If f1x- - x - -2 - and fnx- f n-1 x-2 - for all n- 2- n- N- then number of solution of the equation f2015x-2 is

f(x)=||x| 2| (f_n)x=|f_m 1(x) 2| f_2015(x)=2 #160; #160; #160;No soln=? =1 only solution ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Modulus Function

If f1x=| | ...

Question

If $f_{1} (x) = | | x | - 2 |$ and $f_{n}) x) = ∣ ∣ f_{n - 1} (x) - 2 ∣ ∣$ for all $n \geq 2, n \in N$ , then number of solution of the equation $f_{2015} (x) = 2$ is

Open in App

Solution

Suggest Corrections

Similar questions

f_{1} (x) = \frac{x}{2} + 10 \forall x \in R

f_{n} (x) = f_{1} (f_{n - 1} (x)) \forall n \geq 2, n \in N

{lim}_{n \to \infty} f_{n} (x)

⟨ f_{0} (x), f_{1} (x), f_{2} (x), \dots ⟩

f_{0} (x) = 1, f_{1} (x) = x, {(\begin{matrix} f_{n} (x) \end{matrix})}^{2} - 1 = f_{n + 1} (x) f_{n - 1} (x)

n \geq 1

f_{n} (x)

f_{1} (x) = e^{x}, f_{2} (x) = e^{f_{1} (x)}, . . . . . f_{n + 1} (x) = e^{f_{n} (x)}

n \geq 1

n, \frac{d}{d x} f_{n} (x)

f_{1} (x) = \frac{x}{x - 1}

f_{n} (x) = f_{1} (f_{n - 1} (x))

n \geq 2

x

f_{101} (x) = 3 x

f_{1} (x) = \frac{x}{x - 1}

f_{n} (x) = f_{1} (f_{n - 1} (x))

n \geq 2

x

f_{101} (x) = 3 x

Join BYJU'S Learning Program