Question:easy

If \( f(a,b) = a^2+b^2 \) and \( g(a,b) = \frac{2}{b^2}\left[a^2+b^2\right] \), then what is the value of \( f(6, 3) - g(8, 4) \)?

Show Hint

Substitute the given values directly into each function definition, simplify, then subtract.
Updated On: Jul 21, 2026
  • 30
  • 35
  • 40
  • 45
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Expand f(6, 3) term by term.
$f(a,b) = a^2 + b^2$. With $a=6$ and $b=3$: $a^2 = 36$ and $b^2 = 9$.
Adding these, $f(6,3) = 36+9 = 45$.

Step 2: Expand g(8, 4) term by term.
$g(a,b) = \frac{2}{b^2}(a^2+b^2)$. With $a=8$ and $b=4$: $a^2=64$, $b^2=16$, so $a^2+b^2=80$.
The multiplier is $\frac{2}{b^2} = \frac{2}{16} = \frac{1}{8}$.
So $g(8,4) = \frac{1}{8}\times 80 = 10$.

Step 3: Take the difference.
$f(6,3) - g(8,4) = 45-10 = 35$.
A quick check rules out 40 and 45: 45 ignores g, and 40 would need g to be 5, which the fraction does not give.

Final Answer:
The required value is 35, matching option (b). \[ \boxed{35} \]
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