TS ICET 23 May 2018 Shift 2 Paper

Question : 132 of 200

Marks: +1, -0

[\table\ a, b^2;c^3,0]

[\table\2,9;-8,0]

where a. b, c are real numbers of which b < 0 then 3a +b + c

Solution:

Given

[\table\ a, b^2;c^3,0]=[\table\2,9;-8,0]

⇒ a=2 ; b^{2}=9 ⇒ b=-3

c^{3}=-8 ⇒ c=-2

3 a+b+c=3(2)-3-2=1

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