MCQ Questions for Class 11 Commerce Applied Mathematics Functions Quiz 9

Let f :

$R\rightarrow R$ be a function defined by f(x) =

${ x }^{ 3 }+{ x }^{ 2 }+3x+sin\times .$ Then f is.

0%
one-one & onto
0%
one-one & into
0%
many one & onto
0%
many one & into

Explanation

$f(x)=x^{3}+x^{2}+3 x+\sin x$

$Suppose$

$f(x_{1})=f(x_{2})$

$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}$

$Equating\;above\;equations\;to \;0$

$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0$

$Solving\;them$

$It\; will\; not\; prove\; that$

$x_{1}=x_{2}$

$So,\;f(x)\;is\;one-one$

$Range\;f(x)=Real$

$Co-domain\;f(x)=Real$

$Range=Co-domain$

$Hence,\; f(x) \;is \;onto$

$So,\;option\; C \;is\;correct.$

The domain of the definition of the function

$f(x)=\cfrac { 1 }{ 4-{ x }^{ 2 } } +\log _{ }{ \left( { x }^{ 3 }-x \right) }$ is

0%
$\left( 1,2 \right) \cup \left( 2,\infty \right)$
0%
$\left( -1,0 \right) \cup \left( 1,2 \right) \cup \left( 3,\infty \right)$
0%
$\left( -1,0 \right) \cup \left( 1,2 \right) \cup \left( 2,\infty \right)$
0%
$\left( -2,-1 \right) \cup \left( -1,0 \right) \cup \left( 2,\infty \right)$

Explanation

$4-{ x }^{ 2 }\neq 0;{ x }^{ 3 }-x>0\quad$

$x=\pm 2;x(x-1)(x+1)>0$

$\therefore { D }_{ f }\in \left( -1,0 \right) \cup \left( 1,2 \right) \cup \left( 2,\infty \right)$

If working set window is too large _____________________.

0%
it will not encompass entire locality
0%
it may overlap several localities
0%
it will cause memory problems
0%
none of the mentioned

Time complexity to check if an edge exists between two vertices would be __________.

0%
O(V*V)
0%
O(V+E)
0%
O(1)
0%
O(E)

What is the definition for Ackermann's function?

0%
A(1,i) = i+1 for i>=1
0%
A(i,j) = i+j for i>=j
0%
A(i,j) = i+j for i = j
0%
A(1,i) = i+1 for i<1

Let

$f:R\rightarrow R$ be defined by

$f(x)=\left|\begin{matrix} 2x & x > 3\\ x^2 & 1 < x \leq 3\\ 3x & x\leq 1\end{matrix}\right.$ . Then

$f(-1)+f(2)+f(4)$ is?

0%
$9$
0%
$14$
0%
$5$
0%
$10$

Explanation

$f(-1)+f(2)+f(4)$

$=-3+4+8=9$ .

Following code snippet is the function to insert a string in a trie. Find the missing line.

0%
node = node.children[index];
0%
node = node.children[str.charAt(i + 1)];
0%
node = node.children[index++];
0%
node = node.children[index+++];

Function which is used to read strings is __________.

0%
getch()
0%
getc()
0%
getstr()
0%
gets()

A function inside another function is called a _____ function.

0%
Nested
0%
Sum
0%
Text
0%
None of these

LetN be the set of natural numbers and two functions f and g be defined as

$f, g: N \rightarrow N$ such that

$f ( n ) = \left\{ \begin{array} { l l } { \frac { n + 1 } { 2 } } & { \text { if } n \text { is odd } } \\ { \frac { n } { 2 } } & { \text { if } n \text { is even } } \end{array} \right.$ and

$g ( n ) = n - ( - 1 ) ^ { n } .$ Then fog is :

0%
onto but not one-one.
0%
one-one but not onto.
0%
both one-one and onto.
0%
neither one-one nor onto.

A function

$f$ from the set of natural numbers to integers defined by

$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$ is

0%
neither one-one nor onto
0%
one-one but not onto
0%
onto but not one-one
0%
one-one and onto both

Explanation

$one-one$ test of

$f:$

Let

$x_1$ and

$x_2$ be any two elements in the domain

$(N).$

$Case\,I:$ When both

$x_1$ and

$x_2$ are even.

Let

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$

$\Rightarrow$

$-x_1=-x_2$

$\Rightarrow$

$x_1=x_2$

$Case\,II:$ When both

$x_1$ and

$x_2$ are odd.

Let

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$

$\Rightarrow$

$x_1-1=x_2-1$

$\Rightarrow$

$x_1=x_2$

$Case\,III:$ When

$x_1$ be even and

$x_2$ be odd.

