Question

If $f_1(x) = ||x|-2|, f_n(x) = |f_{n-1}(x)-2|$ for all $n\ge2, n\in N$, and number of solutions of the equation $f_{2015}(x)=2$ is a then the value of $\frac{a}{100}$ is

Answer 1

20.17

Answer 2

Let $N_n$ be the number of solutions to the equation $f_n(x)=2$.

The function is defined by $f_1(x) = ||x|-2|$ and $f_n(x) = |f_{n-1}(x)-2|$ for $n \ge 2$. We want to find the number of solutions for $f_{2015}(x)=2$, which is $a = N_{2015}$.

Let's analyze the solutions for small values of $n$.

For $n=1$: $f_1(x)=2 \implies ||x|-2|=2$. This gives $|x|-2 = 2$ or $|x|-2 = -2$. $|x|=4 \implies x=\pm 4$ (2 solutions). $|x|=0 \implies x=0$ (1 solution). So, $N_1 = 2+1=3$.

For $n=2$: $f_2(x)=2 \implies |f_1(x)-2|=2$. This gives $f_1(x)-2 = 2$ or $f_1(x)-2 = -2$. $f_1(x)=4$ or $f_1(x)=0$.

Let's find the number of solutions for $f_1(x)=4$ and $f_1(x)=0$. $f_1(x)=4 \implies ||x|-2|=4$. $|x|-2 = 4$ or $|x|-2 = -4$. $|x|=6 \implies x=\pm 6$ (2 solutions). $|x|=-2$ (No solution since $|x| \ge 0$). Number of solutions for $f_1(x)=4$ is 2.

$f_1(x)=0 \implies ||x|-2|=0$. $|x|-2 = 0 \implies |x|=2 \implies x=\pm 2$ (2 solutions). Number of solutions for $f_1(x)=0$ is 2.

The total number of solutions for $f_2(x)=2$ is $N_2 = (\text{solutions for } f_1(x)=4) + (\text{solutions for } f_1(x)=0) = 2+2=4$.

For $n=3$: $f_3(x)=2 \implies |f_2(x)-2|=2$. This gives $f_2(x)-2 = 2$ or $f_2(x)-2 = -2$. $f_2(x)=4$ or $f_2(x)=0$.

Let's find the number of solutions for $f_2(x)=4$ and $f_2(x)=0$. $f_2(x)=4 \implies |f_1(x)-2|=4$. $f_1(x)-2 = 4$ or $f_1(x)-2 = -4$. $f_1(x)=6$ or $f_1(x)=-2$. $f_1(x)=6 \implies ||x|-2|=6$. $|x|-2=6$ or $|x|-2=-6$. $|x|=8 \implies x=\pm 8$ (2 solutions). $|x|=-4$ (No solution). Number of solutions for $f_1(x)=6$ is 2. $f_1(x)=-2 \implies ||x|-2|=-2$ (No solution). The number of solutions for $f_2(x)=4$ is 2.

$f_2(x)=0 \implies |f_1(x)-2|=0$. $f_1(x)-2=0 \implies f_1(x)=2$. The number of solutions for $f_2(x)=0$ is the same as the number of solutions for $f_1(x)=2$, which is $N_1=3$.

The total number of solutions for $f_3(x)=2$ is $N_3 = (\text{solutions for } f_2(x)=4) + (\text{solutions for } f_2(x)=0) = 2+3=5$.

Let's summarize the number of solutions for $f_n(x)=2$: $N_1=3$ $N_2=4$ $N_3=5$

It appears that $N_n = n+2$. Let's verify this pattern. $f_n(x)=2 \implies |f_{n-1}(x)-2|=2 \implies f_{n-1}(x)=4$ or $f_{n-1}(x)=0$. $N_n = (\text{solutions for } f_{n-1}(x)=4) + (\text{solutions for } f_{n-1}(x)=0)$.

Let's find the number of solutions for $f_k(x)=c$ where $c>2$. $f_k(x)=c \implies |f_{k-1}(x)-2|=c \implies f_{k-1}(x)=c+2$ or $f_{k-1}(x)=2-c$. Since $c>2$, $c+2>4$ and $2-c<0$. $f_{k-1}(x)=c+2$. Since $c+2>2$, this equation might have solutions. $f_{k-1}(x)=2-c$. Since $2-c<0$ and $f_{k-1}(x) \ge 0$, this equation has no solutions. So, the number of solutions for $f_k(x)=c$ with $c>2$ is the same as the number of solutions for $f_{k-1}(x)=c+2$.

