Let f ( x )=[2 x 2+1] and g ( x )= begincases2 x -3, x

Question

Let f ( x )=[2 x 2+1] and g ( x )= begincases2 x -3, x <0 2 x +3, x ≥ 0 endcases,, where [ t ] is the greatest integer ≤ t. Then, in the open interval (-1,1), the number of points where fog is discontinuous is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f ( g ( x ))=[2 g 2( x )]+1 = begincases [2(2 x -3)2]+1 ; x < 0 [2(2 x +3)2]+1 ; x ≥ 0 endcases ∴ fog is discontinuous whenever 2(2 x-3)2 or 2(2 x+3)2 belongs to integer except x=0. ∴ 62 points of discontinuity.

Q. Let $f (x) = [2 x^{2} + 1]$ and $g (x) = {2 x - 3, 2 x + 3, x < 0 x \geq 0,$ , where $[t]$ is the greatest integer $\leq t$ . Then, in the open interval $(- 1, 1)$ , the number of points where fog is discontinuous is equal to_____

$f (g (x)) = [2 g^{2} (x)] + 1$
$= {[2 (2 x - 3)^{2}] + 1; x < 0 [2 (2 x + 3)^{2}] + 1; x \geq 0$
$∴$ fog is discontinuous whenever $2 (2 x - 3)^{2}$ or $2 (2 x + 3)^{2}$ belongs to integer except $x = 0$ .
$∴ 62$ points of discontinuity.

Q. Let f(x)=[2x2+1] and g(x)={2x−3,2x+3,​x<0x≥0​,, where [t] is the greatest integer ≤t. Then, in the open interval (−1,1), the number of points where fog is discontinuous is equal to_____

f(g(x))=[2g2(x)]+1 ={[2(2x−3)2]+1;x<0[2(2x+3)2]+1;x≥0​ ∴ fog is discontinuous whenever 2(2x−3)2 or 2(2x+3)2 belongs to integer except x=0. ∴62 points of discontinuity.

Q. Let $f (x) = [2 x^{2} + 1]$ and $g (x) = {2 x - 3, 2 x + 3, x < 0 x \geq 0,$ , where $[t]$ is the greatest integer $\leq t$ . Then, in the open interval $(- 1, 1)$ , the number of points where fog is discontinuous is equal to_____

$f (g (x)) = [2 g^{2} (x)] + 1$
$= {[2 (2 x - 3)^{2}] + 1; x < 0 [2 (2 x + 3)^{2}] + 1; x \geq 0$
$∴$ fog is discontinuous whenever $2 (2 x - 3)^{2}$ or $2 (2 x + 3)^{2}$ belongs to integer except $x = 0$ .
$∴ 62$ points of discontinuity.