CUET Preparation Today
If $f(x) =3x^2 +10x-5,$ then by using the concept of differentials, the approximate value of $f(2.03)$ is : |
37.66 47.66 17.66 27.66 |
27.66 |
The correct answer is Option (4) → 27.66 $2.03=2+0.03$ $x=2$ $Δx=0.03$ so $f(x.Δx)=f(x)+\left.\frac{df}{dx}\right]_x(Δx)$ so $f(2.03)=f(2)+(6x+10)_{x=2}(0.03)$ $=27+22×0.03$ $=27.66$ |
