MCQ
If ${u_n} = \int_0^{\pi /4} {{{\tan }^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $
- ✓$\frac{1}{{n - 1}}$
- B$\frac{1}{{n + 1}}$
- C$\frac{1}{{2n - 1}}$
- D$\frac{1}{{2n + 1}}$
$ = \int_0^{\pi /4} {({{\sec }^2}x - 1){{\tan }^{n - 2}}x\,\,dx} $
$ = \int_0^{\pi /4} {{{\sec }^2}x{{\tan }^{n - 2}}x\,\,dx} - \int_0^{\pi /4} {{{\tan }^{n - 2}}x\,\,dx} $
$ = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\pi /4} - {u_{n - 2}}$
$ \Rightarrow {u_n} + {u_{n - 2}} = \frac{1}{{n - 1}}$.
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$f(t)=\left\{\begin{array}{cc}(-1)^{n+1} 2, & \text { if } t=2 n-1, n \in N , \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in N \end{array}\right.$
Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim _{x \rightarrow 1+} \frac{g(x)}{x-1}$. Then the value of $\alpha+\beta$ is equal to. . . . . .