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...
y=\sin then \left(1-x²\right) y_{n+2}-(2 n+1) x y_{n+1}+\left(n²-m²\right) y_n=0y_nn]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.u(x), v(x)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}} 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