CUET Preparation Today
The maximum value of $2x^3-24x+107$ in the interval [1, 3] is : |
87 89 90 85 |
89 |
The correct answer is option (2) → 89 $y=2x^3-24x+107$ $\frac{dy}{dx}=0⇒6x^2-24=0$ $x=±2$ checking at $x=1,2,3$ as $-2∉[1,3]$ $y(1)=85,y(2)=75,y(3)=89$ $y(3)=89$ → max. value |
