Mathematics — Standard (041) — CBSE Class 10 board question

(d) $-\dfrac{1}{2}$

For equal roots, discriminant $= 0$: $b^2 - 4ac = 0 \Rightarrow 1 - 4a^2 = 0 \Rightarrow a = \pm\dfrac{1}{2}$. For positive roots, $x = \dfrac{-b}{2a} = \dfrac{-1}{2a} > 0$, so $a$ must be negative. Thus $a = -\dfrac{1}{2}$.

Source: Chapter 4, Section 4.4

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Two conditions must both be satisfied:

1. Equal roots → Discriminant $= 0$: $1 - 4a^2 = 0$ gives $a = \pm\tfrac{1}{2}$.
2. Positive roots → The repeated root is $\tfrac{-b}{2a} = \tfrac{-1}{2a}$; for this to be positive, $a$ must be negative → $a = -\tfrac{1}{2}$.

Many students stop at step 1 and pick $+\tfrac{1}{2}$, missing the positivity condition. Always check both conditions.