SATHEE: Application of Derivatives 1 Question 8

Application of Derivatives 1 Question 8

8. The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point, where the curve intersects the $Y$ -axis passes through the point

(a) $- \frac{1}{2}, - \frac{1}{2}$

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

(d) $\frac{1}{2}, \frac{1}{3}$

(2017 Main)

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Answer:

Correct Answer: 8. (c)

Solution:

Given curve is

$y (x - 2) (x - 3) = x + 6$

Put $x = 0$ in Eq. (i), we get

$y (- 2) (- 3) = 6 \Rightarrow y = 1$

So, point of intersection is $(0, 1)$ .

Now, $y = \frac{x + 6}{(x - 2) (x - 3)}$

$\Rightarrow \frac{d y}{d x} = \frac{1 (x - 2) (x - 3) - (x + 6) (x - 3 + x - 2)}{(x - 2)^{2} (x - 3)^{2}}$

$\Rightarrow \frac{d y}{d x}_{(0, 1)} = \frac{6 + 30}{4 \times 9} = \frac{36}{36} = 1$

$∴$ Equation of normal at $(0, 1)$ is given by

$\begin{aligned} y - 1 & = \frac{- 1}{1} (x - 0) \\ \Rightarrow x + y - 1 & = 0 \end{aligned}$

which passes through the point $\frac{1}{2}, \frac{1}{2}$ .

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Application of Derivatives 1 Question 8

8. The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point, where the curve intersects the $Y$ -axis passes through the point

Answer:

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Application of Derivatives 1 Question 8

8. The normal to the curve y(x−2)(x−3)=x+6 at the point, where the curve intersects the Y-axis passes through the point

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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8. The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point, where the curve intersects the $Y$ -axis passes through the point