SATHEE: Application of Derivatives 1 Question 2

Application of Derivatives 1 Question 2

####2. Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f (x) = x^{3} - x^{2} - 2 x$ 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) $\frac{1}{3}, - 1$

(b) $\frac{1}{3}, 1$

(d) $- \frac{1}{3}, - 1$

(2019 Main, 9 April I)

Show Answer

Answer:

Correct Answer: 2. (c)

Solution:

Given curve is $y = f (x) = x^{3} - x^{2} - 2 x$

So, $f (1) = 1 - 1 - 2 = - 2$

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 $m = \tan θ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$∴$ Slope of line joining points $(1, f (1))$ and

$(- 1, f (- 1))$ is $m = \frac{f (1) - f (- 1)}{1 - (- 1)} = \frac{- 2 - 0}{1 + 1} = - 1$

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

According to the question,

[differentiating Eq. (i), w.r.t. ’ $x^{2}$ ]

$\begin{array}{ll} \frac{d y}{d x} = m \\ \Rightarrow & 3 x^{2} - 2 x - 2 = - 1 \Rightarrow 3 x^{2} - 2 x - 1 = 0 \\ \Rightarrow & (x - 1) (3 x + 1) = 0 \Rightarrow x = 1, - \frac{1}{3} \end{array}$

Therefore, set $S = - \frac{1}{3}, 1$ .

Application of Derivatives 1 Question 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Application of Derivatives 1 Question 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu