# GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Question 1.
x² + 3x + 2
Solution:
∴ $$\frac { dy }{ dx }$$ = 2x + 3
So, $$\frac{d^{2} y}{d x^{2}}$$ = $$\frac { d }{ dx }$$(2x + 3) = 2.

Question 2.
x20
Solution:
∴ $$\frac { dy }{ dx }$$ = x20
So, $$\frac{d^{2} y}{d x^{2}}$$ = 20 $$\frac { d }{ dx }$$(x19) = 20 x 19 x18 = 380 x18.

Question 3.
x cos x
Solution:
Let y = x cos x.
∴ $$\frac { dy }{ dx }$$ = x $$\frac { d }{ dx }$$ (cos x) + cos x $$\frac { d }{ dx }$$(x).
= x(- sin x) + cos x . 1
= – x sin x + cos x.
So, $$\frac{d^{2} y}{d x^{2}}$$ = – $$\frac { d }{ dx }$$(x sin x) + $$\frac { d }{ dx }$$ (cos x)
= – (x cos x + sin x . 1) – sin x
= – x cos x – 2 sin x.

Question 4.
log x
Solution:
Let y = log x
∴ $$\frac { dy }{ dx }$$ = $$\frac { 1 }{ x }$$ = x-1
So, $$\frac{d^{2} y}{d x^{2}}$$ = $$\frac { d }{ dx }$$(x-1) = – 1. x-2 = – $$\frac{1}{x^{2}}$$

Question 5.
x³ log x
Solution:
Let y = x³ log x

Question 6.
ex sin 5x
Solution:
Let y = ex sin 5x
∴ $$\frac { dy }{ dx }$$ = ex. $$\frac { d }{ dx }$$(sin 5x) + sin 5x$$\frac { d }{ dx }$$ex
= ex.cos 5x . 5 + sin 5x . ex
= ex(5 cos 5x + sin 5x)
So, $$\frac{d^{2} y}{d x^{2}}$$ = ex . $$\frac { d }{ dx }$$(5 cos 5x + sin 5x) + (5 cos 5x + sin 5x) . $$\frac { d }{ dx }$$(ex)
= ex(- 25 sin 5x + 5 cos 5x) + (5 cos 5x + sin 5x) ex
= ex(- 24 sin 5x + 10 cos 5x)
= 2ex (5 cos 5x – 12 sin 5x).

Question 7.
e6x cos 3x
Solution:
Let y = e6x cos 3x
∴ $$\frac { dy }{ dx }$$ = e6x. $$\frac { d }{ dx }$$(cos 5x) + cos 3x $$\frac { d }{ dx }$$e6x
= e6x(- 3 sin 3x (6 e6x)
= e6x(- 3 sin 3x + 6 cos 3x)
So, $$\frac{d^{2} y}{d x^{2}}$$ = e6x $$\frac { d }{ dx }$$(- 3 sin 3x + 6 cos 3x) + (- 3 sin 3x + 6 cos 3x) $$\frac { d }{ dx }$$e6x
= e6x(- 9 cos 3x – 18 sin 3x) + (- 3 sin 3x + 6 cos 3x) e6x . 6
= 9 e6x(3 cos 3x – 4 sin 3x).

Question 8.
tan-1 x
Solution:
∴ $$\frac { dy }{ dx }$$ = $$\frac{1}{1+x^{2}}$$ = (1 + x²)-1.
So, $$\frac{d^{2} y}{d x^{2}}$$ = $$\frac { dy }{ dx }$$ (1 + x²)-1.
= (-1) . (1 + x²)-2 . 2x = $$\frac{-2 x}{\left(1+x^{2}\right)^{2}}$$.

Question 9.
log (log x)
Solution:
Let y = log (log x)

Question 10.
sin (log x)
Solution:
Let y = sin (log x).

Question 11.
If y = 5 cos x – 3 sin x, prove that $$\frac{d^{2} y}{d x^{2}}$$ + y = 0.
Solution:
We have: y = 5 cos x – 3 sin x.
∴ $$\frac { dy }{ dx }$$ = – 5 sin x – 3 cos x.
So, $$\frac{d^{2} y}{d x^{2}}$$ = – 5 cos x + 3 sin x = – (5 cos x – 3 sin x) = – y.
⇒ $$\frac{d^{2} y}{d x^{2}}$$ + y = 0.

Question 12.
If y = cos-1x, find $$\frac{d^{2} y}{d x^{2}}$$ in terms of y alone.
Solution:

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that x²y2 + xy1 + y = 0.
Solution:

Question 14.
If y = Aemx + Benx, show that $$\frac{d^{2} y}{d x^{2}}$$ – (m + n)$$\frac { dy }{ dx }$$ + mny = 0.
Solution:
We have : y = Aemx + Benx.
∴ $$\frac { dy }{ dx }$$ = A . memx + B . nenx
So, $$\frac{d^{2} y}{d x^{2}}$$ = A . m²emx + B . n² enx.
∴ $$\frac{d^{2} y}{d x^{2}}$$ – (m+n) $$\frac { dy }{ dx }$$ + mny
= A m² emx + B n² enx – (m + n) x (A memx + B nenx) + mn (A emx + B enx)
= A m² emx + B n² enx – A m² emx – B mnenx – A mnemx – B n² enx + A mnemx + B mnenx
= 0.

Question 15.
If y = 500 e7x + 600 e-7x, show that $$\frac{d^{2} y}{d x^{2}}$$ = 49y.
Solution:
We have:

Question 16.
If ey (x + 1) = 1, show that $$\frac{d^{2} y}{d x^{2}}$$ = ($$\frac { dy }{ dx }$$)².
Solution:

Question 17.
If y = (tan-1 x)², show that (x² + 1)² y2 + 2x (x² + 1) y1 = 2.
Solution:
We have : y = (tan-1 x)².
∴ y1 = 2 tan-1x $$\frac { d }{ dx }$$(tan-1x),
⇒ y1 = $$\frac{2 \tan ^{-1} x}{1+x^{2}}$$
⇒ (1 + x²)y1 = 2 tan-1x.
Differentiating again w.r.t. x, we have :
(1 + x²)y2 + y1(2x) = $$\frac{2}{1+x^{2}}$$
⇒ (1 + x²)y2 + 2x(1 + x²)y1 = 2.