JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 18)

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 :
6
3
7
$${1 \over 3}$$

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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