Q77 / 155 · WBJEE 2026 · Mathematics

If $\displaystyle\int_{0}^{1}\left(\sum_{r=1}^{2013}\dfrac{x}{x^2 + r^2}\right)\left(\prod_{r=1}^{2013}(x^2 + r^2)\right)dx = \dfrac{1}{2}\left[\prod_{r=1}^{2013}(1 + r^2) - K^2\right]$, then $K$ is

  • A
    $\dfrac{2013(2014)(4027)}{6}$
  • B
    $(2013)^{2013}$
  • Correct answer
    C
    $(2013)!$
  • D
    $(2013!)^2$

Solution

Let $f(x) = \prod_{r=1}^{2013}(x^2 + r^2)$

Taking the natural logarithm on both sides:

$\ln f(x) = \sum_{r=1}^{2013} \ln(x^2 + r^2)$

Differentiating with respect to $x$:

$\dfrac{f'(x)}{f(x)} = \sum_{r=1}^{2013} \dfrac{2x}{x^2 + r^2}$

$\dfrac{1}{2} f'(x) = f(x) \sum_{r=1}^{2013} \dfrac{x}{x^2 + r^2}$

Substituting this into the given integral:

$\displaystyle\int_{0}^{1} \dfrac{1}{2} f'(x) dx = \dfrac{1}{2} [f(x)]_{0}^{1} = \dfrac{1}{2} [f(1) - f(0)]$

Evaluating $f(1)$ and $f(0)$:

$f(1) = \prod_{r=1}^{2013}(1 + r^2)$

$f(0) = \prod_{r=1}^{2013}(0 + r^2) = \prod_{r=1}^{2013} r^2 = (2013!)^2$

The integral evaluates to:

$\dfrac{1}{2} \left[ \prod_{r=1}^{2013}(1 + r^2) - (2013!)^2 \right]$

Comparing this with the given expression $\dfrac{1}{2}\left[\prod_{r=1}^{2013}(1 + r^2) - K^2\right]$:

$K^2 = (2013!)^2 \Rightarrow K = 2013!$

Answer: $(2013)!$

Hence, the correct answer is option: 'C'.

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