Q61 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
If a straight line passing through a fixed point $(a, b)$, where $a, b > 0$, makes positive intercepts OA and OB on the coordinate axes, then the least value of OA + OB is:
- Correct answerA$\left(\sqrt{a} + \sqrt{b}\right)^2$
- B$a + b$
- C$\left(\sqrt{a} + \sqrt{b}\right)^3$
- D$\left(\sqrt{a} - \sqrt{b}\right)^2$
Solution
Let the intercepts on the coordinate axes be $\alpha$ and $\beta$.
The equation of the line is $\dfrac{x}{\alpha} + \dfrac{y}{\beta} = 1$.
Since the line passes through $(a, b)$, we have $\dfrac{a}{\alpha} + \dfrac{b}{\beta} = 1$.
We need to find the minimum value of $OA + OB = \alpha + \beta$.
Using the Cauchy-Schwarz inequality:
$(\alpha + \beta) \left( \dfrac{a}{\alpha} + \dfrac{b}{\beta} \right) \ge \left( \sqrt{\alpha \cdot \dfrac{a}{\alpha}} + \sqrt{\beta \cdot \dfrac{b}{\beta}} \right)^2$
$(\alpha + \beta) (1) \ge \left(\sqrt{a} + \sqrt{b}\right)^2$
Thus, the least value of $OA + OB$ is $\left(\sqrt{a} + \sqrt{b}\right)^2$.
Answer: $\left(\sqrt{a} + \sqrt{b}\right)^2$
The equation of the line is $\dfrac{x}{\alpha} + \dfrac{y}{\beta} = 1$.
Since the line passes through $(a, b)$, we have $\dfrac{a}{\alpha} + \dfrac{b}{\beta} = 1$.
We need to find the minimum value of $OA + OB = \alpha + \beta$.
Using the Cauchy-Schwarz inequality:
$(\alpha + \beta) \left( \dfrac{a}{\alpha} + \dfrac{b}{\beta} \right) \ge \left( \sqrt{\alpha \cdot \dfrac{a}{\alpha}} + \sqrt{\beta \cdot \dfrac{b}{\beta}} \right)^2$
$(\alpha + \beta) (1) \ge \left(\sqrt{a} + \sqrt{b}\right)^2$
Thus, the least value of $OA + OB$ is $\left(\sqrt{a} + \sqrt{b}\right)^2$.
Answer: $\left(\sqrt{a} + \sqrt{b}\right)^2$
Hence, the correct answer is option: 'A'.