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Value of x satisfying \int_{0}^{2\left [ x+14 \right ]}\left \{ \frac{x}{2} \right \}dx=\int_{0}^{{\left \{ x \right \}}}\left [ x+14 \right ]dx is

      (where [.] denotes greatest integer function and {.} denotes fractional part function)

  • Option 1)

    [-14, 13)

  • Option 2)

    (0, 1)

  • Option 3)

    (-15, -14]

  • Option 4)

    none of these 

 

Answers (1)

best_answer

As we learnt

Properties of Definite Integration -

If f(x) is a periodic function with period T then \int_{a}^{a+T}f(x)dxis independent of a.

-

 

 \int_{0}^{2\left [ x+14 \right ]}\left \{ \frac{x}{2} \right \}dx= \int_{0}^{28}\left \{ \frac{x}{2} \right \}dx+\int_{28}^{28+2\left [ x \right ]}\left \{ \frac{x}{2} \right \}dx

                                      = 14\int_{0}^{2}\left \{ \frac{x}{2} \right \}dx+\int_{0}^{2\left [ x \right ]}\left \{ \frac{x}{2} \right \}dx=14+\left [ x \right ]\: \: and\: \: \int_{0}^{\left \{ x \right \}}\left [ x+14 \right ]dx= \left (14+\left [ x \right ] \right )\left \{ x \right \}

\left [ x \right ]=-14\therefore -14\leq x\leq -13


Option 1)

[-14, 13)

Option 2)

(0, 1)

Option 3)

(-15, -14]

Option 4)

none of these 

Posted by

gaurav

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