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CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

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Question : 19 of 36
Marks: +1, -0
If f(x)=x4−10f(x)=x^4-10, then find the approximate value of f(2.1)f(2.1).
OR
Find the slope of the tangent to the curve y=2sin⁡2(3x)y=2\sin^2(3x) at x=π6x=\frac{\pi}{6}.
Solution:     👈: Video Solution
f(x+Δx)=f(x)+f′(x)Δxf(x+\Delta x)=f(x)+f'(x)\Delta x
We have,
f(x)=x4−10f(x)=x^4-10
f′(x)=4x3f'(x)=4x^3
and,
x=2,Δx=0.1x=2, \Delta x=0.1
then,
f(2+0.1)=f(2)+4(2)3(0.1)f(2+0.1)=f(2)+4(2)^3(0.1)
⇒f(2.1)=24−10+3.2\Rightarrow f(2.1)=2^4-10+3.2
∴f(2.1)=9.2\therefore f(2.1)=9.2
OR
y=2sin⁡2(3x)y=2\sin^2(3x)
⇒dydx=2×2sin⁡(3x)×cos⁡(3x)×3\Rightarrow \frac{dy}{dx}=2 \times 2 \sin(3x) \times \cos(3x) \times 3
(dydx)x=π6=12×sin⁡(π2)×cos⁡(π2)\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{6}}=12 \times \sin\left(\frac{\pi}{2}\right) \times \cos\left(\frac{\pi}{2}\right)
(dydx)x=π6=12×1×0\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{6}}=12 \times 1 \times 0
(dydx)x=π6=0\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{6}}=0
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