MATHS M.C.Q (1 Marks)

3. $\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

Solution:

Given that: $\text{f(x)}=\sqrt{1+\text{x}^2}$

$\Rightarrow\text{f(xy)}=\sqrt{1+\text{x}^2\text{y}^2}$

and $\text{f(x)}.\text{f(y)}=\sqrt{1+\text{x}^2}.\sqrt{1+\text{x}^2}=\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$

$\therefore\sqrt{1+\text{x}^2\text{y}^2}\leq\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$

$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

1. Domain $= [1, \infty),$ Range $= [0, \infty)$

 Solution:

We have, $\text{f(x)}=\sqrt{\text{x}-1}$

Clearly, f(x) is defined if $\text{x}-1\geq0$

$\Rightarrow\text{x}\geq1$

$\therefore$ Domain of $\text{f}=[1, \infty)$

Now for $\text{x}\geq1,\text{x}-1\geq0$

$\Rightarrow\sqrt{\text{x}-1}\geq1$

⇒ Range of $= [0, \infty)$

1. $(–\infty, –1) \cup (1, 4]$

Solution:

We have, $\text{f(x)}=\sqrt{4-\text{x}}+\frac{1}{\sqrt{\text{x}^2-1}}$

f(x) is defined if $4 - \text{x}\geq 0$and $\text{x}^2-1>0$

$\Rightarrow\text{x}\leq4$ and $(\text{x}+1)(\text{x}-1)>0$

$\Rightarrow\text{x}\leq4$ and $(\text{x}<-1 \ \text{or} \ \text{x}>1)$

$\therefore$ Domain of $\text{f}=(-\infty, -1)\cup(1, 4]$

2. [-a, a]

Solution:

We have $\text{f(x)}\sqrt{\text{a}^2-\text{x}^2}$

Clearly f(x) is defined, if ${\text{a}^2-\text{x}^2}\geq0$

$\Rightarrow\text{x}^2\leq\text{a}^2$

$\Rightarrow-\text{a}\leq\text{x}\leq\text{a} \ [\therefore\text{a}>0]$

$\therefore$ Domain of f is [-a, a]

1. R – {3, –2}

Solution:

Given that: $\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}-6}$

f(x) is defined if $\text{x}^2-\text{x}-6\neq0$

$\Rightarrow\text{x}^2-3\text{x}+2\text{x}-6\neq0$

$\Rightarrow(\text{x}-3)(\text{x}+2)\neq0$

$\Rightarrow\text{x}\neq-2,\text{x}\neq3$

So, the domain of f(x) = R - {-2, 3}

3. Domain = R - {4}, Range = {-1}

Solution:

 Given that: $\text{f(x)}=\frac{4-\text{x}}{\text{x}-4}$

We know that f(x) is defined if $\text{x}-4\neq0 \Rightarrow \text{x}\neq4$

So, the domain of f(x) is = R - {4}

Let $\text{f(x)}=\text{y}=\frac{4-\text{x}}{\text{x}-4}$

$\Rightarrow\text{yx}-4\text{y}=4-\text{x}\Rightarrow\text{yx}+\text{x}=4\text{y}+4$

$\Rightarrow\text{x}(\text{y}+1)=4\text{y}+4\Rightarrow\text{x}=\frac{4(1+\text{y})}{1+\text{y}}$

If x is real number, then $1+\text{y}\neq0\Rightarrow\text{x}\neq1$

$\therefore$ Range of f(x) = R - {-1)

1. $\Big\{-1, \frac{4}{3}\Big\}$

Solution:

We have, f(x) = 3x² – 1 and g(x) = 3 + x

f(x) = g(x)

⇒ 3x² – 1 = 3 + x

⇒ 3x² – x - 4 = 0

⇒ (3x - 4)(x + 1) = 0

$\therefore\text{x}=-1, \frac{4}{3}$

3. $(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$

Solution:

We know that, $-1\leq-\cos\text{x}\leq1$

$\Rightarrow-1\leq-\cos\text{x}\leq1$

$\Rightarrow-2\leq-2\cos\text{x}\leq2$

$\Rightarrow-1\leq-2\cos\text{x}\leq3$

Now $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is defined if

$-1\leq-2\cos\text{x}\leq0$ or $0<1-2\cos\text{x}\leq3$

$\Rightarrow-1\geq\frac{1}{1-2\cos\text{x}}>-\infty$ or $\infty>\frac{1}{1-2\cos\text{x}}\geq\frac{1}{3}$

$\Rightarrow\frac{1}{1-2\cos\text{x}}\in(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$

2. Domain = R, Range $= ( –\infty, 2]$

Solution:

We have, f(x) = 2 - |x - 5|

Clearly, f(x) is defined for all $\text{x}\in\text{R}.$

$\therefore$ Domain of f = R

Now, $|\text{x}-5|\geq0,\forall\text{x}\in\text{R}$

$\Rightarrow-|\text{x}-5|\leq0$

$\Rightarrow2-|\text{x}-5|\leq2$

$\therefore\text{f(x)}\leq2$

$\therefore$ Range of $\text{f}=(-\infty, 2]$