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Some functions and their graphs -i - class-XI

Description: some functions and their graphs -i
Number of Questions: 47
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Time complexity to check if an edge exists between two vertices would be __________.

  1. O(V*V)

  2. O(V+E)

  3. O(1)

  4. O(E)


Correct Option: D

If $p = q$ then $px =$ ________

  1. $q$

  2. $qx$

  3. $q + x$

  4. $0$


Correct Option: B
Explanation:

Given $p = q$
multiply both sides by $x$, we get
$px = qx$, which is required value.

Which of the following functions are identity functions?

  1. $f:R\rightarrow R, f(x) = x$

  2. $g : N \rightarrow Z, g(p)= 3$

  3. $h:z \rightarrow z, h(y)=y$

  4. $g:N\rightarrow N, g(z) =z$


Correct Option: A,C,D
Explanation:

An identity function is a function that always returns the same value that was used as its argument.

Hence $f:R\rightarrow R,f(x)=x$ is an identity function

If ${ (x, 2), (4, y) }$ represents an identity function, then $( x, y)$ is :

  1. (2, 4)

  2. (4, 2)

  3. (2, 2)

  4. (4, 4)


Correct Option: A
Explanation:

Identity funtion returns the same value as of input.

Since input $x$ gives output $2$  
$\therefore$ $x=2$
Similarly input of $4$ gives output $y$
$\therefore y=4$
$\Rightarrow (x,y)=(2,4)$

Option (a) is correct.

Which of the following functions is/are constant ?

  1. $f(x)=x^{2}+2$

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

  3. $f(x)=7$

  4. $f(x)=6+x$


Correct Option: C
Explanation:

$f(x)=7$ is constant function as its values do not depend on the variable $x$.

Its value is $7$ for any value of $x$.
Option $C$ is correct.

An identity function is a?

  1. Many to many function

  2. One to One function

  3. Many to one function

  4. None


Correct Option: B
Explanation:

An identity function is of the form $f(x) = cx$


one-to-one function is a function that preserves distinctness; it never maps distinct elements of its domain to the same element of its codomain.

In other words, every element of the function's codomain is the image of at most one element of its domain.

As $f(x) =cx$ has a different value at every value of $x$, it is a One to One function.

State whether the following statement is True or False.
The inverse of an identity function is the identity function itself.

  1. True

  2. False


Correct Option: A
Explanation:

An identity function $f(x)$ is of the form $f(x) = cx$

Let inverse of $f(x)$ be $g(x)$
$\therefore g(x) = \dfrac{x}{c}$ (Identity Function)
Thus inverse of an identity function is the identity function itself.

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that for any irrational number $r,$ and any real number $x$ we have $f(x)=f(x+r)$. Then, $f$ is

  1. an identity function

  2. a constant function

  3. a zero function

  4. onto function


Correct Option: B
Explanation:

A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form y=k where k is a constant(a single value that doesn't change).

So in here our input is $x$ i f we change our input to say $x+r$
and still output does'nt change i.e.
$f(x)=f(x+r)$
it means it is a constant function,
example:
$ f(x)=5=5(x)^{0} $
$ f(x+r)=5(x+r)^{0}$
$ f(x+r)=5\times 1 $
$ f(x+r)=5=f(x) $

The graph of an Identity function is?

  1. A straight line parallel to X axis

  2. A straight line parallel to Y axis

  3. A straight line passing through the origin

  4. None


Correct Option: C
Explanation:

Identify function ,$f(x)=x$

$y=x$
So,the graph of an identity function is a straight line passing through the origin.

Let $f$ be a linear function for which $f (6)  - f (2) = 12$. The value of $f (12) - f(2)$ is equal the 

  1. $12$

  2. $18$

  3. $24$

  4. $30$


Correct Option: D
Explanation:

Let $f(x)=ax+b$ [Since $f$ is given to be linear function] where $a$ and $b$ are constants.

According to the problem 
$f(6)-f(2)=12$
or, $6a-b-2a-b=12$
or, $4a=12$
or, $a=3$.
Now 
$f(12)-f(2)$
$=12a+b-2a-b$
$=10a=30$. [Using value of $a$]

The set values of $x$ for which function $f(x)=x\ln {x}-x+1$

  1. $\left( 1,\infty \right) $

  2. $\left( \cfrac { 1 }{ e } ,\infty \right) $

  3. $[e,\infty )$

  4. $\left( 0,1 \right) \cup \left( 1,\infty \right) $


Correct Option: A
Explanation:
$f(x)=x \ln x -x +1$

we use the formula

$\log _{a}f(x)\Rightarrow f(x)>0$

$\Rightarrow x>0$

$\therefore x>0\Rightarrow (1,\infty )$

Let $f$ be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false :
$f (x) = 1, f (y) \sqrt 1, f (z) \sqrt 2$. The value of $f^{-1} (1)$ is

  1. x

  2. y

  3. z

  4. none of these


Correct Option: B
Explanation:

$f(x)=1,\quad f(y)\neq 1,\quad f(z)\neq 1$

Case 1:
$f(x)=1\ f(z)=2\ f(y)=1$
$\therefore f $ is not injective
Case 2: $f(y)\neq 1,\quad f(z)=2,\quad f(x)=1$
Case 3:
$f(z)\neq 2\quad \quad \quad f(z)=3\ f(x)\neq 1\quad \quad \quad f(x)=2\ f(y)=1\quad \quad \quad f(y)=1\ f(x)=2,f(y)=1,f(z)=3\ f^{ -1 }\left( 2 \right) =x,f^{ -1 }\left( 1 \right) =y,f^{ -1 }\left( 3 \right) =z$ 

