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प्रश्न
The point on the curve y = 6x − x2 at which the tangent to the curve is inclined at π/4 to the line x + y= 0 is __________ .
पर्याय
(−3, −27)
(3, 9)
(7/2, 35/4)
(0, 0)
उत्तर
`(3, 9)`
Let (x1, y1) be the point where the given curve intersect the given line at the given angle.
\[\text { Since, the point lie on the curve } . \]
\[\text { Hence }, y_1 = 6 x_1 - {x_1}^2 \]
\[\text { Now,} y = 6x - x^2 \]
\[ \Rightarrow \frac{dy}{dx} = 6 - 2x\]
\[ \Rightarrow m_1 = 6 - 2 x_1 \]
\[\text { and }\]
\[x + y = 0\]
\[ \Rightarrow 1 + \frac{dy}{dx} = 0 \]
\[ \Rightarrow \frac{dy}{dx} = - 1\]
\[ \Rightarrow m_2 = - 1\]
\[\text { It is given that the angle between them is }\frac{\pi}{4}.\]
\[ \therefore \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\]
\[ \Rightarrow \tan \frac{\pi}{4} = \left| \frac{6 - 2 x_1 + 1}{1 - 6 + 2 x_1} \right|\]
\[ \Rightarrow 1 = \left| \frac{7 - 2 x_1}{2 x_1 - 5} \right|\]
\[ \Rightarrow \frac{7 - 2 x_1}{2 x_1 - 5} = \pm 1\]
\[ \Rightarrow \frac{7 - 2 x_1}{2 x_1 - 5} = 1 \ or\frac{7 - 2 x_1}{2 x_1 - 5}=-1\]
\[ \Rightarrow 7 - 2 x_1 = 2 x_1 - 5 \ or \ 7 - 2 x_1 = - 2 x_1 + 5\]
\[ \Rightarrow 4 x_1 = 12 \ or \ 2 = 0 (\text {It is not true }.)\]
\[ \Rightarrow x_1 = 3\]
\[\text { and }\]
\[ y_1 = 6 x_1 - {x_1}^2 = 18 - 9 = 9\]
\[\therefore\left( x_1 , y_1 \right)=\left( 3, 9 \right)\]
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