If 8A $△$ 6B and 3B $△$ 4C, therefore

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2C + A

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Solution

The correct option is A C $□$ A
As per the question
8A<6B and 3B <4C i.e. 4A<3B and 3B<4C i.e. 4A<3B<4C- Equation 1
Option A. C>A. This is true from the equation 1.
Option B. C<A. This is not true
Answer is Option A

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Similar questions

\to A = (\to a - 2 \to b + 3 \to c)

\to B = (- 2 \to a + 3 \to b + 2 \to c)

\to C = (- 8 \to a + 13 \to b)

\to A B

\to A C

Add
( 2a + 3b - 4c ) ; ( 8c + 4a - 5b ) and ( 6b + 5c - 2a )

3 a - 4 b + 4 c, 2 a + 3 b - 8 c, a - 6 b + c

2 a + 6 b - 3 (a + 3 b)^{2}

2 a + 6 b - 3 (a + 3 b)^{2}

= 2 (a + 3 b) - 3 (a + 3 b)^{2} [T a k i n g 2 c o m m o n f r o m 2 a + 6 b]

= (a + 3 b) {\begin{matrix} 2 - 3 (a + 3 b) \end{matrix}} [T a k i n g (a + 3 b) c o m m o n]

= (a + 3 b) (2 - 3 a - 9 b)

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