TS EAMCET · Maths · Application of Derivatives
For a given function \(y=f(x), \delta y\) denotes the actual error in \(y\) corresponding to actual error \(\delta x\) in \(x\) and \(d y\) denotes the approximate value of \(\delta y\). If \(y=f(x)=2 x^2-3 x+4\) and \(\delta x=0.02\), then the value of \(\delta y-d y\) when \(x=5\) is
- A \(0.0008\)
- B \(0.008\)
- C \(0.0004\)
- D \(0.004\)
Answer & Solution
Correct Answer
(A) \(0.0008\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \quad \delta_y=\left(2 \times 5.02^2-3 \times 5.02+4\right)-\left(2 \times 5^2-3 \times 5+4\right) \\ & =2 \times 0.02 \times 10.02-3 \times 0.02=0.4008-0.06=0.3408 . \\ & d y=\left(\frac{d y}{d x}\right) \times d x=(4 x-3) d x=(4 \times 5-3) \times 0.02 \\ &…
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