## Solve the equation x^3 – [x]= 5 , where [x] denotes the integral part of the number x.

Q: Solve the equation x3 – [x]= 5 , where [x] denotes the integral part of the number x.

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.

## Find the period of f(x) = sinx + tan(ax) where a ∈ R+, If it exists.

Q: Find the period of f(x) = sinx + tan(ax) where a ∈ R+, If it exists.

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.

## Let the function f(x) = x^2 + x + sinx – cosx + log(1 + |x|) be defined on the interval [0, 1]…..

Q: Let the function f(x) = x2 + x + sinx – cosx + log(1 + |x|) be defined on the interval [0, 1]. Define functions g(x) and h(x) in [-1, 0] satisfying g(-x) = -f(x) and h(-x) = f(x) ∀x  ∈ [0 , 1].

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.

## Let f: R→R where f(x) = sinx . Show that f is into.

Q: Let f: R→R where f(x) = sinx . Show that f is into.