Then,

$f(x_1)=\dfrac{-x_1}{2}$ and

$f(x_2)=\dfrac{x_2-1}{2}$

Then clearly,

$\Rightarrow$

$x_1\ne x_2$

$\Rightarrow$

$f(x_1)\ne f(x_2)$

From, all the cases, we can say that,

$f$ is one-one.

$onto$ test of

$f:$

Co-domain of

$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$

Range of

$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$

Range of

$f=\{...,-2,1,-1,0,0,-1,1,..\}$

$\Rightarrow$ Co-domain of

$f=$ Range of

$f$

$\therefore$

$f$ is onto.

The function

$f:A\rightarrow B$ defined by

$f(x)=-{ x }^{ 2 }+6x-8$ is a bijection, if

0%
$A=(-\infty ,3];B=(-\infty ,1]$
0%
$A=[-3,\infty );B=(-\infty ,1]$
0%
$A=(-\infty ,3];B=[1,\infty )$
0%
$A=[3,\infty );B=[1,\infty )$

Explanation

$f(x)=-x^2+6x-8$

We can see function is a polynomial and the domain of polynomial function is real number.

$\therefore$

$x\in R$

$f(x)=-x^2+6x-8$

$=-(x^2-6x+8)$

$=-(x^2-6x+9-1)$

$=-(x-3)^2+1$

Maximum value of

$-(x-3)^2$ would be

$0$

$\therefore$ Maximum value of

$-(x-3)^2+1$ would be

$1.$

$\therefore$

$f(x)\in(-\infty,1]$

We can see from the given graph that function is symmetrical about

$x=3$ and the given function is bijective.

So,

$x$ would be either

$(-\infty,3]$ or

$[3,\infty)$

The correct option which satisfy

$A$ and

$B$ both is:

$A=(-\infty,3]$ and

$B=(-\infty,1]$

1407102_1659305_ans_9061dba9d28c4dc38d8f5272ce18fb4e.jpeg

Which of the following functions from

$Z$ to itself are bijections?

0%
$f(x)={x}^{3}$
0%
$f(x)=x+2$
0%
$f(x)=2x+1$
0%
$f(x)={X}^{2}+x$

Explanation

$(A)$

$f(x)=x^3$

$one-one$ test:

Let

$x_1$ and

$x_2$ be the elements in the domain such that,

$f(x_1)=f(x_2)$

$\Rightarrow$

$x_1^3=x_2^3$

$\Rightarrow$

$x_1=x_2$

$\therefore$

$f$ is one-one.

$onto$ test :

Let

$y$ be an element in the co-domain

$(Z),$ such that,

$f(x)=y$

$\Rightarrow$

$x^3=y$

Consider

$y=3$

$\Rightarrow$

$x^3=3$

$\Rightarrow$

$x=\sqrt[3]{3}\notin Z$

$\therefore$

$f$ is not onto.

$\therefore$

$f$ is not bijection.

$(B)$

$f(x)=x+2$

$one-one$ test:

Let

$x_1$ and

$x_2$ be the elements in the domain such that,

$f(x_1)=f(x_2)$

$\Rightarrow$

$x_1+2=x_2+2$

$\Rightarrow$

$X_1=x_2$

$\therefore$

$f$ is one-one.

$onto$ test :

Let

$y$ be an element in the co-domain

$(Z),$ such that,

$f(x)=y$

$\Rightarrow$

$x+2=y$

$\Rightarrow$

$x=y-2\in Z(Domain)$

$\therefore$

$f$ is onto

$\therefore$

$f$ is bijection.

$(C)$

$f(x)=2x+1$

$one-one$ test:

Let

$x_1$ and

$x_2$ be the elements in the domain such that,

$f(x_1)=f(x_2)$

$\Rightarrow$

$2x_1+1=2x_2+1$

$\Rightarrow$

$2x_1=2x_2$

$\Rightarrow$

$x_1=x_2$

$\therefore$

$f$ is one-one.

$onto$ test :

Let

$y$ be an element in the co-domain

$(Z),$ such that,

$f(x)=y$

$\Rightarrow$

$2x+1=y$

Consider

$y=4$

$\Rightarrow$

$2x+1=4$

$\Rightarrow$

$2x=3$

$\Rightarrow$

$x=\dfrac{3}{2}\notin Z$

$\therefore$

$f$ is not onto.

$\therefore$

$f$ is not bijection.

$(D)$

$f(x)=x^2+x$

$\Rightarrow$

$f(0)=(0)^2+0=0$

$\Rightarrow$

$f(-1)=(-1)^2+(-1)=1-1=0$

We can see that

$0$ and

$-1$ have the same image.

