Question

Let S be the set of all the natural numbers, for which the line 
\(\frac{x}{a}+\frac{y}{b}=2 \)
is a tangent to the curve
\((\frac{x}{a})^n+(\frac{y}{b})^n=2 \)
at the point (a, b), ab ≠ 0. Then :

• A. S=Φ
• B. n(S)=1
• C. \(S=\left\{2k:k∈N\right\}\)
• D. \(S=N\)

Answer 1

The Correct Option is D

The correct answer is (D) : S=N
\((\frac{x}{a})^n+(\frac{y}{b})^n=2 \)
\(⇒\frac{n}{a}(\frac{x}{a})^{n-1} +\frac{n}{b}(\frac{y}{b})^{n-1}\frac{dy}{dx} =0\)
\(⇒\frac{dy}{dx}=-\frac{b}{a}(\frac{bx}{ay})^{n-1}\)
\(\frac{dy}{dx_{(a,b)}}=-\frac{b}{a}\)
So line always touches the given curve.