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Let S be the set of all values of x for ...

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

`{(1)/(3), -1}`

B

`{(1)/(3), 1}`

C

`{-(1)/(3), 1}`

D

`{-(1)/(3), -1}`

Text Solution

Verified by Experts

The correct Answer is:
C
Given curve is `y = f(x) = x^(3)- x^(2)- 2x " "`…(i)
So, `f(-1) = -1- 1+2 =0`
and `f(-1)=-1 - 1 +2 =0`
Since, slope of a line passing through `(x_1, y_1) and (x_2, y_2)` is given by `m = tan theta = (y_2-y_1)/(x_2 - x_1)`
`therefore ` Slope of line joining points `(1, f(1)) and (-1, f(-1))` is `m = (f(1)- f(-1))/(1-(-1))= (-2-0)/(1+ 1) =-1`
Now, `(dy)/(dx) = 3x^(2) - 2x - 2 `
`" " ` [differentiating Eq. (i), w.r.t. 'x']
According to the question,
`" " (dy)/(dx) = m`
`rArr " " 3x^(2) - 2x - 2 =-1 rArr 3x^(2) - 2x - 1 =0`
`rArr " " ( x- 1) ( 3x + 1) = 0 rArr x =1 , - (1)/(3)`
Therefore, set `S= {-(1)/(3), 1}`.
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