Q107 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

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

  • Correct answer
    A
    $0.0008$
  • B
    $0.008$
  • C
    $0.0004$
  • D
    $0.004$

Solution

$\begin{aligned} & \quad \delta_y=\left(2 \times 5.02^2-3 \times 5.02+4\right)-\left(2 \times 5^2-3 \times 5+4\right) \\ & =2 \times 0.02 \times 10.02-3 \times 0.02=0.4008-0.06=0.3408 . \\ & d y=\left(\frac{d y}{d x}\right) \times d x=(4 x-3) d x=(4 \times 5-3) \times 0.02 \\ & =17 \times 0.02=0.34=d y \\ & \therefore \delta_y-d y=0.3408-0.34=0.0008 .\end{aligned}$

Hence, the correct answer is option: 'A'.

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