WBJEE 2024 Math Paper

To find the values of $a$ that satisfy the inequality $\frac{1}{\sqrt{a}} \int\limits_{1}^{a} \left( \frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}} \right) dx < 4$, we need to first evaluate the integral and then solve the inequality. Consider the integral $\int\limits_{1}^{a} \left( \frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}} \right) dx$ We will solve this integral term by term:For the term $\frac{3}{2} \sqrt{x}$ :For the term 1 :$\int 1 \, dx = x$For the term $-\frac{1}{\sqrt{x}}$ :$\int -\frac{1}{\sqrt{x}} \, dx = -\int x^{-1/2} \, dx = -2 x^{1/2}$Now, combining these results, we have: Evaluating this at the bounds $x=1$ and $x=a$ $\left( x^{3/2}+x-2x^{1/2} \right) \big|_{1}^{a} = \left( a^{3/2+a-2a^{1/2}} \right) - \left( 1^{3/2}+1-2 \cdot 1^{1/2} \right)$ $= a^{3/2} + a - 2a^{1/2} - (1+1-2)$$= a^{3/2} + a - 2a^{1/2} - 0$$= a^{3/2} + a - 2a^{1/2}$ Now, plugging this back into the inequality:$\frac{1}{\sqrt{a}} (a^{3/2}+a-2a^{1/2}) < 4$ Simplify the expression inside the inequality:Now, let $\sqrt{a} = t$. Hence, $a = t^2$. Substituting $t$ into the inequality:$t^2 + t - 2 < 4$Rearranging the inequality: $t^2 + t - 6 < 0$We solve the quadratic equation $t^2 + t - 6 = 0$ by factoring it:$(t+3)(t-2) = 0$The roots are $t = -3$ and $t = 2$. The quadratic inequality $t^2 + t - 6 < 0$ is satisfied between the roots: $-3 < t < 2$Since $\sqrt{a} = t$, this translates to:$-3 < \sqrt{a} < 2$However, since $\sqrt{a}$ represents a real, non-negative quantity, the inequality reduces to:$0 \leq \sqrt{a} < 2$Squaring both sides to return to $a$-terms:$0 \leq a < 4$The appropriate interval for $a$ is:Therefore, the values of $a$ that satisfy the original inequality lie in the interval $[0,4)$. Since $a$ cannot be exactly 0 (as it would make the original integral undefined), the interval we need is:Option C: $(0,4)$