JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q9

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MathematicsSingle correct+4 / -1Question 9 of 90

Let S be the set of all the natural numbers, for which the line $\dfrac xa+\dfrac yb=2$ is a tangent to the curve $\left(\dfrac xa\right)^n+\left(\dfrac yb\right)^n=2$ at the point $(a,b)$ , $ab\ne0$ . Then :

S=\phi

n(S)=1

S=\{2k:k\in\mathbb N\}

S=\mathbb N

Answer and worked solution

Official answer: D

Tangent at $(a,b)$ . Differentiating, $\dfrac na\left(\dfrac xa\right)^{n-1}dx+\dfrac nb\left(\dfrac yb\right)^{n-1}dy=0$ ; at $(a,b)$ the tangent is $\dfrac xa+\dfrac yb=2$ .

For every $n$ , so $S=\mathbb N$ .

(D)

Topics:Tangents and normals

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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 9