$\therefore$

$f$ is not on-one.

$\therefore$

$f$ is not bijection.

Let

$M$ be the set of all

$2\times 2$ matrices with entries from the set

$R$ of real numbers. Then the function

$f:M\rightarrow R$ defined by

$f(A)=\left| A \right|$ for every

$A\in M$ , is

0%
one-one and onto
0%
neither one-one nor onto
0%
one-one but not onto
0%
onto but not one-one

Explanation

$M=\left\{A=\begin{bmatrix} a & b\\ c& d\end{bmatrix}: a,b,c,d\in R\right\}$

$f:M\rightarrow R$ is given by

$f(A)=|A|$

$Injectivity:$

$f\left(\begin{bmatrix} 0 & 0\\0& 0\end{bmatrix} \right )=\begin{vmatrix} 0&0\\0&0\end{vmatrix}=0$

and

$f\left(\begin{bmatrix} 1 & 0\\0& 0\end{bmatrix} \right )=\begin{vmatrix} 1&0\\0&0\end{vmatrix}=0$

$\Rightarrow$

$f\left(\begin{bmatrix} 0&0\\0&0\end{bmatrix}\right)=f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)=0$

So,

$f$ is not one-one.

$Surjectivity:$

Let

$f(A)=y,\,A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$

$\Rightarrow$

$\begin{vmatrix} a& b\\c&d\end{vmatrix}=y$

$\Rightarrow$

$ad-bc=y$

$\Rightarrow$

$a,b,c,d\in R$ implies

$y\in R$

$\Rightarrow$ CoDomain=Range.

$\therefore$

$f$ is onto.

If a function

$f:(2,\infty )\rightarrow B$ defined by

$f(x)={ x }^{ 2 }-4x+5$ is a bijection, then

$B=$

0%
$R$
0%
$[1,\infty)$
0%
$(0,1]$
0%
$[0,1)$

Explanation

Since,

$f$ is a bijection,

Co-domain of

$f=$ Range of

$f$

$\Rightarrow$

$B=$ range of

$f$

$\Rightarrow$

$f(x)=x^2-4x+5$

Let

$f(x)=y$

$\Rightarrow$

$y=x^2-4x+5$

$\Rightarrow$

$x^2-4x+(5-y)=0$

$\because,\,D=b^2-4ac\ge 0,$

$\Rightarrow$

$(-4)^2-4\times 1\times (5-y)\ge 0$

$\Rightarrow$

$16-20+4y\ge 0$

$\Rightarrow$

$4y\ge 4$

$\Rightarrow$

$y\ge 1$

$\Rightarrow$

$y\in [1,\infty)$

$\Rightarrow$ Range of

$f=[1,\infty)$

$\Rightarrow$

$B=[1,\infty)$

Let

$f:Z\rightarrow Z$ be given by

$f(x)=\begin{cases} \cfrac { x }{ 2 } ,\quad \text{if}\ x \ \text{is even} \\ 0,\quad \text{if }\ x \ \text{is odd} \end{cases}$ . Then,

$f$ is

0%
onto but not one-one
0%
one-one but not onto
0%
one-one and onto
0%
neither one-one nor onto

Explanation

$Injectivity:$

Let

$x_1$ and

$x_2$ be two elements in the domain

$(Z)$ , such that,

$f(x_1)=f(x_2)$

$Case\,I:$

Let both

$x_1$ and

$x_2$ be even.

Then,

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{x_1}{2}=\dfrac{x_2}{2}$

$\Rightarrow$

$x_1=x_2$

$Case\,II:$

Let both

$x_1$ and

$x_2$ be odd.

Then,

$f(x_1)=f(x_2)$

$\Rightarrow$

$0=0$

Here, we cannot determine whether

$x_1=x_2.$

So,

$f$ is not one-one.

$Surjectivity:$

Let

$y$ be an element in the co-domain

$(Z)$ , such that

Co-domain of

$f=Z=\{0,\pm 1,\pm 2,\pm 3,\pm 4,...\}$

Range of

$f=\left\{0,0,\dfrac{\pm 2}{2},0,\dfrac{\pm 4}{2},...\right\}=\{0,\pm 1,\pm 2,....\}$

$\Rightarrow$ Co-domain of

$f=$ Range of

$f$

$\therefore$

$f$ is onto.

Which of the following functions from

$A=\left\{ x:-1\le x\le 1 \right\}$ to itself are bijections?

0%
$f(x)=\cfrac { x }{ 2 }$
0%
$g(x)=\sin { \left( \cfrac { \pi x }{ 2 } \right) }$
0%
$h(x)=\left| x \right|$
0%
$k(x)={ x }^{ 2 }$

Explanation

$(A)$ Range of

$f=\left[\dfrac{-1}{2},\dfrac{1}{2}\right]\ne A$

So,

$f$ is not a bijection.

