Official PYQ · 2021Set 2 · Q4MCQ1 mark

GATE CS 2021 Set 2 Q4 — General Aptitude

If $\left( x – \dfrac{1}{2} \right)^2 – \left( x- \dfrac{3}{2} \right) ^2 = x+2$ , then the value of $x$ is:

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Answer: B

We are given the algebraic equation:

\left(x - \frac{1}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2 = x + 2

We can simplify the left-hand side (LHS) using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$ :

Let $a = x - \frac{1}{2}$ and $b = x - \frac{3}{2}$ .

Difference ( $a - b$ ): $a - b = \left(x - \frac{1}{2}\right) - \left(x - \frac{3}{2}\right) = -\frac{1}{2} + \frac{3}{2} = 1$
Sum ( $a + b$ ): $a + b = \left(x - \frac{1}{2}\right) + \left(x - \frac{3}{2}\right) = 2x - 2$

Substituting these back into the identity:

\text{LHS} = (a - b)(a + b) = 1 \cdot (2x - 2) = 2x - 2

Now, equate the simplified LHS to the right-hand side (RHS):

2x - 2 = x + 2

Solving for $x$ :

2x - x = 2 + 2 \implies x = 4

Thus, the value of $x$ is $4$ .

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Published 10 Oct 2026

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