IGNOU MTE 1 2026 - If , then show that . Hence using Leibnitz's formu

Question

If $y = e^{m \sin^{-1} x}$ , then show that $(1 - x^2)y_2 - xy_1 - m^2y = 0$ . Hence using Leibnitz's formula, find the value of (1 - x²)y_n+2 - (2n + 1)xy_n+1.
b) Find the largest subset of R on which the function $f : \mathbf{R} \to \mathbf{R}$ defined as:
$$f(x) = \begin{cases} 2x, & x > 5 \\ x + 5, & 1 \le x \le 5 \\ |x|, & x < 1 \end{cases}$
is continuous.

09 Jan 2026

Answer :

Word Count : 374

यदि (y = e^{m \sin^{-1} x}), तो इसे (x) के अनुसार अलग करें। पहली अवकल: [ y_1 = \frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{m}{\sqrt{1 - x^2}} = \frac{m y}{\sqrt{1 - x^2}} ] दूसरी अवकल: [ y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{m y}{\sqrt{1 - x^2}}\right) = m \left(\frac{y_1}{\sqrt{1 - x^2}} + y \frac{d}{dx}\left(\frac{1}{\sqrt{1 - x^2}}\right)\right) ] [ \frac{d}{dx}\left(\frac{1}{\sqrt{1 - x^2}}\right) = \frac{x}{(1 - x^2)^{3/2}} ] इसलिए: [ y_2 = m \left(\frac{m y}{1 - x^2} + y \frac{x}{(1 - x^2)^{3/2}}\right) = \frac{m^2 y}{1 - x^2} + \frac{m x y}{(1 - x^2)^{3/2}} ] दोनों पक्षों को ((1 - x^2)) से गुणा करें: [ (1 - x^2)y_2 = m^2 y + \frac{m x y}{\sqrt{1 - x^2}} (1 - x^2) = m^2 y + m x y \sqrt{1 - x^2} ] लेकिन (y_1 = \frac{m y}{\sqrt{1

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