UPSC NDA II 2024 Mathematics Paper

Concept:A quadratic equation has equal roots when its discriminant is zero. This condition leads to a relation among the coefficients $a^{2}, b^{2}, c^{2}$ that determines whether they are in arithmetic, geometric, or harmonic progression.Explanation:Given the quadratic equation:$a^{2}(b^{2}-c^{2})x^{2}+b^{2}(c^{2}-a^{2})x+c^{2}(a^{2}-b^{2})=0$For equal roots, the discriminant $D$ must be zero:$D = [b^{2}(c^{2}-a^{2})]^{2} - 4a^{2}c^{2}(b^{2}-c^{2})(a^{2}-b^{2}) = 0$Simplify $D=0$ step by step:$b^{4}(c^{2}-a^{2})^{2} - 4a^{2}c^{2}(b^{2}-c^{2})(a^{2}-b^{2}) = 0$Expanding and rearranging leads to:$b^{4}(c^{2}+a^{2})^{2} = 4a^{4}c^{4}$Taking square roots (positive value since all terms are squares):$b^{2}(a^{2}+c^{2}) = 2a^{2}c^{2}$Divide both sides by $a^{2}b^{2}c^{2}$ (assuming none are zero):$\frac{2}{b^{2}} = \frac{1}{a^{2}} + \frac{1}{c^{2}}$This relation means the reciprocals $\frac{1}{a^{2}}, \frac{1}{b^{2}}, \frac{1}{c^{2}}$ are in arithmetic progression.Hence $a^{2}, b^{2}, c^{2}$ are in harmonic progression (HP).Answer:$a^{2}, b^{2}, c^{2}$ are in HP. Therefore, option C is correct.