<p>A tangent to the curve \(y = f(x)\) cuts the line \(y = x\) at a point which is at a distance of 1 unit from Y-axis. The equation of the curve is</p>
Step-by-Step Solution
Key Concept: Set up a differential equation by using the condition that a tangent at point (x,y) on the curve intersects y=x at a point (1,1) which is 1 unit from the Y-axis. The slope of the tangent and the geometry of the intersection point determine the differential equation.
<p><strong>Step 1: Identify the intersection point.</strong></p><p>A point at distance 1 unit from Y-axis lies on the line y = x at point (1, 1).</p><p><strong>Step 2: Set up the tangent line equation.</strong></p><p>Let the point of tangency be P(x, y) on the curve y = f(x). The tangent line at P has slope dy/dx.</p><p>Equation of tangent: Y - y = (dy/dx)(X - x)</p><p><strong>Step 3: Use the condition that tangent passes through (1,1).</strong></p><p>Substituting point (1, 1) into the tangent equation:</p><p>1 - y = (dy/dx)(1 - x)</p><p>Therefore: dy/dx = (1 - y)/(1 - x)</p><p><strong>Step 4: Rearrange as a separable differential equation.</strong></p><p>dy/(1 - y) = dx/(1 - x)</p><p><strong>Step 5: Integrate both sides.</strong></p><p>∫dy/(1 - y) = ∫dx/(1 - x)</p><p>-ln|1 - y| = -ln|1 - x| + C'</p><p>ln|1 - y| = ln|1 - x| + ln K (where K is a constant)</p><p><strong>Step 6: Solve for the curve equation.</strong></p><p>|1 - y| = K|1 - x|</p><p>1 - y = C(1 - x), where C = ±K</p><p>(1 - y)/(1 - x) = C</p><p>Or equivalently: (x - 1)/(y - 1) = 1/C = C (renaming the constant)</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a