Consider $f_n(x)=c$ for $c>2$. $f_n(x)=c \implies f_{n-1}(x)=c+2 \implies f_{n-2}(x)=c+4 \implies ... \implies f_1(x)=c+2(n-1)$. The number of solutions for $f_n(x)=c$ (with $c>2$) is the same as the number of solutions for $f_1(x)=c+2(n-1)$. Since $c>2$, $c+2(n-1) > 2+2(n-1) = 2n > 2$ for $n \ge 1$. $f_1(x)=c+2(n-1) \implies ||x|-2|=c+2(n-1)$. $|x|-2 = c+2(n-1)$ or $|x|-2 = -(c+2(n-1))$. $|x| = c+2n$ or $|x| = 2-c-2(n-1) = 2-c-2n+2 = 4-c-2n$. Since $c>2$, $c+2n > 2n > 0$. So $|x|=c+2n$ gives $x=\pm (c+2n)$ (2 solutions). Since $c>2$ and $n \ge 1$, $c \ge 2+\epsilon$ for some $\epsilon>0$. $4-c-2n \le 4-(2+\epsilon)-2 = 2-\epsilon-2n < 0$ for $n \ge 1$. So $|x|=4-c-2n$ has no solutions. Thus, for any $c>2$, the number of solutions for $f_n(x)=c$ is always 2 for $n \ge 1$. In particular, the number of solutions for $f_{n-1}(x)=4$ is 2 for $n-1 \ge 1$, i.e., $n \ge 2$.

Now let's find the number of solutions for $f_k(x)=0$. Let this be $M_k$. $f_k(x)=0 \implies |f_{k-1}(x)-2|=0 \implies f_{k-1}(x)-2=0 \implies f_{k-1}(x)=2$. So, $M_k = N_{k-1}$ for $k \ge 2$.

Now we can write the recurrence for $N_n$: For $n \ge 2$, $N_n = (\text{solutions for } f_{n-1}(x)=4) + (\text{solutions for } f_{n-1}(x)=0)$. Since $n \ge 2$, $n-1 \ge 1$, so the number of solutions for $f_{n-1}(x)=4$ is 2. $N_n = 2 + (\text{solutions for } f_{n-1}(x)=0) = 2 + M_{n-1}$. For $n \ge 3$, $M_{n-1} = N_{n-2}$. So, for $n \ge 3$, $N_n = 2 + N_{n-2}$.

We have $N_1=3$ and $N_2=4$. Using the recurrence $N_n = N_{n-2} + 2$ for $n \ge 3$: $N_3 = N_1 + 2 = 3+2=5$. (Matches our calculation) $N_4 = N_2 + 2 = 4+2=6$. $N_5 = N_3 + 2 = 5+2=7$. The pattern $N_n = n+2$ holds for $n=1, 2, 3, 4, 5$.

Let's prove $N_n=n+2$ by induction for $n \ge 1$. Base cases: $N_1=3=1+2$, $N_2=4=2+2$. The formula holds for $n=1, 2$. Assume $N_k=k+2$ for all $1 \le k \le m$, where $m \ge 2$. Consider $N_{m+1}$. Since $m \ge 2$, $m+1 \ge 3$. Using the recurrence relation $N_{m+1} = N_{(m+1)-2} + 2 = N_{m-1} + 2$. By the induction hypothesis, $N_{m-1} = (m-1)+2 = m+1$ since $m-1 \ge 1$. So, $N_{m+1} = (m+1) + 2 = m+3$. The formula $N_n=n+2$ holds for $n=m+1$. By induction, $N_n=n+2$ for all $n \ge 1$.

We need to find the number of solutions for $f_{2015}(x)=2$, which is $a = N_{2015}$. Using the formula $N_n=n+2$, we have $a = N_{2015} = 2015+2 = 2017$.

We are asked to find the value of $\frac{a}{100}$. $\frac{a}{100} = \frac{2017}{100} = 20.17$.