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  2. Please help! - Sets, Relations and Functions - JEE Main-5
# Engineering

The domain of the definition of the function 

f(x)=\frac{1}{4-x^{2}}+ \log_{10}(x^{3}-x)  is :

  • Option 1)

    (-1,0)\cup (1,2)\cup (3,\infty )

  • Option 2)

    (-2,-1)\cup (-1,0)\cup (2,\infty )

  • Option 3)

    (-1,0)\cup (1,2)\cup (2,\infty )

  • Option 4)

    (1,2)\cup (2,\infty )

 

Answers (1)
A Aadil Khan

\\f(x)=\frac{1}{4-x^{2}}+\log_{10}(x^{3}-x)\\\\\\\:(4-x^{2} )\neq 0\\\\\\\:(2-x)(2+x) \neq 0\\\\\\\:  

    x\neq 2,-2

                                                            x\epsilon R-\left \{ 2,-2 \right \}

(x^{3}-x)\:\:so                                                 \downarrow

=x(x^{2}-1)                             

=x(x+1)(x-1)>0

x\epsilon (-1,0)\cup (1,2)\cup (2,\infty )

Domain :  x\epsilon (-1,0)\cup (1,2)\cup (2,\infty )


Option 1)

(-1,0)\cup (1,2)\cup (3,\infty )

Option 2)

(-2,-1)\cup (-1,0)\cup (2,\infty )

Option 3)

(-1,0)\cup (1,2)\cup (2,\infty )

Option 4)

(1,2)\cup (2,\infty )

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