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Question

The number of real solutions of (x1)(x+1)(2x+1)(2x3)=15 is
  1. 0
  2. 2
  3. 3
  4. 4


Solution

The correct option is B 2
(x1)(x+1)(2x+1)(2x3)=15(x1)(2x+1)(x+1)(2x3)=15
(2x2x1)(2x2x3)=15   
Assuming 2x2x=y
(y1)(y3)=15y24y+3=15
y24y12=0(y6)(y+2)=0
y=6,2
Now,
when y=6
2x2x=62x2x6=0(2x+3)(x2)=0
x=32,2
and
when y=2
2x2x=22x2x+2=0
D=14(2)(2)<0
No real roots.

Hence, the equation has 2 real roots.

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