Set, Relations and FunctionsMTP Oct 21Question 1981 of 136
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Find gf\displaystyle g \circ f for the functions f(x)=x\displaystyle f(x) = \sqrt{x}, g(x)=2x2+1\displaystyle g(x) = 2x^2+1.

Options

A2x2+1\displaystyle 2x^2+1
B2x+1\displaystyle 2x+1
C(2x2+1)(x)\displaystyle (2x^2+1)(\sqrt{x})
D10\displaystyle \sqrt{10}
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Correct Answer

Option a2x2+1\displaystyle 2x^2+1

All Options:

  • A2x2+1\displaystyle 2x^2+1
  • B2x+1\displaystyle 2x+1
  • C(2x2+1)(x)\displaystyle (2x^2+1)(\sqrt{x})
  • D10\displaystyle \sqrt{10}

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