Q.

Let f(x)={ax2+2ax+34x2+4x-3,x≠-32,12b      ,x=-32,12

be continuous at x=-32. If fof(x)=75, then x is equal to:                          [2026]

1 1.4  
2 0  
3 1  
4 2  

Ans.

(3)

f(x)={ax2+2ax+3(2x-1)(2x+3);x≠-32, 12b  ;x=-32, 12

For continuity at x=-32

LHL = RHL

⇒ limx→-32(ax2+2ax+3)(2x-1)(2x+3)

At x=-32⇒Numerator =0

a(-32)2+2a(-32)+3=0

94a-3a+3=0

3a4=3⇒a=4

∴f(x)={4x2+8x+3(2x-1)(2x+3);x≠-32, 12b   ;x=-32, 12

f(x)={(2x+1)(2x+3)(2x-1)(2x+3);x≠-32, 12b    ;x=-32, 12
fof(x)=f(2x+12x-1)=2(2x+12x-1)+12(2x+12x-1)-1

=4x+22x-1+14x+22x-1-1=6x+12x-12x+32x-1

=6x+12x+3=75

⇒5(6x+1)=7(2x+3)

30x+5=14x+21

16x=16

⇒x=1

Ans. x=1