MCQ

MCQ 511 Mark

Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and let $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}.$ Then,

✓
$S$ defines a function from $A$ to $B.$
B
$S_0$ defines a function from $A$ to $C.$
C
$S_0$ defines a function from $A$ to $B.$
D
$S$ defines a function from $A$ to $C.$

Answer

Correct option: A.

$S$ defines a function from $A$ to $B.$

Given that $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}$
$\text{x}^2+\text{y}^2=1$
$\Rightarrow\ \text{y}^2=1-\text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{1-\text{x}^2}$
$\text{y}\in\text{B}$
Hence, S defines a function from $A$ to $B$.

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MCQ 521 Mark

Let $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}.$ Then,

✓
$f^{-1}(x) = f(x)$
B
$f^{-1}(x) = -f(x)$
C
$fof(x) = -x$
D
$\text{f}^{-1}(\text{x})=\frac{1}{19}\text{f(x)}$

Answer

Correct option: A.

$f^{-1}(x) = f(x)$

Given function is $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}$
$fof(x) = f(f(x))$
$=\text{f}\Big(\frac{3\text{x}+2}{5\text{x}-3}\Big)$
$=\frac{3\big(\frac{3\text{x}+2}{5\text{x}-3}\big)+2}{5\big(\frac{3\text{x}+2}{5\text{x}-3}\big)-3}$
After solving you will get
$f(f(x)) = x$
Also,$f^{-1}(x) = f(x)$ you can check.

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MCQ 531 Mark

If $f : A \rightarrow B$ given by $3^{f(x)} + 2^{-x} = 4$ is a bijection, then

A
$\text{A}=\{\text{x}\in\text{R}:-1<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
B
$\text{A}=\{\text{x}\in\text{R}:-3<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
C
$\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
✓
$\text{None of these.}$

Answer

Correct option: D.

$\text{None of these.}$

$\text{f}:\text{A}\rightarrow\text{B}$
$3^\text{f(x)}+2^{-\text{x}}=4$
$\Rightarrow\ 3^{\text{f(x)}}=4-2^{-\text{x}}$
Taking log on both the sides,
$\text{f(x)}\log3=\log(4-2^{-\text{x}})$
$\Rightarrow\ \text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
Logaritmic function will only be defined if $4-2^{-\text{x}}>0$
$\Rightarrow\ 4>2^{-\text{x}}$
$\Rightarrow\ 2^2>2^{-\text{x}}$
$\Rightarrow\ 2>-\text{x}$
$\Rightarrow-2<\text{x}$
$\Rightarrow\ \text{x}\in(-2,\infty)$
That means $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\}$
As we know that, $\text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
We take $\text{x}=0\in(-2,\infty)$
$\Rightarrow f(x) = 1$ which does not belong to any of the options.

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MCQ 541 Mark

If $\text{g(f(x))}=|\sin\text{x}|$ and $\text{f(g(x))}=(\sin\sqrt{\text{x}})^2,$ then

✓
$\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
B
$\text{f(x)}=\sin\text{x},\ \text{g(x)}=|\text{x}|$
C
$\text{f(x)}=\text{x}^2,\ \text{g(x)}=\sin\sqrt{\text{x}}$
D
$\text{f and g cannot be determined.}$

Answer

Correct option: A.

$\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$

If we solve it by the trial$-$and$-$error method, we can see that $(a)$ satisfies the given condition.
From $(a):$
$\text{f(x)}=\sin^2\text{x}$ and $\text{g(x)}=\sqrt{\text{x}}$
$\Rightarrow\ \text{f(g(x))}=\text{f}(\sqrt{\text{x}})=\sin^2\sqrt{\text{x}}$
$=(\sin\sqrt{\text{x}})^2$

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MCQ 551 Mark

Let $\text{f}:[2,\infty)\rightarrow\ \text{X}$ be defined by $f(x) = 4x - x^2$. Then, f is invertible if $X =$

A
$[2,\infty)$
B
$(-\infty,2]$
✓
$(-\infty,4]$
D
$[4,\infty)$

Answer

Correct option: C.

$(-\infty,4]$

Since f is invertible, range of f = co-domain of $f = X$
So, we need to find the range of f to find $X$.
For finding the range, let
$f(x) = y$
$\Rightarrow 4x - x^2 = y$
$\Rightarrow x^2 - 4x = -y$
$\Rightarrow x^2 - 4x + 4 = 4 - y$
$\Rightarrow (x - 2)^2 = 4 - y$
$\Rightarrow\ \text{x}-2=\pm\sqrt{4-\text{y}}$
$\Rightarrow\ \text{x}=2\pm\sqrt{4-\text{y}}$
This is defined only when
$4-\text{y}\geq0$
$\Rightarrow\ \text{y}\leq4$
$X =$ Range of $\text{f}=(-\infty,4]$

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Maths MCQ