MHTCET 5 May 2025 Shift 2 Paper

Concept:The slope of the tangent to a curve is given by its derivative. To find the minimum slope, we differentiate the slope function and set its derivative to zero.Explanation:The curve is $y = x^{3} - 3x^{2} + 2x + 93$.Differentiate $y$ with respect to $x$ to get the slope $T$:$$T = \frac{dy}{dx} = 3x^{2} - 6x + 2$$Now find $\frac{dT}{dx}$ to locate the critical point:$$\frac{dT}{dx} = 6x - 6$$Set $\frac{dT}{dx} = 0$:$$6x - 6 = 0 \Rightarrow x = 1$$Check that this is a minimum by finding $\frac{d^{2}T}{dx^{2}}$:$$\frac{d^{2}T}{dx^{2}} = 6 > 0$$Since the second derivative is positive, $x = 1$ gives the minimum slope.Substitute $x = 1$ into $T$:$$T = 3(1)^{2} - 6(1) + 2 = 3 - 6 + 2 = -1$$Therefore, the minimum value of the slope is $-1$.Answer:The minimum value of the slope is $-1$, which corresponds to option B.