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Limits — JEE Maths practice question

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

Question

Range of the function

f(x)=\left[\dfrac{1}{\ln(x^{2}+e)}\right]+\dfrac{1}{\sqrt{1+x^{2}}}

[\cdot]

denotes greatest integer function and

e=\displaystyle\lim_{\alpha\to 0}(1+\alpha)^{1/\alpha}

, is

\left(0,\dfrac{e+1}{e}\right]\cup\{2\}

(0,1)

(0,1]\cup\{2\}

(0,1)\cup\{2\}

correct

Solution

Step 1: Note

e\approx 2.718

, so

\ln e=1

. For

x=0

\ln(0+e)=\ln e=1

, so

\dfrac{1}{\ln(x^{2}+e)}=1

, and

[1]=1

. Step 2: For

x\neq 0

x^{2}+e>e

, so

\ln(x^{2}+e)>1

, making

\dfrac{1}{\ln(x^{2}+e)}<1

(and positive). So

\left[\dfrac{1}{\ln(x^{2}+e)}\right]=0

for

x\neq 0

. Step 3: At

x=0

f(0)=1+\dfrac{1}{1}=2

. Step 4: For

x\neq 0

f(x)=0+\dfrac{1}{\sqrt{1+x^{2}}}

. Since

x\neq 0

\sqrt{1+x^{2}}>1

, so

\dfrac{1}{\sqrt{1+x^{2}}}\in(0,1)

. Step 5: Range:

(0,1)\cup\{2\}

. Correct answer: (4)

(0,1)\cup\{2\}

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