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Limits: Range Function Denotes Greatest Integer Function

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

Question
Range of the function f(x)=[1ln⁡(x2+e)]+11+x2f(x)=\left[\dfrac{1}{\ln(x^{2}+e)}\right]+\dfrac{1}{\sqrt{1+x^{2}}}, [⋅][\cdot] denotes greatest integer function and e=lim⁡α→0(1+α)1/αe=\displaystyle\lim_{\alpha\to 0}(1+\alpha)^{1/\alpha}, is
A(0,e+1e]∪{2}\left(0,\dfrac{e+1}{e}\right]\cup\{2\}
B(0,1)(0,1)
C(0,1]∪{2}(0,1]\cup\{2\}
D(0,1)∪{2}(0,1)\cup\{2\}correct
Solution
Step 1: Note e≈2.718e\approx 2.718, so ln⁡e=1\ln e=1. For x=0x=0: ln⁡(0+e)=ln⁡e=1\ln(0+e)=\ln e=1, so 1ln⁡(x2+e)=1\dfrac{1}{\ln(x^{2}+e)}=1, and [1]=1[1]=1. Step 2: For x≠0x\neq 0: x2+e>ex^{2}+e>e, so ln⁡(x2+e)>1\ln(x^{2}+e)>1, making 1ln⁡(x2+e)<1\dfrac{1}{\ln(x^{2}+e)}<1 (and positive). So [1ln⁡(x2+e)]=0\left[\dfrac{1}{\ln(x^{2}+e)}\right]=0 for x≠0x\neq 0. Step 3: At x=0x=0: f(0)=1+11=2f(0)=1+\dfrac{1}{1}=2. Step 4: For x≠0x\neq 0: f(x)=0+11+x2f(x)=0+\dfrac{1}{\sqrt{1+x^{2}}}. Since x≠0x\neq 0, 1+x2>1\sqrt{1+x^{2}}>1, so 11+x2∈(0,1)\dfrac{1}{\sqrt{1+x^{2}}}\in(0,1). Step 5: Range: (0,1)∪{2}(0,1)\cup\{2\}. Correct answer: (4) (0,1)∪{2}(0,1)\cup\{2\}
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