$(B)$ Range

$=\left[\sin\left(\dfrac{-\pi}{2}\right),\sin\left(\dfrac{\pi}{2}\right)\right]=[-1,1]=A$

So,

$g$ is a bijection.

$(C)$

$h(-1)=|-1|=1$

And

$h(1)=|1|=1$

$\Rightarrow$

$-1$ and

$1$ have the same images

So,

$h$ is not a bijection.

$(D)$

$k(-1)=(-1)^2=1$

And

$k(1)=(1)^2=1$

$\Rightarrow$

$-1$ and

$1$ have the same images.

So,

$k$ is not a bijection.

The function

$f:\left[ -\dfrac {1}{2},\dfrac {1}{2} \right] \rightarrow \left[ -\dfrac {\pi }{2},\dfrac {\pi }{2} \right]$ defined by

$f(x)=\sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) }$ is

0%
bijection
0%
injection but not a surjection
0%
surjection but not an injection
0%
neither injection nor a surjection

Explanation

$f(x)=\sin^{-1}(3x-4x^3)$

$f(x)=3\sin^{-1}x$

$one-one$ test of

$f:$

Let

$x_1$ and

$x_2$ be two elements in the domain

$\left[\dfrac{-1}{2},\dfrac{1}{2}\right],$ such that,

$f(x_1)=f(x_2)$

$\Rightarrow$

$3\sin^{-1}x=3\sin^{-1}x_2$

$\Rightarrow$

$\sin^{-1}x=\sin^{-1}x_2$

$\Rightarrow$

$x_1=x_2$

$\therefore$

$f$ is one-one.

$onto$ test :

Let

$y$ be element in the co-domain, such that

$f(x)=y$

$\Rightarrow$

$3sin^{-1}(x)=y$

$\Rightarrow$

$\sin^{-1}(x)=\dfrac{y}{3}$

$\Rightarrow$

$x=\sin\dfrac{y}{3}\in\left[\dfrac{-1}{2},\dfrac{1}{2}\right]$

$\therefore$

$f$ is onto.

Since it is one-one and onto both then,

$f$ is a bijection.

$g(x)={ x }^{ 2 }+x-2$ and

$\cfrac { 1 }{ 2 } (g\circ f(x))=2{ x }^{ 2 }-5x+2$ , then

$f(x)$ is equal to

0%
$2x-3$
0%
$2x+3$
0%
$2{x}^{2}+3x+1$
0%
$2{x}^{2}-3x-1$

Explanation

We will solve this problem by the trial and error method.

Let us check option

$A$ first.

$f(x)=2x-3$

$g(x)=x^2+x-2$ [ Given ]

$\Rightarrow$

$\dfrac{1}{2}(g\circ f)(x)=g[f(x)]$

$=\dfrac{1}{2}g(2x-3)$

$=\dfrac{1}{2}[(2x-3)^2+(2x-3)-2]$

$=\dfrac{1}{2}[4x^2+9-12x+2x-3-2]$

$=\dfrac{1}{2}[4x^2-10x+4]$

$=2x^2-5x+2$

The given condition is satisfied by

$A.$

A function

$f$ from the set of natural numbers to the set of integers defined by

$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$

0%
neither one-one nor onto
0%
one-one but not onto
0%
onto but not one-one
0%
one-one and onto both

Explanation

$one-one$ test of

$f:$

Let

$x_1$ and

$x_2$ be any two elements in the domain

$(N).$

$Case\,I:$ When both

$x_1$ and

$x_2$ are even.

Let

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$

$\Rightarrow$

$-x_1=-x_2$

$\Rightarrow$

$x_1=x_2$

$Case\,II:$ When both

$x_1$ and

$x_2$ are odd.

Let

$f(x_1)=f(x_2)$

$\Rightarrow$

$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$

$\Rightarrow$

$x_1-1=x_2-1$

$\Rightarrow$

$x_1=x_2$

$Case\,III:$ When

$x_1$ be even and

$x_2$ be odd.

Then,

$f(x_1)=\dfrac{-x_1}{2}$ and

$f(x_2)=\dfrac{x_2-1}{2}$

Then clearly,

$\Rightarrow$

$x_1\ne x_2$

$\Rightarrow$

$f(x_1)\ne f(x_2)$

From, all the cases, we can say that,

$f$ is one-one.

$onto$ test of

$f:$

Co-domain of

$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$

Range of

$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$

Range of

$f=\{...,-2,1,-1,0,0,-1,1,..\}$

$\Rightarrow$ Co-domain of

$f=$ Range of

$f$

$\therefore$

$f$ is onto.

