Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(l) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(l)

Question 1.

If y = tan^{-1} x Prove that (1 + x^{2}) y_{2} + 2xy_{1} = 0

Solution:

y = tan^{-1} x ⇒ y_{1} = \(\frac{1}{1+x^2}\)

⇒ (1 + x^{2}) y_{1} = 1

⇒ (1 + x^{2}) y_{2} + 2xy_{1} = 0

Question 2.

If 2y = x (1 + \(\frac{d y}{d x}\)) show that y_{2} is a constant.

Solution:

Question 3.

If y = ax sin x show that x^{2}y_{2} – 2xy_{1} + (x^{2} + 2) y = 0

Solution:

y = ax sin x

⇒ y_{1} = a sin x + ax cos x

⇒ y_{2} = a cos x + a cos x – ax sin x

= 2a cos x – ax sin x

Now x^{2}y_{2} – 2xy_{1} + (x^{2} + 2)y

= 2ax^{2} cos x – ax^{3} sin x – 2ax sin x – 2ax^{2} cos x + ax^{3} sin x + 2ax sin x = 0

Question 4.

If y = \(e^{m \cos ^{-1} x}\) (1 – x^{2}) y_{2} – xy_{1}= m^{2}y

Solution:

y = \(e^{m \cos ^{-1} x}\)

⇒ In y_{1 }= m cos^{-1} x

⇒ \(\frac{1}{y}\) . y = \(\frac{-m}{\sqrt{1-x^2}}\)

(\(\sqrt{1-x^2}\)) . y_{1} = -my

⇒ (1 – x^{2}) (y_{1})^{2} = m^{2}y^{2}

⇒ (1 – x^{2}) 2y_{1}y_{2} – 2x (y_{1})^{2} = 2m^{2}yy_{1}

⇒ (1 – x^{2}) y_{2} – xy_{1} = m^{2}y

Question 5.

If x = sin t, y = sin 2t then prove that, (1 – x^{2})\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) + 4y = 0

Solution:

x= sin t, y= sin 2t

Question 6.

If y = \(\left(\sin ^{-1} x\right)^2\) , prove that (1 – x^{2}) y_{2} – xy_{1} – 2 = 0

Solution:

Question 7.

If y = tan^{-1} x, Prove that (1 + x^{2}) y_{2} + 2xy_{1} = 0

Solution:

Same as No. 1