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If the tangent to the curve, y = x^(3) +...

If the tangent to the curve, `y = x^(3) + ax -b ` at the point `(1, -5)` is perpendicular to the line, `-x +y + 4 =0`, then which one of the following points lies on the curve ? (A) `(-2, 2)` (B) `(2, -2)` (C) `(-2, 1)` (D) `(2, -1)`

A

`(-2, 2)`

B

`(2, -2)`

C

`(-2, 1)`

D

`(2, -1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given curve is `y = x^(3) + ax - b" " `…(i)
passes through point `P (1, 5)`
`therefore " " -5 = 1 + a - b rArr b - a = 6" "`…(ii)
and slope of tangent at point `P(1, -5)` to the curve (i), is
`" "m_1= (dy)/(dx) :|_("("1, -5")")= [3x^(2) + a] _("("1, -5")")= a + 3`
`because` The tangent having slope `m_1 = a + 3 ` at point `P(1, -5)` is perpendicular to line `-x = y = 4 = 0`, whose slope is `m_2 =1`.
`therefore " " a + 3 = - 1 rArr a = -4 " " [because m _1 m_2 = -1]`
Now, on substituting `a= - 4` in Eq. (ii), we get `b = 2`
On putting `a = - 2 and b = 2 ` in Eq. (i), we get
`" " y = x^(3)- 4x - 2 `
Now, from option `(2, -2)` is the required point which lie on it.
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