$n \geq 2$ then the number of surjections that can be defined from

$\{1, 2, 3, ....... n\}$ onto

$\{1, 2\}$ is

0%
$2n$
0%
$^nP_2$
0%
$2^n$
0%
$2^{n}-2$

Explanation

Let

$A=\{1,2,3,...,n\}$ and

$B=\{1,2\}$ .

Therefore, total number of mappings from

$A$ to

$B$ is

$2^n$ of which two functions

$f(x)=1$ for all

$x\in A$ and

$g(x)=2$ for all

$x\in A$ are not surjective.
Thus, the total number of surjections from

$A$ to

$B$ is

$2^n-2$ .

The function

$f:R\rightarrow R$ defined by

$f(x)=(x-1)(x-2)(x-3)$ is

0%
one-one but not onto
0%
onto but not one-one
0%
both one and onto
0%
neither one-one nor onto

Explanation

$f(x)=(x-1)(x-2)(x-3)$

$one-one$ test:

$\Rightarrow$

$f(1)=(1-1)(1-2)(1-3)=0$

$\Rightarrow$

$f(2)=(2-1)(2-2)(2-3)=0$

$\Rightarrow$

$f(3)=(3-1)(3-1)(3-3)=0$

$\Rightarrow$

$f(1)=f(2)=f(3)=0$

We can see,

$1,2,3$ has same image

$0.$

$\therefore$

$f$ is not one-one.

$onto$ test:

Let

$y$ be an element in the co-domain

$R,$ such that

$y=f(x)$

$\Rightarrow$

$y=(x-1)(x-2)(x-3)$

Since,

$y\in R$ and

$x\in R$ .

$\therefore$

$f$ is onto.

$g(x)=x^2+x-1$ and

$(gof)(x)=4x^2-10x+5$ , then

$f\left(\dfrac{5}{4}\right)$ is equal to:

0%
$\dfrac{3}{2}$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{3}{2}$
0%
$-\dfrac{1}{2}$

Explanation

$g(x)=x^2+x-1$

$g\left(f\left(\dfrac{5}{4}\right)\right)=4\left(\dfrac{5}{4}\right)^2-10\dfrac{5}{4}+5=-\dfrac{5}{4}$

$g\left(f\left(\dfrac{5}{4}\right)\right)=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$

$-\dfrac{5}{4}=f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)-1$

$f^2\left(\dfrac{5}{4}\right)+f\left(\dfrac{5}{4}\right)+\dfrac{1}{4}=0$

$\left(f\left(\dfrac{5}{4}\right)+\dfrac{1}{2}\right)^2=0$

$\boxed{f\left(\dfrac{5}{4}\right)=\dfrac{-1}{2}}$

Let

$S$ be the set of all real roots of the equation,

${ 3 }^{ x }\left( { 3 }^{ x }-1 \right) +2=\left| { 3 }^{ x }-1 \right| +\left| { 3 }^{ x }-2 \right|$ . Then

$S$ :

0%
contains at least four elements
0%
is a singleton
0%
is an empty set
0%
contains exactly two elements

Explanation

$3^x (3^x - 1) + 2 = |3^x - 1| + |3^x - 2|$

Let

$3^x = t$

So,

$t (t - 1) + 2 = | t - 1| + |t - 2|$

$\Rightarrow t^2 - t + 2 = |t - 1|+ |t - 2|$

$f(t)=t^2-t+2 \quad\quad\text{[green curve]}$

$g(t)=|t-1|+|t+2|\quad\quad\text{[blue curve]}$

we plot

$f(t)=t^2-t+2$ and

$g(t)=|t-1|+|t+2|$

$\text{these two curves intersect at two points but only one root is positive.}$

$\text{As} \,3^x \,\text{is always positive, therefore only positive value of} \,t\, \text{will be the solution.}$

$\text{Therefore, we have only one solution.}$

1476851_1697590_ans_47abda3d5a504345a33f51f3a516d4c1.png

$f(x) = \dfrac{x+1}{x-1}$ , then the valueof

$f(f(x))$ is equal to

0%
$x$
0%
$0$
0%
$-x$
0%
$1$

Explanation

$f(x)=\dfrac{x+1}{x-1}$

$\therefore f(f(x))=f\left(\dfrac{x+1}{x-1}\right)$

$\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}$

$=\dfrac{x+1+x-1}{x+1-x+1}$

$=\dfrac{2x}{2}$

$=x$

Let

$f : x \rightarrow y$ be such that

$f(1) = 2$ and

$f(x + y) = f(x) f(y)$ for all natural numbers x and y. If

$\displaystyle \sum_{k= 1}^n f(a + k) = 16 (2^n - 1)$ , then a is equal to

0%
$3$
0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

We have,

$f(1)=2$ and

$f(x+y)=f(x).f(y)$

Now,

$f(2)=f(1+1)=f(1).f(1)=2.2=2^2$

$f(3)=f(2+1)+f(2).f(1)=2^2.2=2^3$

and so on

$\therefore f(x)=2^n$ ......(i)

