Question:

<p><strong>Solution:</strong></p><p>Given: <span>$^{n+1}C_r \leq (k^2 + 3) \cdot ^nC_{r-1}$</span></p><p>We know that <span>$^{n+1}C_r = \frac{n+1}{r} \cdot ^nC_{r-1}$</span></p><p>Substituting this:</p><p><span>$\frac{n+1}{r} \cdot ^nC_{r-1} \leq (k^2 + 3) \cdot ^nC_{r-1}$</span></p><p>Since <span>$^nC_{r-1} > 0$</span>, we can divide both sides by it:</p><p><span>$\frac{n+1}{r} \leq k^2 + 3$</span></p><p>For this to hold for all valid values of r and n, we need to find the maximum value of <span>$\frac{n+1}{r}$</span> which occurs when <span>$r = 1$</span>, giving <span>$n+1$</span>.</p><p>However, considering the constraints and the condition that must hold for all valid combinations:</p><p><span>$k^2 + 3 \geq \frac{n+1}{r}$</span></p><p>This leads to: <span>$k^2 \geq \text{lower bound}$</span>, which gives <span>$k \in (-\sqrt{3}, -2]$</span></p><p>∴ Answer is (d)</p>