## 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.

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.

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, ∞).

## Prove that the number 0.01001000100001…… is an irrational number.

Q: Prove that the number 0.01001000100001…… is an irrational number.

Solution: Since this number is neither terminating nor non terminating and repeating (i.e. this number is non terminating and non repeating decimal). It is an irrational number.

## Prove irrationality of the number tan5°

Q: Prove irrationality of the number tan5°.

Solution: Let tan5° be a rational number than $\displaystyle tan10^o = \frac{2 tan5^o }{1-tan^2 5^o}$ is also rational

⇒ $\displaystyle tan30^o = \frac{3 tan10^o – tan^3 10^o}{1-3 tan^2 10^o}$ is also rational which is not true

⇒ tan5° is an irrational number.

## If f(x) = sin^2x + sin^2(x+π/3) + cosx. cos(x+π/3) and g(5/4) = 1, then

Q: If $\displaystyle f(x) = sin^2 x + sin^2(x+\frac{\pi}{3}) + cosx. cos(x+\frac{\pi}{3})$ and g(5/4) = 1, then

$(A) f(\frac{\pi}{12}) = \frac{5}{4}$

$(B) gof(\frac{\pi}{10}) = 1$

(C) gof is a constant function

(D) fog is a constant function

Ans: (A),(B),(C),(D)

Sol: $\displaystyle f(x) = sin^2 x + sin^2(x+\frac{\pi}{3}) + cosx. cos(x+\frac{\pi}{3})$

$\displaystyle = \frac{1}{2}[1-cos2x + 1- cos2(x+\frac{2\pi}{3}) + cos2(x+\frac{\pi}{3}) + cos(\frac{\pi}{3})]$

$\displaystyle = \frac{1}{2}[\frac{5}{2}-2cos(2x+\frac{\pi}{3})cos\frac{\pi}{3} +cos(2x+\frac{\pi}{3}) ]$

$\displaystyle f(x) = \frac{5}{4}$ ∀ x

$\displaystyle gof(x) = g(f(x))= \frac{5}{4} = g(\frac{5}{4}) = 1$

Hence (A), (B), (C) and (D) are the correct answers.

## f(x) is a real valued function , satisfying f(x + y) + f(x – y) = 2f(x).f(y) for all y ∈ R . Then

Q: f(x) is a real valued function, satisfying f(x + y) + f(x – y) = 2f(x).f(y) for all y ∈ R . Then

(A) f(x) is an even function

(B) f(x) is even if f(0) = 1

(C) f(x) is odd if f(0) = 0

(D) f(x) is even if f(0) = 0

Ans: (A),(B)(C),(D)

Sol: f(x + y) + f(x – y) = 2 f(x). f(y)

f(0+0) + f(0-0) = 2 f(0). f(0)

⇒ 2f(0) = 2[f(0)]2

⇒ f(0) = 1 or f(0) = 0

If f(0) = 0, then f(0 + y) + f(0-y) = 2 f(0). f(y)

⇒ f(y) + f(-y) = 0

⇒ f is an odd function.

Also replacing y = 0 in the given expression , f(x) = 0 , ∀ x ∈ R

⇒ f(x) would be odd and even simultaneously.

If f(0) = 1, then f(0+y) + f(0 – y) = 2 f(0). f(y)

f(y) + f(- y) = 2f(y)

⇒ f is an even function.

Hence (A), (B), (C) and (D) are the correct answers

## If f’ (x) = 1-2sin^2 x/f(x)  , ( f(x) ≥ 0, ∀ x ∈ R and f(0) = 1) then f (x) is a periodic function with the period

Q: If $\displaystyle f’ (x) = \frac{1-2sin^2x}{f(x)}$ , ( f(x) ≥ 0, ∀ x ∈ R and f(0) = 1) then f (x) is a periodic function with the period

(A) π

(B) 2π

(C) π/2

(D) none of these

Ans: (A), (B)

Sol: $f'(x).f(x) = cos2x$

$\displaystyle \int f'(x).f(x) dx = \int cos2x dx$

$\displaystyle f(x) = \sqrt{sin2x + c}$

$\displaystyle f(x) = \sqrt{sin2x + 1}$

⇒ f(x) is periodic with period π . Clearly 2π is also a period.
Hence (A) and (B) are the correct answers.