_{1} = \(\frac{\cos (\log x)}{x}\)

xy_{1} = cos (log x).

Again differentiate w.r.t. x

xy_{2} + y_{1}1 = \(\frac{-\sin (\log x)}{x}\)

x^{2}y_{2} + xy_{1} = -y

⇒ x^{2}y_{2} + xy_{1} + y = 0.