STATEMENT 1: If y= (m ^ -1 x) then (1-x²) y_ n+2 -(2 n+1) x y_ n+1 +(n²-m²) y_n=0 [ y_n is n th derivative of y w.r.t. x ] STATEMENT 2: (d^n)/(d x^n)[ u(x) v(x)]=u_n v_0+ ^n C_1 u_ n-1 v_1+ ^n C_2 u_ n-2 v_2+ . .+ ^n

Correct answer: D. Statement 1 is false and Statement 2 is true. y_1= (m ^ -1 x) · m 1-x^2 aligned & 1-x^2 y_1=m (m ^ -1 x) and (1-x²) y_2-x y_1+m² y=0 & [(1-x²) y_ n+2 +n(-2 x) y_ n+1 +(n(n-1))/(2)(-2) y_n]-[x y_ n+1 +n ·1 _n]+m² y_n=0 aligned Simplifying we get, (1-x^2) y_ n+2 -(2 n+1) y_ n+1 -(n^2-m^2) y_n=0

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Differentiation

JEE Advanced · Mathematics · Differentiation

STATEMENT 1: If y= (m ^ -1 x) then (1-x²) y_ n+2 -(2 n+1) x y_ n+1...

STATEMENT 1: If

y=\sin

\left(m \sin ^{-1} x\right)

then \left(1-x²\right) y_{n+2}-(2 n+1) x y_{n+1}+\left(n²-m²\right) y_n=0

[

y_n

is

th derivative of y w.r.t. x

]

STATEMENT 2: \frac{d^n}{d x^n}[

u(x) v(x)]=u_n v_0+{ }^n C_1 u_{n-1} v_1+{ }^n C_2 u_{n-2} v_2+\ldots . .+{ }^n C_n u_0 v_n \

quad\left[

u_n, v_n\right.

denote nth derivative of

u(x), v(x)

w.r.t.

x ]$

Options

ABoth the statements are true and Statement 2 is correct explanation of Statement 1.
BBoth the Statements are true and Statement 2 is not the correct explanation of Statement 1.
CStatement 1 is true and Statement 2 is false.
DStatement 1 is false and Statement 2 is true.

Correct answer

D. Statement 1 is false and Statement 2 is true.

Step-by-step solution

y_1=\cos \left(m \sin ^{-1} x\right) \cdot \frac{m}{\sqrt{1-x^2}}

\begin{aligned}& \sqrt{1-x^2}y_1=m \cos \left(m \sin ^{-1} x\right) \\ \text{ and } \Rightarrow\left(1-x²\right) y_2-x y_1+m² y=0 \\ & {\left[\left(1-x²\right) y_{n+2}+n(-2 x) y_{n+1}+\frac{n(n-1)}{2}(-2) y_n\right]-\left[x y_{n+1}+n \cdot1 \cdoty_n\right]+m² y_n=0} \end{aligned}

Simplifying we get, \left(1-x^2\right) y_{n+2}-(2 n+1) y_{n+1}-\left(n^2-m^2\right) y_n=0

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