@albiesriram said:Here 8th is unsolved . OA for 7 is B solved by@Subhashdec2New set
is it option B for 8??
(-4,4)
albiesriram
8. option d
@Tusharrr said:If a,b and c are non-zero reals such that a+b+c=11and 1/a+1/b+1/c=0, what is the value of a^2+b^2+c^2?
121.
albiesriram
9 . option d
10. option a
@albiesriram said:Here 8th is unsolved . OA for 7 is B solved by@Subhashdec2New set
Q1 a1 + a3 + a5 = -12
=> a3 = -4
a1 = (-4 - 2d), a2 = (-4 - d)
a1*a2*a3 = 8
a1*a2 = -2
(4 + 2d)(4 + d) = -2
2d^2 + 12d + 18 = 0
d^2 + 6d + 9 = 0
d = -3
a2 = -1, a4 = -7, a6 = -13
a2 + a4 + a6 = -21
Q2 if x > 3, then
(x - 3) + ˆš(x^2 - 6x + 6) = 1
x^2 - 6x + 6 = x^2 - 8x + 16
x = 5 (doesn;t satisfy)
if x is less than or equal to 3, then
(3 - x) + ˆš(x^2 - 6x + 6) = 1
x^2 - 6x + 6 = x^2 - 4x + 4
x = 1 (doesn't satisfy)
No solution
how many integers satisfy the inequality
∣ 10(x+1)/(x^2+2x+3) ∣≥1?
∣ 10(x+1)/(x^2+2x+3) ∣≥1?
Details : ∣a∣ this line sign is of absolute of a
Tusharrr
18 integers????
OA for 9,10 is D n A 
11 &12
albiesriram
for quad. x^2+2x+3 -->no real roots hence always +ve(mod fn. useless)
check the solutions for X^2+2X+3 = 10(x+1) X >= -1
X^2+2X+3 = -10(x+1) X
@albiesriram said:OA for 9,10 is D n A11 &12
11. b?
substitute x=1, 2, 0...we get unique values for a and b
12. d?
x2 =0
x2 -x = 0
x2 -2x + 1 = 0
x2 +x + 1 = 0
albiesriram
11. option b
12. option c