Now, we have

$\displaystyle \sum^n_{k=1}f(a+k)=16(2^n-1)$

$\Rightarrow f(a+1)+f(a+2)+----f(a+n)=16(2^n-1)$

$\Rightarrow f(a).f(1)+f(a).f(2)+-----f(a).f(n)=16(2^n-1)$

$\Rightarrow f(a)=[f(1)+f(2)+---f(n)]=16(2^n-1)$

$\Rightarrow f(a)[2+2^2+----+2^n]=16(2^n-1)$

$\Rightarrow f(a).[2\dfrac{(2^n-1)}{2-1}]=16(2^n-1)$

$\Rightarrow 2f(a).(2^n-1)=16.(2^n-1)\Rightarrow f(a)=8$

$\Rightarrow 2^a=8[\because f(x)=2^n\Rightarrow f(a)=2^a]$

$\Rightarrow 2^a=2^3=a=3$

$f(x)=\dfrac{(4x+3)}{(6x-4)}, x\neq \dfrac{2}{3}$ then

$(f o f)(x)=?$

0%
$x$
0%
$(2x-3)$
0%
$\dfrac{4x-6}{3x+4}$
0%
None of these

$f(x)=\sqrt[3]{3-x^3}$ then

$(f o f)(x)=?$

0%
$x^{1/3}$
0%
$x$
0%
$(1-x^{1/3})$
0%
None of these

Let

$f : R \rightarrow R : f(x) =x +1$ and

$g : R \rightarrow R : g(x) = 2x -3$ .
Find

$(f +g) (x)$ .

0%
$3x -2$
0%
$4x -5$
0%
$3x -4$
0%
$2x -3$

Explanation

Given ,

$f(x)=x+1,g(x)=2x-3$

$\implies (f+g)x=f(x)+g(x)=x+1+2x-3=3x-2$

$\displaystyle f(x) = | x - 2 |$ and

$g(x) = fof\,(x)$ , then for

$x > 20 , {g}\,'(x) =$

0%
$2$
0%
$1$
0%
$3$
0%
None of these

$\displaystyle {f}'(x) = g\,(x)$ and

$\displaystyle {g}'(x) = - f\,(x)$ for all

$x$ and

$f\,(2) = 4 = {f}'(2)$ then

$\displaystyle f^{2}\,(19) + g^{2} \,(19)$ is

0%
$16$
0%
$32$
0%
$64$
0%
None of these

The value of f(0), so that the function
f(x) =

$\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x}$ is continuous at each point in its domain, is equal to

0%
2
0%
1/3
0%
2/3
0%
-1/3

Explanation

The function f is clearly continuous at each point in its domain except possibly at x=0 Given that f(x) is continuous at x=0
Therfore,f (0) =

$\underset{x\rightarrow 0}{lim}f(x)$

$=\underset{x\rightarrow 0}{lim}\dfrac{2x-sin^{-1}x}{2x+tan^{-1}x}$

$= \underset{x\rightarrow 0}{lim}\dfrac{2-\frac{(sin^{-1}x)}x}{2+\frac{(tan^{-1}x)}x}$

$\dfrac{2-1}{2+1}=\dfrac13$

let

$f(x) = sin^2 x/2 + cos ^2 x/2$ and

$g(x) = sec^2 x - tan ^2 x.$ The two functions are equal over the set

0%
$\phi$
0%
$R$
0%
$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$
0%
None of these

Let

$f(n)$ denote the number of different ways in which the positive integer

$n$ can be expressed as the sum of

$1s$ and

$2s$ . For example,

$f(4) = 5$ , since

$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$ . Note that order of

$1s$ and

$2s$ is important.

