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

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, ∞).

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.

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

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.