Q57 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $f(x) = x^2$ and $g(x) = [x^2]$ where $[\cdot]$ represents the greatest integer function then, $(f \circ g)\left(\dfrac{3}{2}\right) + (g \circ f)\left(\dfrac{3}{2}\right)$ is equal to ...
- A$2$
- B$5$
- Correct answerC$9$
- D$10$
Solution
Given $f(x) = x^2$ and $g(x) = [x^2]$.
First, evaluating $(f \circ g)\left(\dfrac{3}{2}\right)$:
$g\left(\dfrac{3}{2}\right) = \left[\left(\dfrac{3}{2}\right)^2\right] = \left[\dfrac{9}{4}\right] = [2.25] = 2$
$f\left(g\left(\dfrac{3}{2}\right)\right) = f(2) = 2^2 = 4$
Next, evaluating $(g \circ f)\left(\dfrac{3}{2}\right)$:
$f\left(\dfrac{3}{2}\right) = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$
$g\left(f\left(\dfrac{3}{2}\right)\right) = g\left(\dfrac{9}{4}\right) = \left[\left(\dfrac{9}{4}\right)^2\right] = \left[\dfrac{81}{16}\right] = [5.0625] = 5$
Adding both values:
$(f \circ g)\left(\dfrac{3}{2}\right) + (g \circ f)\left(\dfrac{3}{2}\right) = 4 + 5 = 9$
Answer: $9$
First, evaluating $(f \circ g)\left(\dfrac{3}{2}\right)$:
$g\left(\dfrac{3}{2}\right) = \left[\left(\dfrac{3}{2}\right)^2\right] = \left[\dfrac{9}{4}\right] = [2.25] = 2$
$f\left(g\left(\dfrac{3}{2}\right)\right) = f(2) = 2^2 = 4$
Next, evaluating $(g \circ f)\left(\dfrac{3}{2}\right)$:
$f\left(\dfrac{3}{2}\right) = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$
$g\left(f\left(\dfrac{3}{2}\right)\right) = g\left(\dfrac{9}{4}\right) = \left[\left(\dfrac{9}{4}\right)^2\right] = \left[\dfrac{81}{16}\right] = [5.0625] = 5$
Adding both values:
$(f \circ g)\left(\dfrac{3}{2}\right) + (g \circ f)\left(\dfrac{3}{2}\right) = 4 + 5 = 9$
Answer: $9$
Hence, the correct answer is option: 'C'.