The point at the curve y = 12x − x² where the slope of the tangent is zero will be

(a) (0, 0)
(b) (2, 16)
(c) (3, 9)
(d) none of these

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Solution

(d) none of these

$Let (x_{1}, y_{1}) be the required point. Since, the point lies on the curve, Hence, y_{1} = 12 x_{1} - {x_{1}}^{2} Now, y = 12 x - x^{2} \Rightarrow \frac{d y}{d x} = 12 - 2 x S lope of the tangent = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 12 - 2 x_{1} Given: Slope of the tangent=0 12 - 2 x_{1} = 0 \Rightarrow x_{1} = 6 Now, y_{1} = 12 x_{1} - {x_{1}}^{2} = 72 - 36 = 36 ∴ (x_{1}, y_{1}) = (6, 36)$

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The point at the curve y = 12x − x2 where the slope of the tangent is zero will be (a) (0, 0) (b) (2, 16) (c) (3, 9) (d) none of these

(d) none of these Let x1, y1 be the required point.Since, the point lies on the curve,Hence, y1=12x1-x12Now,y=12x-x2⇒dydx=12-2xSlope of the tangent = dydxx1, y1=12-2x1Given:Slope of the tangent=012-2x1=0⇒x1=6Now,y1=12x1-x12=72-36=36∴x1, y1=6, 36

The point at the curve y = 12x − x² where the slope of the tangent is zero will be

(a) (0, 0)
(b) (2, 16)
(c) (3, 9)
(d) none of these