JEE Main 2022 · 26 June · morning · Single correct · +4 / −1
Let S be the set of all the natural numbers, for which the line x/a + y/b = 2 is a tangent to the curve ( x/a )ⁿ + ( y/b )ⁿ = 2 at the point (a, b), ab ≠ 0. Then :
- (a) S = φ
- (b) n(S) = 1
- (c) S = 2k : k ∈ N
- (d) S = N
Answer and solution
Answer (d) S = N
Solution
Maths sir · 4 steps
Find the slope at (a, b) for a general n. Watch the n cancel, so every natural number works.
- Point is on curve. Put x = a, y = b into the curve: 1ⁿ + 1ⁿ = 2. This holds for every n, so (a, b) always lies on the curve.
- Differentiate implicitly. Differentiate (x/a)ⁿ + (y/b)ⁿ = 2 with respect to x. At (a, b) both brackets equal 1, so the powers n - 1 drop out.
n/a(x/a)ⁿ⁻¹ + n/b(y/b)ⁿ⁻¹dy/dx = 0 ⇒ n/a + n/bdy/dx = 0 ⇒ dy/dx = -b/a
- Write the tangent. The tangent at (a, b) is y - b = -b/a(x - a). Dividing by b gives y/b - 1 = -x/a + 1, that is x/a + y/b = 2. The slope does not depend on n at all.
- So. The given line is the tangent for every natural number n. For n = 1 the curve is this same line, and a line is its own tangent. Hence S = ℕ, option (d).
The idea to keep. At a point where every bracket equals 1, the power n drops out of the slope. Always check whether the parameter actually survives before you restrict its values.
Answer from the NTA final answer key. Solution by PrepWiser.