$f : N\rightarrow N$ is

0%
One-one and onto
0%
One-one and into
0%
Many-one and onto
0%
Many-one and into

Explanation

$6 = 0(2) + 6(1) = 1(2) + 4(1) = 2(2) + 2(1) = 3(2) + 0(1)$

Number of $2s$	Number of $1s$	Number of permutations
$0$	$6$	$1$
$1$	$4$	$\dfrac {5!}{4!} = 5$
$2$	$2$	$\dfrac {4!}{2!2!} = 6$
$3$	$0$	$\dfrac {3!}{3!} = 1$
		$Total = 13$

$\therefore f(6) = 13$

Now,

$f(f(6)) = f(13)$

Number of $1s$	Number of $2s$	Number of permutations
$13$	$0$	$1$
$11$	$1$	$\dfrac {12!}{11!} = 12$
$9$	$2$	$\dfrac {11!}{9!2!} = 55$
$7$	$3$	$\dfrac {10!}{7!3!} = 120$
$5$	$4$	$\dfrac {9!}{5!4!} = 126$
$3$	$5$	$\dfrac {8!}{3!5!} = 56$
$1$	$6$	$\dfrac {7!}{6!} = 7$
		$Total = 377$

$\therefore f(f(6)) = f(13) = 377$

$f(1) = 1(1)$

$f(2) = 2 (1, 1$ or

$2)$

$f(3) = 3(1, 1, 1\ or\ 2, 1\ or\ 1, 2)$

$f(4) = 5$ (explained in the paragraph)

By taking higher value of

$n$ in

$f(n)$ , we always get more value of

$f(n)$ . Hence,

$f(x)$ is one-one. Clearly,

$f(x)$ is into.

The domain of the function

$\sqrt { (\log\ 5x) }$ is

0%
$(1, \infty )$
0%
$(0, \infty )$
0%
$(0, 1 )$
0%
$(0.5 , 1 )$

The function

$f(x)= \dfrac{(3^{x}-1^{})^2}{\sin x. \ln(1+x)}, x\neq 0$ , is continuous at

$x=0$ . Then the value of

$f(0)$ is

0%
2log 3
0%
$(\log_{e}3)^{2}$
0%
$\log_{e} 6$
0%
None of these

Explanation

Given f(x) is continuous at

$x=0$

$\Rightarrow \underset{x\rightarrow 0}{\lim}f(x)=f(0)$

$\Rightarrow \underset{x\rightarrow 0}{\lim}\dfrac{(3^{x}-1)^{2}}{\sin x\ln(1+x)}=f(0)$

$\Rightarrow f(0)=\underset{x\rightarrow 0}{\lim} \dfrac{\bigg({\dfrac{3^x-1}{x}}\bigg)^2}{\dfrac{\sin x}{x}\dfrac{\ln(1+x)}{x}}=$

$(\log_e3)^{2}$

The domain of the function

$f(x) = \left [ log_{10} \left ( \frac{5x - x^{2}} {4} \right ) \right]^{1/2}$ is

0%
$- \infty < x < \infty$
0%
$1 \leqslant x \leqslant 4$
0%
$4 \leqslant x \leqslant 16$
0%
$-1 \leqslant x \leqslant 1$

Explanation

We have

$f(x) = \left [ log_{10} \left ( \frac{5x - x^{2}} {4} \right ) \right]^{1/2}$

From (1), clearly f(x) is defined for those values of x for

which

$log_{10} \left [\frac{5x - x^{2}} {4} \right] \geq 0$

$\Rightarrow \left (\frac {5x - x^{2}} {4} \right ) \geq 10^{0}$

$\Rightarrow \left (\frac {5x - x^{2}} {4} \right ) \geq 1$

$\Rightarrow x^{2} - 5x + 4 \leq 0$

$\Rightarrow (x - 1) (x - 4) \leq 0$

Hence the domain of the function is [1, 4]

The domain of

$f(x) = \log| \log {x}|$ is

0%
$\left(0, \infty\right )$
0%
$\left(1, \infty\right )$
0%
$\left(0, 1\right )\cup \left(1, \infty \right )$
0%
$\left(-\infty, 1\right )$

Explanation

$f(x) = \log| \log {x}|$ , f(x) is defined if

$|log \space x| > 0$ and x > 0, i.e.,

if x > 0 and

$x \neq 1$

$(\therefore |log \space x| > 0 \space if \space x \neq 1)$

$\Rightarrow x \epsilon (0, 1) \cup (1, \infty)$

Which of the following is not true about

$h_1(x)$ ?

0%
It is periodic function with period $\pi$
0%
Range is [0,1]
0%
Domain if R
0%
None of these

If f:

$R\rightarrow R$ be given by

$f(x) = 3 + 4x$ and

$a_n = A + Bx$ , then which of the following is not true?

