SATHEE: Application of Derivatives 1 Question 3

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) $(2, - 2)$

(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 $P (1, - 5)$ .

$∴ - 5 = 1 + a - b \Rightarrow b - a = 6$

and slope of tangent at point $P (1, - 5)$ to the curve (i), is

$m_{1} = {\frac{d y}{d x} |}_{(1, - 5)} = {[3 x^{2} + a]}_{(1, - 5)} = a + 3$

$∵$ The tangent having slope $m_{1} = a + 3$ at point $P (1, - 5)$ is perpendicular to line $- x + y + 4 = 0$ , whose slope is $m_{2} = 1$ .

$∴ a + 3 = - 1 \Rightarrow a = - 4 [∵ m_{1} m_{2} = - 1]$

Now, on substituting $a = - 4$ in Eq. (ii), we get $b = 2$

On putting $a = - 2$ and $b = 2$ in Eq. (i), we get

$y = x^{3} - 4 x - 2$

Now, from option $(2, - 2)$ is the required point which lie on it.

Application of Derivatives 1 Question 3

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Application of Derivatives 1 Question 3

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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