Question

If $y=2x^3-2x^2+3x-5$, then for $x=2$ and $∆x=0.1$, value of $∆y$ is

• A. $2.002$
• B. $1.9$
• C. $0$
• D. $0.9$

Answer 1

The correct option is B

$1.9$

Explanation for the correct option.

Find the value of $∆y$.

Differentiate the equation $y=2x^3-2x^2+3x-5$ with respect to $x$.

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}(2x^3-2x^2+3x-5) \\ \Rightarrow \frac{∆y}{∆x}=2\times 3x^2-2\times 2x+3-0 \\ \Rightarrow \frac{∆y}{∆x}=6x^2-4x+3 \\ \Rightarrow ∆y=\left(6x^2-4x+3\right)∆x \end{gathered}$$

Now substitute $2$ for $x$ and $0.1$ for $∆x$ to find the value of $∆y$.

$\begin{array}{rcl}∆y& =& \left(6\times 2^2-4\times 2+3\right)\times 0.1\\ & =& \left(24-8+3\right)\times 0.1\\ & =& 19\times 0.1\\ & =& 1.9\end{array}$

Hence, the correct option is B.