If the tangent to the curve,

If the tangent to the curve, $y=x^{3}+a x-b$ 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?

  1. (1) $(-2,1)$

  2. (2) $(-2,2)$

  3. (3) $(2,-1)$

  4. (4) $(2,-2)$

Correct Option: 4,


$y=x^{3}+a x-b$

Since, the point $(1,-5)$ lies on the curve.

$\Rightarrow 1+a-b=-5$

$\Rightarrow a-b=-6$ ………..(1)

Now, $\frac{d y}{d x}=3 x^{2}+a$

$\Rightarrow\left(\frac{d y}{d x}\right)_{\text {at } x=1}=3+a$

Since, required line is perpendicular to $y=x-4$,

then slope of tangent at the point $P(1,-5)=-1$

$\therefore 3+a=-1$

$\Rightarrow \mathrm{a}=-4$

$\Rightarrow \mathrm{b}=2$

$\therefore$ the equation of the curve is $\mathrm{y}=\mathrm{x}^{3}-4 \mathrm{x}-2$

$\Rightarrow(2,-2)$ lies on the curve




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