Question:
The following operations are defined for real numbers.
\(a \,\#\, b = a + b\), if \(a\) and \(b\) both are positive, else \(a \,\#\, b = 1\).
\(a \,\triangledown\, b = (a \times b)^{a+b}\) if \(a \times b\) is positive, else \(a \,\triangledown\, b = 1\).
\[ \dfrac{(2 \,\#\, 1)}{(1 \,\triangledown\, 2)} = \]
The following operations are defined for real numbers.
\(a \,\#\, b = a + b\), if \(a\) and \(b\) both are positive, else \(a \,\#\, b = 1\).
\(a \,\triangledown\, b = (a \times b)^{a+b}\) if \(a \times b\) is positive, else \(a \,\triangledown\, b = 1\).
\[ \dfrac{(2 \,\#\, 1)}{(1 \,\triangledown\, 2)} = \]
\(a \,\#\, b = a + b\), if \(a\) and \(b\) both are positive, else \(a \,\#\, b = 1\).
\(a \,\triangledown\, b = (a \times b)^{a+b}\) if \(a \times b\) is positive, else \(a \,\triangledown\, b = 1\).
\[ \dfrac{(2 \,\#\, 1)}{(1 \,\triangledown\, 2)} = \]
Show Hint
Evaluate the numerator and the denominator separately with the two rules, after checking whether the conditions (both positive / product positive) hold.
Updated On: Nov 4, 2026
- \(\dfrac{1}{8}\)
- \(1\)
- \(\dfrac{3}{8}\)
- \(3\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Numerator \(2 \,\#\, 1\):
Both 2 and 1 are positive, so \(2 \,\#\, 1 = 2 + 1 = 3\).
Step 2: Denominator \(1 \,\triangledown\, 2\):
\(1 \times 2 = 2\) is positive, so \(1 \,\triangledown\, 2 = (1 \times 2)^{1+2} = 2^3 = 8\).
Step 3: Divide:
\[ \dfrac{2 \,\#\, 1}{1 \,\triangledown\, 2} = \dfrac{3}{8} \]
Final Answer:
The correct option is (3): \(\dfrac{3}{8}\)
Both 2 and 1 are positive, so \(2 \,\#\, 1 = 2 + 1 = 3\).
Step 2: Denominator \(1 \,\triangledown\, 2\):
\(1 \times 2 = 2\) is positive, so \(1 \,\triangledown\, 2 = (1 \times 2)^{1+2} = 2^3 = 8\).
Step 3: Divide:
\[ \dfrac{2 \,\#\, 1}{1 \,\triangledown\, 2} = \dfrac{3}{8} \]
Final Answer:
The correct option is (3): \(\dfrac{3}{8}\)
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