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When do sqrt(x + 3) and sqrt(1 - x) + f(k) meet | JEE Advanced

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Question
The graphs of y=x+3y = \sqrt{x+3} and y=1−x+f(k)y = \sqrt{1-x}+f(k) intersect, where kk is a real parameter.
Aif f(k)=k2−7k+12f(k) = k^2-7k+12, then the maximum value of kk is 55correct
Bif f(k)=kf(k) = k, then the number of integral values of kk is 55correct
Cif f(k)=k2−7k+12f(k) = k^2-7k+12, then the minimum value of kk is 22correct
Dif f(k)=kf(k) = k, then the number of integral values of kk is 22
Solution
Question attachment
x+3≥0  and  1−x≥0⟹x∈[−3,1]x+3 \ge 0 \ \text{ and }\ 1-x \ge 0 \quad\Longrightarrow\quad x \in \left[-3,1\right]
 Step 2: The graphs meet iff
x+3=1−x+f(k)for some x∈[−3,1]\sqrt{x+3} = \sqrt{1-x}+f(k) \quad\text{for some } x \in \left[-3,1\right]
 i.e. iff f(k)f(k) lies in the range of
h(x)=x+3−1−xh(x) = \sqrt{x+3}-\sqrt{1-x}
 Step 3: hh is continuous on [−3,1]\left[-3,1\right] and strictly increasing (the first term increases, the second decreases), with
h(−3)=0−2=−2,h(1)=2−0=2h\left(-3\right) = 0-2 = -2 ,\qquad h\left(1\right) = 2-0 = 2
 So the range is
[−2, 2]\left[-2,\ 2\right]
 and the condition is
−2≤f(k)≤2-2 \le f(k) \le 2
 Step 4: Case-I: f(k)=k2−7k+12f(k) = k^2-7k+12.
k2−7k+12≥−2 ⟺ k2−7k+14≥0k^2-7k+12 \ge -2 \ \Longleftrightarrow\ k^2-7k+14 \ge 0
 whose discriminant is 49−56=−7<049-56 = -7<0, so this holds for every real kk.
k2−7k+12≤2 ⟺ k2−7k+10≤0 ⟺ (k−2)(k−5)≤0 ⟺ 2≤k≤5k^2-7k+12 \le 2 \ \Longleftrightarrow\ k^2-7k+10 \le 0 \ \Longleftrightarrow\ \left(k-2\right)\left(k-5\right) \le 0 \ \Longleftrightarrow\ 2 \le k \le 5
 So the maximum of kk is 55, (1) TRUE, and the minimum is 22, (3) TRUE. Step 5: Case-II: f(k)=kf(k) = k.
−2≤k≤2⟹k∈{−2,−1,0,1,2}-2 \le k \le 2 \quad\Longrightarrow\quad k \in \left\{-2,-1,0,1,2\right\}
 which is 55 integral values. (2) TRUE, (4) FALSE. Answer: (1), (2), (3)
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