MHTCET 26 Apr 2025 Shift 2 Paper

Concept:For the angle between two vectors to be obtuse, their dot product must be negative.Explanation:Given vectors:$$\overline{a}=2 x^{2} \hat{\mathrm{i}}+4 x \hat{\mathrm{j}}+\hat{\mathrm{k}}$$$$\overline{b}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+x \hat{\mathrm{k}}$$Compute the dot product:$$\overline{a} \cdot \overline{b} = (2x^2)(7) + (4x)(-2) + (1)(x)$$$$= 14x^2 - 8x + x = 14x^2 - 7x$$For an obtuse angle, we require:$$14x^2 - 7x < 0$$Factorize the expression:$$7x(2x - 1) < 0$$Since $7 > 0$, divide both sides by 7:$$x(2x - 1) < 0$$The critical points are $x = 0$ and $x = \frac{1}{2}$.Check the sign of $x(2x - 1)$ in the intervals:For $x < 0$, the product is positive.For $0 < x < \frac{1}{2}$, the product is negative.For $x > \frac{1}{2}$, the product is positive.Thus, the inequality holds for:$$0 < x < \frac{1}{2}$$Answer:The required values of $x$ are $0 < x < \frac{1}{2}$.Therefore, Option A is correct.