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CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

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Question : 14 of 36
Marks: +1, -0
The slope of the tangent to the curve y=x3−xy=x^3-x at the point (2,6)(2,6) is____________
OR
The rate of change of the area of a circle with respect to its radius rr, when r=3 cmr=3 \text{ cm}, is________
Solution:     👈: Video Solution
y=x3−xy=x^3-x
dydx=3x2−1\frac{dy}{dx}=3x^2-1
slope of tangent=dydx\text{slope of tangent} = \frac{dy}{dx}
=3x2−1=3x^2-1
x=2x=2
Slope of tangent=3(2)2−1\text{Slope of tangent} = 3(2)^2-1
=12−1=12-1
=11=11
OR
6π cm2/cm6\pi \text{ cm}^2/\text{cm}
Explanation: Area of circle, A=πr2A=\pi r^2
∴dAdr=2πr\therefore \frac{dA}{dr}=2\pi r
[dAdr]r=3 cm=2×π×3\left[ \frac{dA}{dr} \right]_{r=3\,\text{cm}} = 2 \times \pi \times 3
=6π cm2/cm=6\pi \text{ cm}^2/\text{cm}
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