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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)=x4−10, then find the approximate value of f(2.1).
OR
Find the slope of the tangent to the curve y=2sin‌2(3x) at x=‌
π
6
.
Solution:     👈: Video Solution
f(x+∆x)=f(x)+f′(x)∆x
We have,
f(x)=x4−10
f′(x)=4x3
and,
x=2,∆x=0.1
then,
f(2+0.1)=f(2)+4(2)3(0.1)
⇒f(2.1)=24−10+3.2
∴f(2.1)=9.2
OR
y=2sin‌2(3x)
⇒‌
dy
dx
=2×2sin‌(3x)×cos(3x)×3
(‌
dy
dx
)x=‌
π
6
=12×sin‌(‌
π
2
)×cos(‌
π
2
)
(‌
dy
dx
)x=‌
π
6
=12×1×0
(‌
dy
dx
)x=‌
π
6
=0
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