0%
A + B + 1 = $2^{2n + 1}$
0%
| A - B| = 1`
0%
$\displaystyle \lim_{n \to \infty} \dfrac{A}{B} = -1$
0%
None of these

Explanation

Since

$a_1 = g(x) = 3 + 4x$

$\therefore a_2 = g\{g^2(x)\} = g(3+4x) = 3 + 4(3+4x) = (4^2 - 1) + 4^x$

$a_3 = g\{g^(x)\} = g(15 + 4^2x) = 3 + 4 (15 + 4^2x) = 63 + 4^3x = (4^3 - 1) + 4^3 x$

Similarly, we get

$a_n = (4^n - 1) + 4^n x$

$\Rightarrow A = 4^n - 1 \space and B = 4^n$

$\Rightarrow A + B + 1 = 2^{2n + 1}, |a - b| = 1\space and\space lim_{n \to \infty}\dfrac{4^n - 1}{4^n}$

$= lim_{n \to \infty}\left(1 - \dfrac{1}{4^n}\right) = 1$

Let

$f(x) + f(y) = f(x\sqrt{1 - y^2} + y\sqrt{1 - x^2})$ (f(x) is not identically zero) the.

0%
$f(4x^3 - 3x) + 3f(x) = 0$
0%
$f(4x^3 - 3x) = 3f(x)$
0%
$f(2x\sqrt{1-x^2}) + 2f(x) = 0$
0%
$f(2x\sqrt{1-x^2}) = 2f(x)$

Consider two functions

$f(x) = \begin{cases} [x], -2 \leq x \leq -1\\ |x| + 1, -1 < x \leq 2 \end{cases}$ and g(x) =

$\begin{cases} [x], -\pi \leq x < 0 \\ sin x, 0 \leq x \leq \pi \end{cases}$ where [.] denotes thegreatest integer function

The exhaustive domain of g(f(x)) is

0%
[0,2]
0%
[-2, 0]
0%
[-2,2]
0%
[-1,2]

The graph of y = g(x) in its domain is broken at

0%
1 point
0%
2 point
0%
3 point
0%
None of these

Let

$g(x) = f(x) - 1$ . If

$f(x) + f(1 - x) = 2 \space \forall \space x \space \epsilon \space R$ , then

$g(x)$ is symmetrical about

0%
Origin
0%
The line $x = \frac{1} {2}$
0%
The point $\left (1, 0 \right )$
0%
The point $\left (\frac {1} {2}, 0 \right )$

Domain (D) and range (R) of

$f(x) = \sin^{-1}\left (\cos^{-1} [x] \right )$ where [.] denotes the greatest integer function is

0%
$D \equiv x \epsilon \left [ 1, 2 \right ), R \epsilon \left \{0 \right \}$
0%
$D \equiv x \epsilon \left [ 0, 1 \right ], R \equiv \left \{-1, 0, 1 \right \}$
0%
$D \equiv x \epsilon \left [ -1, 1 \right ), R \epsilon \left \{0, \sin^{-1} \left (\frac{\pi} {2}\right), \sin^{-1} (\pi) \right \}$
0%
$D \equiv x \epsilon \left [-1, 1 \right ), R \epsilon \left \{-\frac{\pi} {2}, 0 , \frac{\pi} {2} \right \}$

g(f(x)) is not defined if

0%
$a\space \epsilon ( 10, \infty) b\space \epsilon (5, \infty)$
0%
$a\space \epsilon ( 4, 10) b\space \epsilon (5, \infty)$
0%
$a\space \epsilon ( 10, \infty) b\space \epsilon (0, 1)$
0%
$a\space \epsilon ( 4, 10) b\space \epsilon (1, 5)$

Let

$A=\left\{0,1\right\}$ and

$N$ be the set of natural numbers. Then, the mapping

$f:N\to A$ defined by

$f(2n-1)=0, f(2n)=1, \forall n \in R$ , is onto.

0%
True
0%
False

For set

$A, B$ and

$C$ , let

$f:A\to B, g:B\to C$ be functions such that

$gof$ is surjective.
Then

$g$ is surjective function.

0%
True
0%
False

Explanation

Suppose that

$g\circ f$ is surjective.

Let

$z∈C$ .

Then since

$g\circ f$ is surjective, there exists

$x\in A$ such that

$(g\circ f)(x) =g(f(x)) =z$ .

Therefore if we assume

$y=f(x)\in B$ ,

then

$g(y) =z$ .

Thus

$g$ is surjective

Let

$A$ be a finite set. Then, each injective function from

$A$ into itself is not surjective.

0%
True
0%
False

Let

$f:A\to B$ and

$g:B\to C$ be the bijective function. Then

$(gof)^{-1}$ is

0%
$f^{-1}og^{-1}$
0%
$fog$
0%
$g^{-1} of^{-1}$
0%
$gof$

Explanation

Given that,

$f:A\to B$ and

$g:B\to C$ be the bijective functions.

$(gof)^{-1}=f^{-1}og^{-1}$

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Functions Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.