Piecewise Functions - 4

Algebra Level 5

Let a function f : R [ 0 , 1 ) R + f : \mathbb{R} - [0, 1) \rightarrow \mathbb{R}^+ be defined as

f ( x ) = { g ( x ) if x Z + h ( x ) if x Z x x otherwise f(x) = \begin{cases} g(x) & \text{if }x \in \mathbb{Z}^+\\ h(x) & \text{if } x \in \mathbb{Z}^- \\ x\lfloor x \rfloor & \text{otherwise} \end{cases}

If f 1 : R + R [ 0 , 1 ) f^{-1}: \mathbb{R}^+ \mapsto \mathbb{R} - [0, 1) is a function, and g g and h h are polynomials of degree 2 2 ,

Find 5 g ( 3.5 ) h ( 3.5 ) 5 \lfloor g(3.5) - h(3.5)\rfloor .

Details and assumptions

  • R [ 0 , 1 ) \mathbb{R} - [0,1) is the interval ( , 0 ) [ 1 , ) (-\infty, 0) \cup [1, \infty)
  • R + \mathbb{R}^+ is the set of all positive reals, that is ( 0 , ) (0, \infty)
  • Z + \mathbb{Z}^+ is the set of all integers > 0 > 0
  • Similarly, Z \mathbb{Z}^- is the set of all integers < 0 < 0
  • \lfloor \cdot \rfloor is the floor function

This problem is part of the set - Piecewise-defined Functions


The answer is 15.

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1 solution

Pranshu Gaba
Oct 19, 2014

For f 1 f^{-1} to be a function, f f must be bijective. It must pass the horizontal line test, and its range must be equal to codomain. The range of f f must be R + \mathbb{R}^+ .

If x ∉ Z , f ( x ) = x x x \not \in \mathbb{Z}, ~ f(x) = x \lfloor x \rfloor .

For any non-zero integer n n , we can define this as

f ( x ) = n x , if n < x < ( n + 1 ) f(x) = nx, \text{ if } n < x < (n+ 1)


For any general n n , range of f f is

If n > 0 n > 0 , f f is an increasing function,

f ( n + ) < f ( x ) < f ( ( n + 1 ) ) \implies f(n^+) < f(x) < f((n+1)^-)

If n < 0 n < 0 , f f is a decreasing function,

f ( ( n + 1 ) ) < f ( x ) < f ( n + ) \implies f((n+1)^-) < f(x) < f(n^+)

The limiting values can be found by substitution.

lim x n + f ( x ) = n 2 \displaystyle\lim_{x \rightarrow n^{+}} f(x) = n^2 and lim x ( n + 1 ) f ( x ) = n ( n + 1 ) \displaystyle\lim_{x \rightarrow (n+1)^{-}} f(x) = n(n + 1)

Since n + 1 n + 1 is an integer as well, we can replace n + 1 n+1 by n n

lim x n f ( x ) = ( n 1 ) n = n 2 n \displaystyle\lim_{x \rightarrow n^{-}} f(x) = (n-1)n =n^2 -n


On observation, for some positive integer k k ,

n = k k ( k 1 ) < f ( x ) < k 2 n = k k 2 < f ( x ) < k ( k + 1 ) n = ( k + 1 ) k ( k + 1 ) < f ( x ) < ( k + 1 ) 2 n = k + 1 ( k + 1 ) 2 < f ( x ) < ( k + 1 ) ( k + 2 ) \begin{array}{l|l} n = -k & k(k-1) < f(x) < k^2 \\ n = k & k^2 < f(x) < k(k+1) \\ n = -(k + 1) & k(k+1) < f(x) < (k+1)^2 \\ n = k +1 & (k+1)^2 < f(x) < (k+1)(k+2)\\ \end{array}

We see that if we extend this table for all k 1 k \geq 1 , then the range of f f is all positive real numbers except the limiting values.

If g ( x ) g(x) and h ( x ) h(x) were x 2 x^2 and x 2 x x^2 -x , or x 2 + x x^2 + x and x 2 x^2 , then range of f f will become R + \mathbb{R}^+ , and it will become bijective.

So the functions are g ( x ) = x 2 g(x) = x^2 and h ( x ) = x 2 x h(x) = x^2 - x , or x 2 + x x^2 + x and x 2 x^2 .

In both cases, we get g ( x ) h ( x ) = x g(x) - h(x) = x .

We get the result 5 × g ( 3.5 ) h ( 3.5 ) = 5 3.5 = 5 × 3 = 15 5 \times \lfloor g(3.5) - h(3.5) \rfloor = 5 \lfloor 3.5 \rfloor = 5 \times 3 = \boxed{15}

I think u forgot something ...f(x) will be one one onto if g(x)=x^2+x and f(x)=x^2 also.by that u get different answer.... @Pranshu Gaba

Rushikesh Joshi - 6 years, 1 month ago

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You are right, I didn't see that. Thanks for pointing out the mistake... I have edited the question accordingly. Can you check whether it is correct now?

Pranshu Gaba - 6 years, 1 month ago

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