Applications of Derivatives
Tangent Lines
Grade 12
Question:
<p><strong>Ex. 13 Statement I:</strong> Tangent drawn at the point \((0, 1)\) to the curve \(y = x^3 - 3x + 1\) meets the curve thrice at one point only.</p>
<p>(a) Statement I is true and Statement II is false</p>
<p>(b) Statement I is false and Statement II is true</p>
<p>(c) Both statements are true</p>
<p>(d) Both statements are false</p>
Step-by-Step Solution
Key Concept: A tangent at a point meets the curve at that point with multiplicity at least 2. Check if there are other intersection points by solving the equation formed by equating the tangent with the curve.
<p><strong>Step 1:</strong> Find $\frac{dy}{dx} = 3x^2 - 3$</p><p><strong>Step 2:</strong> At $(0,1)$: $\frac{dy}{dx}\bigg|_{(0,1)} = -3$</p><p><strong>Step 3:</strong> Equation of tangent: $y - 1 = -3(x - 0)$, i.e., $y = -3x + 1$</p><p><strong>Step 4:</strong> Find intersections: $-3x + 1 = x^3 - 3x + 1 \Rightarrow x^3 = 0 \Rightarrow x = 0$</p><p><strong>Step 5:</strong> The tangent meets the curve at one point only (at $x = 0$ with multiplicity 3).</p><p><strong>Step 6:</strong> Statement I is TRUE.</p><p><strong>Step 7:</strong> At $(1, -1)$: $\frac{dy}{dx}\bigg|_{(1,-1)} = 0$</p><p><strong>Step 8:</strong> Equation of tangent: $y = -1$</p><p><strong>Step 9:</strong> Find intersections: $-1 = x^3 - 3x + 1 \Rightarrow x^3 - 3x + 2 = 0 \Rightarrow (x-1)(x^2 + x - 2) = 0$</p><p><strong>Step 10:</strong> The tangent meets the curve at two distinct points.</p><p><strong>Step 11:</strong> Statement II is FALSE.</p><p>∴ Answer is (b).</p>
Correct Answer: B