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Question

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

(a) (−3, −27)
(b) (3, 9)
(c) (7/2, 35/4)
(d) (0, 0)

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Solution

(b) (3, 9)

Let (x1, y1) be the point where the given curve intersect the given line at the given angle.
Since, the point lie on the curve.Hence, y1=6x1-x12Now, y=6x-x2dydx=6-2xm1=6-2x1andx+y=01+dydx=0 dydx=-1m2=-1It is given that the angle between them is π4.tan θ=m1-m21+m1m2tan π4=6-2x1+11-6+2x11=7-2x12x1-57-2x12x1-5=±17-2x12x1-5=1 or 7-2x12x1-5=-17-2x1=2x1-5 or 7-2x1=-2x1+54x1=12 or 2 = 0 (It is not true.)x1=3andy1=6x1-x12=18-9=9x1, y1=3, 9

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