<p><strong>Question nos. 663 to 665</strong><br>Match the condition of column-I with corresponding number of real roots of \(f(x) = 0\) in column-II and number of points of non-derivability of \(y = f(|x|)\) in column-III, where \(f(x) = ax^2 + bx + c\).</p><p><strong>Column-I</strong><br>(I) \(a^2 + b^2 + c^2 - ab - bc - ca \leq 0\)<br>(II) \(a^2 + b^2 + c^2 + ab + bc + ca \leq 0\)<br>(III) \(3(a^2 + b^2 + c^2 + 1) \leq 2(a + b + c + ab + bc + ca)\)<br>(IV) \(a^2 + b^2 + c^2 \leq 2a + 6b + 4c + 14\)</p><p><strong>Column-II</strong><br>(i) 0 (ii) 1 (iii) 2 (iv) \(\infty\)</p><p><strong>Column-III</strong><br>(P) 0 (Q) 1 (R) 3 (S) 5</p><p><strong>Q665.</strong> Which of the following is <strong>correct</strong> combination?</p>
Step-by-Step Solution
Key Concept: Recognize that algebraic expressions involving a, b, c can be rewritten as sums of squares to determine constraints on the coefficients. Then analyze the number of real roots using the discriminant and count non-derivability points of y = f(|x|) by examining where the derivative fails to exist.
<p><strong>Step 1: Analyze Condition (IV)</strong></p><p>Given: $a^2 + b^2 + c^2 \leq 2a + 6b + 4c + 14$</p><p>Rearrange: $a^2 - 2a + b^2 - 6b + c^2 - 4c \leq 14$</p><p>Complete the square: $(a-1)^2 - 1 + (b-3)^2 - 9 + (c-2)^2 - 4 \leq 14$</p><p>$(a-1)^2 + (b-3)^2 + (c-2)^2 \leq 28$</p><p>This defines a sphere in (a,b,c) space with center (1,3,2) and radius $2\sqrt{7}$.</p><p><strong>Step 2: Find discriminant constraint</strong></p><p>For $f(x) = ax^2 + bx + c$, discriminant $\Delta = b^2 - 4ac$</p><p>From the sphere constraint, we need to find if $\Delta$ can equal specific values.</p><p>Testing the center (1,3,2): $\Delta = 9 - 4(1)(2) = 9 - 8 = 1 > 0$</p><p>This gives 2 real roots for $f(x) = 0$ (Column-II: (iii)).</p><p><strong>Step 3: Count non-derivability points of y = f(|x|)</strong></p><p>The function $y = f(|x|) = a|x|^2 + b|x| + c$ is an even function.</p><p>Non-derivability occurs at:</p><p>• $x = 0$ (always, due to the corner from |x|) — 1 point</p><p>• Points where $f(|x|) = 0$, if they exist at $x = \pm\alpha$ with $\alpha \neq 0$ — adds 2 more points</p><p>Total: $1 + 2 = 3$ points of non-derivability (Column-III: (R))</p><p>When $f(x) = 0$ has two distinct positive roots, symmetry about the y-axis in the |x| domain gives 3 non-derivability points total.</p><p><strong>Step 4: Match with options</strong></p><p>Condition (IV) gives:</p><p>• Number of real roots of $f(x) = 0$: 2 roots → (iii) in Column-II</p><p>• Points of non-derivability of $y = f(|x|)$: 3 points → (R) in Column-III</p><p>Therefore: (IV) (iii) (R)</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C