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JEE Mains · Maths · STD 12 - 6. Application of derivatives

Let \(S\) be the set of all values of \(x\) for which the tangent to the curve \(y = f(x) = x^3 -x^2 -2x\) at \((x, y)\) is parallel to the line segment joining the points \((1,f(1))\) and \(( - 1,f( - 1)),\) then \(S\) is equal to

  1. A \(\left\{ {\frac{1}{3}, - 1} \right\}\)
  2. B \(\left\{ {-\frac{1}{3}, - 1} \right\}\)
  3. C \(\left\{ {\frac{1}{3},  1} \right\}\)
  4. D \(\left\{ {-\frac{1}{3}, 1} \right\}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left\{ {-\frac{1}{3}, 1} \right\}\)

Step-by-step Solution

Detailed explanation

\(f(1)=1-1-2=-2\) \(f(-1)=-1-1+2=0\) \(m = \frac{{f\left( 1 \right) - f\left( { - 1} \right)}}{{1 + 1}} = \frac{{ - 2 - 0}}{2} = - 1\) \(\frac{{dy}}{{dx}} = 3{x^2} - 2x - 2\) \(3{x^2} - 2x - 2 = - 1\) \( \Rightarrow 3{x^2} - 2x - 1 = 0\)…