SATHEE: Application of Derivatives 1 Question 4

Application of Derivatives 1 Question 4

####4. The tangent to the curve $y = x^{2} - 5 x + 5$ , parallel to the line $2 y = 4 x + 1$ , also passes through the point

(a) $\frac{1}{4}, \frac{7}{2}$

(b) $\frac{7}{2}, \frac{1}{4}$

(d) $\frac{1}{8}, - 7$

(2019 Main, 12 Jan II)

Show Answer

Answer:

Correct Answer: 4. (b)

Solution:

The given curve is $y = x^{2} - 5 x + 5$

Now, slope of tangent at any point $(x, y)$ on the curve is

$\frac{d y}{d x} = 2 x - 5$

[on differentiating Eq. (i) w.r.t. $x$ ]

$∵$ It is given that tangent is parallel to line

$2 y = 4 x + 1$

So, $\frac{d y}{d x} = 2 [∵$ slope of line $2 y = 4 x + 1$ is 2 $]$

$\Rightarrow 2 x - 5 = 2 \Rightarrow 2 x = 7 \Rightarrow x = \frac{7}{2}$

On putting $x = \frac{7}{2}$ in Eq. (i), we get

$y = \frac{49}{4} - \frac{35}{2} + 5 = \frac{69}{4} - \frac{35}{2} = - \frac{1}{4}$

Now, equation of tangent to the curve (i) at point

$\frac{7}{2}, - \frac{1}{4}$ and having slope 2 , is

$\begin{aligned} y + \frac{1}{4} & = 2 x - \frac{7}{2} \Rightarrow y + \frac{1}{4} = 2 x - 7 \\ \Rightarrow y & = 2 x - \frac{29}{4} \end{aligned}$

On checking all the options, we get the point $\frac{1}{8}, - 7$ satisfy the line (iii).

Application of Derivatives 1 Question 4

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 4

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu