Q. 1264.6( 5 Votes )

If ∆ is an operation such that for integers a and b we have

a ∆ b = a × b – 2 × a × b + b × b (–a) × b + b × b

then find (i) 4 (– 3)

(ii) (– 7) ∆ (– 1)

Also show that 4 ∆ (– 3) ≠ (– 3) ∆ 4

and (– 7) ∆ (– 1) ≠ ( – 1) ∆ (– 7)

Answer :

We have been given that,

a ∆ b = a × b – 2 × a × b + b × b (-a) × b + b × b


Simplifying it, we get


a ∆ b = a × b – 2 × a × b – a × b3 + b2 …(A)


Apply the same formula in the questions that follows:


(i). We have 4 ∆ (-3).


Put a = 4 and b = -3 in equation (A), we get


4 ∆ (-3) = 4 × (-3) – 2 × 4 × (-3) – 4 × (-3)3 + (-3)2


4 ∆ (-3) = -12 + 24 + 108 + 9


4 ∆ (-3) = 129


Thus, answer is 129.


(ii). We have (-7) ∆ (-1).


Put a = -7 and b = -1 in equation (A), we get


(-7) ∆ (-1) = (-7) × (-1) – 2 × (-7) × (-1) – (-7) × (-1)3 + (-1)2


(-7) ∆ (-1) = 7 – 14 – 7 + 1


(-7) ∆ (-1) = -13


Thus, the answer is -13.


To show: 4 ∆ (-3) ≠ (-3) ∆ 4


LHS: 4 ∆ (-3) = 129 [from answer of part (i)]


RHS: (-3) ∆ 4 = (-3) × 4 – 2 × (-3) × 4 – (-3) × (4)3 + (4)2 [by putting a = -3 and b = 4 in equation A]


(-3) ∆ 4 = -12 + 24 + 192 + 16


(-3) ∆ 4 = 220


Comparing LHS and RHS, we see that


LHS ≠ RHS [ 129 ≠ 220]


To show: (-7) ∆ (-1) ≠ (-1) ∆ (-7)


LHS: (-7) ∆ (-1) = -13 [from answer of part (ii)]


RHS: (-1) ∆ (-7) = (-1) × (-7) – 2 × (-1) × (-7) – (-1) × (-7)3 + (-7)2


(-1) ∆ (-7) = 7 – 14 – 343 + 49


(-1) ∆ (-7) = -301


Clearly, LHS ≠ RHS [ -13 ≠ -301]


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