Application of Derivatives 1 Question 3
3. 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?
(a) $(-2,2)$
(b) $(2,-2)$
(c) $(-2,1)$
(d) $(2,-1)$
(2019 Main, 9 April I)
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Answer:
Correct Answer: 3. (c)
Solution:
- Given curve is $y=x^{3}+a x-b$
passes through point $P(1,-5)$.
$$ \therefore \quad-5=1+a-b \Rightarrow b-a=6 $$
and slope of tangent at point $P(1,-5)$ to the curve (i), is
$$ m _1=\left.\frac{d y}{d x}\right| _{(1,-5)}=\left[3 x^{2}+a\right] _{(1,-5)}=a+3 $$
$\because$ The tangent having slope $m _1=a+3$ at point $P(1,-5)$ is perpendicular to line $-x+y+4=0$, whose slope is $m _2=1$.
$\therefore \quad a+3=-1 \Rightarrow a=-4 \quad\left[\because m _1 m _2=-1\right]$
Now, on substituting $a=-4$ in Eq. (ii), we get $b=2$
On putting $a=-2$ and $b=2$ in Eq. (i), we get
$$ y=x^{3}-4 x-2 $$
Now, from option $(2,-2)$ is the required point which lie on it.
