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Solution for Find the Approximate Value of F (2.01), Where F (X) = 4x2 + 5x + 2 ? - CBSE (Science) Class 12 - Mathematics

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Question

Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2 ?

Solution

\[\text { Let }: \]

\[ x = 2\]

\[x + ∆ x = 2 . 01\]

\[ \Rightarrow ∆ x = 0 . 01\]

\[f\left( x \right) = 4 x^2 + 5x + 2\]

\[ \Rightarrow f\left( x = 2 \right) = 16 + 10 + 2 = 28\]

\[\text { Now,} y = f\left( x \right)\]\[ \Rightarrow \frac{dy}{dx} = 8x + 5\]

\[ \therefore dy = ∆ y = \frac{dy}{dx}dx = \left( 8x + 5 \right) \times 0 . 01 = \left( 16 + 5 \right) \times 0 . 01 = 0 . 21\]

\[ \therefore f\left( 2 . 01 \right) = y + ∆ y = 28 . 21\]

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Solution for question: Find the Approximate Value of F (2.01), Where F (X) = 4x2 + 5x + 2 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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