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

To solve the given problem, we need to determine the set \( S \) of all natural numbers for which the line \( \frac{x}{a} + \frac{y}{b} = 2 \) is a tangent to the curve \( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \) at the point \((a, b)\) with \( ab \neq 0 \).

Let's perform a systematic analysis:

1. The line equation is \( \frac{x}{a} + \frac{y}{b} = 2 \).
2. The curve equation is \( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \).
3. The tangency condition at the point \((a, b)\) implies that the gradient (derivative) of the curve at this point is equal to the slope of the line. Calculate the derivative of the curve using implicit differentiation:

Differentiate the curve equation implicitly:

\[ n \left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a} \cdot \frac{dx}{dy} + n \left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} = 0 \]

4. At point \((a, b)\), substitute \( x = a \) and \( y = b \):

\[ n \cdot \left(1\right)^{n-1} \cdot \frac{1}{a} \cdot \frac{dx}{dy} + n \cdot \left(1\right)^{n-1} \cdot \frac{1}{b} = 0 \]

Simplify: \[ \frac{1}{a} \cdot \frac{dx}{dy} = -\frac{1}{b} \]

5. The slope of the tangent of the line is \(-\frac{b}{a}\).

For tangency, the slopes must be equal:

\[ -\frac{b}{a} = \frac{1}{a}\left(-\frac{b}{1}\right) \]

6. Select natural number \( n \) so that the tangency condition is satisfied for any \((a, b)\) where \(a, b \neq 0\).

Since the given condition holds true irrespective of the power \( n \) (i.e., the structural form involves equal terms on both sides for any positive \( n \)), the set \( S \) is all positive integers, which implies \( S = \mathbb{N} \).

Therefore, the answer is: \( S = \mathbb{N} \).