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The value of the integral $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,$$ where [x] denotes the greatest integer less than or equal to x, is :

JEE · Math · previous-year question

  1. A.6
  2. B.3correct
  3. C.7
  4. D.$${1 \over 3}$$

Answer

B. 3

Explanation

Let I = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} $$ = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)$$ Using, $$\int\limits_a^b {f\left( {a + b - x} \right)dx\,} $$ = $$\,\,\int\limits_a^b {f(x)\,\,dx} $$ I = $$\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)$$ Adding (1) and (2) 2I = $$\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx$$ $$ \Rightarrow $$$$\,\,\,$$ 2I = $$\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}$$ = 6 = I = 3

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