The equation of the normal to the curve
y = (1+x)²ʸ + cos ²(sin –1 x) at x = 0 is :

Solution

Maths sir · 5 steps

Find the point first, then use logs on the power term; at x = 0 the messy y' term drops out.

1. Tidy the curve. Put θ = sin⁻¹ x, so sin θ = x and cos² θ = 1 - x². The curve becomes y = (1+x)²ʸ + 1 - x².
2. Find the point. At x = 0 the power term is 1²ʸ = 1, so y = 1 + 1 = 2. The normal passes through (0, 2).
3. Differentiate the power. Let u = (1+x)²ʸ, so ln u = 2y ln(1+x). Differentiating both sides with respect to x gives the equation below.

   u'/u = 2y'ln(1+x) + 2y/(1+x)

4. Slope at the point. At (0, 2): u = 1 and ln 1 = 0, so the y' term vanishes and u' = (2 × 2)/1 = 4. The 1 - x² part has derivative -2x = 0, so y' = 4 and the normal slope is -1/4.
5. So. The normal is y - 2 = -1/4x, that is 4y - 8 = -x, or x + 4y = 8. That is option (b).

The idea to keep. When a variable sits in both the base and the exponent, take logs before differentiating. Always find the point first, because known values often wipe out awkward terms.

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