If log ₑ y=3 sin ⁻¹ x, then (1-x²) y^(′ ′)-x y′ at x=1/2 is equal to

Solution

Maths sir · 4 steps

Clear the square root from y' first, then differentiate once more and the whole expression becomes 9y.

1. First derivative. Differentiate ln y = 3 sin⁻¹ x to get y'/y = 3/√(1-x²). Cross-multiply to remove the fraction: √(1-x²) y' = 3y.
2. Differentiate again. Use the product rule on the left: √(1-x²) y'' - x/√(1-x²) y' = 3y'. Multiply through by √(1-x²) and replace √(1-x²) y' with 3y.

   (1-x²)y'' - xy' = 3√(1-x²) y' = 3(3y) = 9y

3. Value of y. At x = 1/2, sin⁻¹ 1/2 = π/6, so ln y = 3 × π/6 = π/2 and y = e^(π/2).
4. So. The expression equals 9y = 9e^(π/2), which is option (a).

The idea to keep. When y' carries a √(1-x²) in the denominator, multiply it across before differentiating again. The second-order expression then collapses into a multiple of y.

Answer from the NTA final answer key. Solution by PrepWiser.