Math Solutions

Q: Find the domain and range of the function $\displaystyle y = \frac{1}{\sqrt{x-[x]}} $

Solution : The function is defined for all x ∈ R except when x – [x] = 0

⇒ x is an integer i.e. x ∈ I

Domain = R – I , where I is set of integers .

Also x – [x] > 0

Again $\displaystyle x – [x] = \frac{1}{y^2}$ from given relation

By the definition of x-[x] we know that

0 < x-[x] < 1

$\displaystyle 0 < \frac{1}{y^2} < 1 $

y² > 1

⇒ y² – 1 > 0

(y+1)(y-1) > 0

y < -1 , y > 1 , but y is +ve

Hence y > 1

Hence Range = (1 , ∞)

Q:  Find the domain and Range of $\displaystyle y = \frac{1}{\sqrt{4 + 3 cosx}}$

Solution: Domain = R , Since  – 1 ≤ cosx ≤ 1

$\displaystyle y \in [\frac{1}{\sqrt{7}} , 1 ]$

The function is many-one but not one-one as for so many values of x we will have same image as

cos(2nπ+x) = cos x

Q: What is the domain and Range of the function $\displaystyle y = f(x) = \frac{x^2 – 3x + 2}{x^2+ x -6}$ .Find the limit of f(x) as x approaches 2 .

Solution : $\displaystyle f(x) = \frac{x^2 – 3x + 2}{x^2 + x -6}$

$\displaystyle y = f(x) = \frac{(x-1)(x-2)}{(x+3)(x-2)}$

Therefore function is not defined at x = 2 and x= -3

Hence domain of f = {x: x ∈R , x ≠-3 , x ≠2}

$\displaystyle lim_{x \rightarrow 2} \frac{x^2 – 3x + 2}{x^2+ x -6}$

$\displaystyle = lim_{\rightarrow 2}\frac{(x-1)(x-2)}{(x+3)(x-2)}$

$\displaystyle = lim_{\rightarrow 2}\frac{(x-1)}{(x+3)} = \frac{1}{5}$

To find the range of f we first observe that it can not take the value of 1/5 since it is not defined at x = 2 . For x ≠ 2 we have

$\displaystyle y = f(x) = \frac{x-1}{x+3} $

x y + 3 y = x -1

$\displaystyle x = -\frac{3y+1}{y-1}$

Hence y ≠1 in the domain of x . Also at x = -3 , y = ∞

Thus f(x) takes all real values in the domain of x except y =1/5 and y = 1

Hence range of f = {y : y∈R , y ≠1/5 , y ≠ 1}

Q: If f(x) = 1 + x² and f[g(x)] = 1 + x² – 2x³ + x⁴ , then determine the function g(x) along with its domain and range .

Solution : f[g(x)] = 1 + [g(x)]²

1 + x² – 2x³ + x⁴ = 1 + [g(x)]²

[g(x)]² = x² – 2x³ + x⁴

= x² (x² – 2 x + 1 )

= x² (x-1)²

g(x) = ± x (x-1)

g(x) = ± (x² – x)

$\displaystyle = \pm (x^2 – x + \frac{1}{4} – \frac{1}{4}) $

$\displaystyle = \pm [(x-\frac{1}{2})^2 – \frac{1}{4}] $

$\displaystyle g(x) = (x-\frac{1}{2})^2 – \frac{1}{4} $ ….(i)

or , $\displaystyle g(x) = \frac{1}{4} – (x-\frac{1}{2})^2 $ …(ii)

In either case domain of g(x) is R .

In (i) of g(x) it is always ≥ -1/4

∴ Range is [-1/4 , ∞)

In the form (ii) of g(x) it is always ≤ 1/4

∴ Range is (-∞ , 1/4]

Q: Find the range of the function y = 3 sinx + 4 cos(x + π/3) + 7

Solution : $\displaystyle cos (x + \frac{\pi}{3}) = cos x cos\frac{\pi}{3} – sinx sin\frac{\pi}{3} $

$\displaystyle cos (x + \frac{\pi}{3}) = \frac{1}{2} cos x – \frac{\sqrt{3}}{2} sinx $

$\displaystyle y = 3 sinx + 4 (\frac{1}{2} cos x – \frac{\sqrt{3}}{2} sinx) + 7 $

$\displaystyle y = 3 sinx + 2 cos x – 2 \sqrt{3} sinx + 7 $

$\displaystyle y = ( 3 – 2 \sqrt{3} ) sinx + 2 cos x + 7 $

Put $\displaystyle 3 – 2\sqrt{3} = r cos \alpha$ …(i)

and $ 2 = r sin\alpha $  …(ii)

$\displaystyle y = r sinx . cos \alpha + r cos x . sin\alpha + 7 $

$\displaystyle y = r sin(x + \alpha) + 7 $

Range of sin(x+α) is -1 to 1

Range of y = -r + 7 , r + 7

Now by squaring & adding equation (i) & (ii) we can get the value of r .

Q: Find the range of the function $\displaystyle y = sin^{-1}[\frac{1}{2} + x^2]$ ; Where [] denotes the greatest integer function .

Solution : $\displaystyle siny = [\frac{1}{2} + x^2] \le 1 $

Since $\frac{1}{2} + x^2$ is +ve for all x .

$\displaystyle [\frac{1}{2} + x^2] $ is a +ve integer ≤ 1 i.e. 0 and 1 only .

Therefore siny = 0 or 1

y = 0 , π/2 ;

Range of y = {0 , π/2} only .