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Range and limit of h(x) = (f(x) + 2)/g(x) | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let α\alpha, β\beta, γ\gamma be the roots of the equation f(x)=3x3−13x2+14x−2=0f(x) = 3x^3-13x^2+14x-2 = 0. Let [α]\left[\alpha\right], [β]\left[\beta\right], [γ]\left[\gamma\right] be the roots of the cubic polynomial equation g(x)=0g(x) = 0; then for h(x)=f(x)+2g(x)h(x) = \dfrac{f(x)+2}{g(x)} (where [ . ][\,.\,] represents the greatest integer function)
Alim⁡x→∞h(x)\displaystyle\lim_{x\to\infty}h(x) does not exist
Brange of h(x)h(x) is R−{−1,3,7}\mathbb R-\left\{-1,3,7\right\}correct
Crange of h(x)h(x) is R−{0,1,2}\mathbb R-\left\{0,1,2\right\}
Dlim⁡x→∞h(x)=3\displaystyle\lim_{x\to\infty}h(x) = 3correct
Solution
Step 1:
3x3−13x2+14x=x(3x2−13x+14)=x(x−2)(3x−7)3x^3-13x^2+14x = x\left(3x^2-13x+14\right) = x\left(x-2\right)\left(3x-7\right)
 so
f(x)=x(x−2)(3x−7)−2,f(x)+2=x(x−2)(3x−7)f(x) = x\left(x-2\right)\left(3x-7\right)-2 ,\qquad f(x)+2 = x\left(x-2\right)\left(3x-7\right)
 Step 2:
f(0)=−2<0,f(1)=3−13+14−2=2>0,f(2)=24−52+28−2=−2<0f(0) = -2<0 ,\quad f(1) = 3-13+14-2 = 2>0 ,\quad f(2) = 24-52+28-2 = -2<0
 f(3)=4>0f(3) = 4>0 so ff has a root in each of (0,1)\left(0,1\right), (1,2)\left(1,2\right), (2,3)\left(2,3\right); a cubic has at most three roots, so these are α\alpha, β\beta, γ\gamma (numerically 0.160.16, 1.471.47, 2.692.69). Hence
[α]=0,[β]=1,[γ]=2\left[\alpha\right] = 0 ,\qquad \left[\beta\right] = 1 ,\qquad \left[\gamma\right] = 2
 Step 3:
g(x)=x(x−1)(x−2)g(x) = x\left(x-1\right)\left(x-2\right)
h(x)=x(x−2)(3x−7)x(x−1)(x−2)=3x−7x−1(x≠0,1,2)h(x) = \frac{x\left(x-2\right)\left(3x-7\right)}{x\left(x-1\right)\left(x-2\right)} = \frac{3x-7}{x-1} \qquad \left(x \ne 0,1,2\right)
 The factors xx and x−2x-2 cancel, but x=0x = 0 and x=2x = 2 remain excluded from the domain, since there hh is 00\tfrac00. Step 4:
h(x)=3x−7x−1=3−4x−1 ⟶ 3as x→∞h(x) = \frac{3x-7}{x-1} = 3-\frac4{x-1} \ \longrightarrow\ 3 \qquad\text{as } x\to\infty
 (4) is TRUE, (1) is FALSE. Step 5: y=3−4x−1y = 3-\dfrac4{x-1} is a bijection from R∖{1}\mathbb R\setminus\left\{1\right\} onto R∖{3}\mathbb R\setminus\left\{3\right\}. Removing the two extra excluded points,
x=0 ↦ 3−4−1=7,x=2 ↦ 3−41=−1x = 0 \ \mapsto\ 3-\frac4{-1} = 7 ,\qquad x = 2 \ \mapsto\ 3-\frac41 = -1
 so those two values are not attained either. Hence
range(h)=R−{−1, 3, 7}.\text{range}\left(h\right) = \mathbb R-\left\{-1,\ 3,\ 7\right\} .
 (2) is TRUE, (3) is FALSE.
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