If a∆b=a^{2}+b^{2}-3ab for any real number a and b, then \{1∆(-1)\}∆\\sqrt{2}∆\sqrt{2}\}=

AP ICET 2017 Shift 1 Paper

Question : 74 of 200

Marks: +1, -0

a∆b=a^{2}+b^{2}-3ab

for any real number a and b, then

\{1∆(-1)\}∆\{√{2}∆√{2}\}=

Solution:

Given :

a∆b=a^2+b^2-3ab=(a–b)^2-ab

∴1∆(-1)={1-(-1)}^2-1(-1)=4+1=5

∴\{5-(-2)^2\}-5(-2)=7^2+10=59

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