Function

Solution: Q x = [x] + f where 0 ≤ f < 1

given equation becomes x³ – (x – f) = 5

i.e. x³ – x = 5 – f ⇒ 4 < x³ – x ≤ 5

Now, x³ – x is negative for x ∈ (-∞, -1) ∪ (0, 1)

So, possible values of x lie in the interval [-1, 0] ∪ [1, ∞)

for –1 ≤ x ≤ 0, we have x³ – x < 1 < 4 ;

for x = 1, we have x³ – x = 0 < 4

further for x ≥ 2 we have

x³ – x = x(x² – 1) ≥ 2(4 – 1) = 6 > 5;

Therefore, 1 < x < 2, ⇒ [x] = 1

Now the original equation can be written as x³ – 1 = 5 hence x³ = 6,

i.e. x = (6)^1/3.

Solution: f(x) = sinx + tan(ax) , where a ∈ R⁺

Period of sinx = 2 π and period of tanax = π/a

If a is irrational, then f(x) is non-periodic

If a is rational the period will be L.C.M. of 2π and π/a.

Solution: Clearly g(x) is the odd extension of the function f(x) and h(x) is the even extension.

Since x² , cosx , log(1 + |x|) are even functions and x , sinx and odd functions.

g(x) = -x² + x + sinx + cosx – log(1 + |x|)

and h(x) = x² – x – sinx – cosx + log(1 + |x|)

Clearly this function satisfy the restriction of the problem.

Solution: Since co-domain of f is the set R whereas the range of f is the interval [-1, 1], hence f is into.

Can you make it onto?

The answer is ‘yes’, if you redefine the co-domain.

Let f be defined from R to another set Y = [-1, 1] i.e. f: R→Y where

f(x) = sinx, then f is onto as range f(x) = [-1, 1] = Y