CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

Question : 19 of 36

Marks: +1, -0

{\f}(x)=x^4-10

, then find the approximate value of

{\f}(2.1)

.
OR
Find the slope of the tangent to the curve

{\y}=2 sin \;^2(3 x)

x=\;{π}/{6}

Solution:

{\f}( {\x}+∆ {\x})= {\f}( {\x})+ {\f}^{'}( {\x}) ∆ {\x}

We have,

f(x)=x^4-10

f^{'}(x)=4 x^3

and,

{\x}=2, ∆ {\x}=0.1

then,

{\f}(2+0.1)= {\f}(2)+4(2)^3(0.1)

⇒ {\f}(2.1)=2^4-10+3.2

∴ {\f}(2.1)=9.2

y=2 sin \;^2(3 x)

⇒ \;{d y}/{d x}=2 × 2 sin \;(3 x) × \cos (3 x) × 3

(\;{d y}/{d x}) _{x=\;{π}/{6}}=12 × sin \; (\;{π}/{2}) × \cos (\;{π}/{2})

(\;{d y}/{d x}) _{x=\;{π}/{6}}=12 × 1 × 0

(\;{d y}/{d x}) _{x=\;{π}/{6}}=0

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