Let the domain of the function be (a, b). If , , gcd (p, q, r) = 1, where is the greatest integer function, then p + q + r is equal to [2025]

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Solution

Q.
Let the domain of the function $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$ be (a, b). If $\int_{0}^{b - a} [x^{2}] d x = p - \sqrt{q} - \sqrt{r}$ , $p, q, r \in ℕ$ , gcd (p, q, r) = 1, where $[\cdot]$ is the greatest integer function, then p + q + r is equal to [2025]

1 8

2 11

3 9

4 10

Ans.
(4)

We have, $f (x) = \log_{2} \log_{4} \log_{6} (3 + 4 x - x^{2})$

$∴$ f(x) is define when $\log_{4} \log_{6} (3 + 4 x - x^{2}) > 0$

$\Rightarrow \log_{6} (3 + 4 x - x^{2}) > 1$

$\Rightarrow 3 + 4 x - x^{2} > 6$

$\Rightarrow - x^{2} + 4 x - 3 > 0$

$\Rightarrow x^{2} - 4 x + 3 < 0$

$\Rightarrow (x - 1) (x - 3) < 0$

$\Rightarrow x \in (1, 3)$

$∴$ a = 1 and b = 3

Now, $\int_{0}^{2} [x^{2}] d x = \int_{0}^{1} [0] d x + \int_{1}^{\sqrt{2}} [1] d x + \int_{\sqrt{2}}^{\sqrt{3}} [2] d x + \int_{\sqrt{3}}^{\sqrt{4}} [3] d x$

   $= 0 + | x |_{1}^{\sqrt{2}} + 2 | x |_{\sqrt{2}}^{\sqrt{3}} + 3 | x |_{\sqrt{3}}^{\sqrt{4}}$

   $= (\sqrt{2} - 1) + 2 (\sqrt{3} - \sqrt{2}) + 3 (2 - \sqrt{3})$

   $= 5 - \sqrt{2} - \sqrt{3}$

$\Rightarrow$ p = 5, q = 2, r = 3

$∴$ p + q + r = 5 + 2 + 3 = 10.

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