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.

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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.

 

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].

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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.

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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

 

If f : X→[1, ∞) be a function defined as f(x) = 1 + 3x^3 . Find the super set of all the sets X such that f(x) is one-one.

Q: If f : X→[1, ∞) be a function defined as f(x) = 1 + 3x3 . Find the super set of all the sets X such that f(x) is one-one.

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Solution: Note that f(x) ≥ 1

⇒ 1 + 3x3 ≥ 1

⇒ x3 ≥ 0

⇒ x ∈ [0 , ∞ )

Moreover for x1 , x2 ∈ [0 , ∞), x1 ≠ x2

⇒ 1 + 3x13 ≠ 1 + 3x23

⇒ f(x1) ≠ f(x2)

Thus f : [1, ∞) is one-one for x ∈ [0, ∞).