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
- A \(\left\{ {\frac{1}{3}, - 1} \right\}\)
- B \(\left\{ {-\frac{1}{3}, - 1} \right\}\)
- C \(\left\{ {\frac{1}{3}, 1} \right\}\)
- D \(\left\{ {-\frac{1}{3}, 1} \right\}\)
Answer & Solution
Correct Answer
(D) \(\left\{ {-\frac{1}{3}, 1} \right\}\)
Step-by-step Solution
Detailed explanation
\(f(1)=1-1-2=-2\) \(f(-1)=-1-1+2=0\) \(m = \frac{{f\left( 1 \right) - f\left( { - 1} \right)}}{{1 + 1}} = \frac{{ - 2 - 0}}{2} = - 1\) \(\frac{{dy}}{{dx}} = 3{x^2} - 2x - 2\) \(3{x^2} - 2x - 2 = - 1\) \( \Rightarrow 3{x^2} - 2x - 1 = 0\)…
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