$c \to c\,\,is\,defined\,as\,f\left( x \right) = \frac{{ax + b}}{{cx + d}}\,\,bd \ne 0$.then f is a constant function when

  1. a=c

  2. b=d

  3. ad=bc

  4. ab=cd


Correct Option: B

$f:c \to c$ is defined as $f(x) = \dfrac{{ax + b}}{{cx + d}},bd \ne 0$ then $f$ is a constant function when,

  1. a=c

  2. b=d

  3. ad=bc

  4. ab=cd


Correct Option: C
Explanation:

f($x$)=$\frac{ax+b}{cx+d}$ is a constant function,

 then lets say it equal to same constant m. 
$m(cx+d)=ax+b$ 
$a=mc $
$b=md $
$\frac{a}{c}$ =$\frac{b}{d}=m$
$\frac{a}{b}$ =$\frac{c}{d}$
 $ad=bc$
C is correct.

If $f(n+1)=f(n)$ for all $n\in N, f(7)=5$  then  $f(35)=$

  1. $25$

  2. $49$

  3. $35$

  4. $5$


Correct Option: D
Explanation:

$f(n+1)=f(n)$
$\Rightarrow f\left ( n \right )=f\left ( n-1 \right )=f\left ( n-2 \right )=...........=f\left ( 1 \right )$
$\therefore f\left ( 7 \right )=5\Rightarrow f\left ( 35 \right )=5$

Let $f(x)$ is a cubic polynomial with real coefficients, $x\ \in R$ such that $f"(3)=0,\ f'(5)=0$  
If $f(3)=1$ and $f(5)=-3$, then $f(1)$ is equal to

  1. $2$

  2. $3$

  3. $5$

  4. $6$


Correct Option: A

The complete set of values of $x$ for which the function $f(x)=2\tan^{-1}x+\sin^{-1} \dfrac{2x}{1+x^{2}}$ behaves like a constant function with positive output is equal to

  1. $x \in [-1,1]$

  2. $[1,\infty)$

  3. $(-\infty,1]$

  4. $(-\infty, -1] \cup [1,\infty)$


Correct Option: A

Let f be a polynomial function such that $f(3x)=f'(x).f"(x)$, for all $x\epsilon R$. Then :

  1. $f(2)+f'(2)=28$

  2. $f"(2)-f'(2)=0$

  3. $f"(2)-f(2)=4$

  4. $f(2)-f'(2)+f"(2)=10$


Correct Option: A

If  $f \left( \dfrac { x + y } { 2 } \right) = \dfrac { f ( x ) + f ( y ) } { 2 }$  for all  $x , y \in R$  and  $f ^ { \prime } ( o ) = - 1 , f ( o ) = 1$  then  $f(2)=$

  1. $\dfrac { 1 } { 2 }$

  2. $1$

  3. $-1$

  4. $\dfrac { -1 } { 2 }$


Correct Option: A
Explanation:

let $f(x)=ax+b$

$f(0)=1\implies b=1$
$f'(0)=-1 \implies a=-1$
$\implies f(x)=1-x$
$\implies f(2)=-1$

let $f(x)$ be a polynomial of degree $4$ having extreme values at $x=2$.if $\underset { x\rightarrow 0 }{ lim } \left( \frac { f\left( x \right)  }{ { x }^{ 2 } } +1 \right) =3$ then $f(1)$

  1. $\frac { 1 }{ 2 } $

  2. $\frac { 3 }{ 2 } $

  3. $\frac { 5 }{ 2 }$

  4. $\frac { 9 }{ 2 } $


Correct Option: A

If $\alpha$ and $\beta$ are the polynomial  $f(x)=x^2-5x+k$ such that $\alpha-\beta=1$, then value of k is 

  1. $8$

  2. $6$

  3. $\dfrac{13}{2}$

  4. $4$


Correct Option: A

If $y^2 = ax^2 +bx+c$, then $y^2 \dfrac{d^2y}{dx^2}$ is

  1. a constant function

  2. a function of x only

  3. a function of y only

  4. a function of both x and y


Correct Option: A

If $fxln\left(1+\dfrac{1}{x}\right)dx=p(x)ln\left(1+\dfrac{1}{x}\right)+\dfrac{1}{2}x-\dfrac{1}{2}ln(1+x)+c$, being arbitary costant, then

  1. $p(X)=\dfrac{1}{2}x^{2}$

  2. $p(x)=0$

  3. $p(x)=1$

  4. $none\ of\ these$


Correct Option: A

Let $f(x)$ is cubic polynomial with real coefficient such that $f''(3) = 0, f'(5) = 0$. If $f(3) = 1$ and $f(5) = -3$, then $f(1)$ is equal to

  1. $2$

  2. $3$

  3. $5$

  4. $6$


Correct Option: A

$f (x) = x^4 - 10x^3 + 35x^2 - 50x + c$ is a constant. the number of real roots of . f (x) = 0 and 
f'' (x) = 0 are respectively 

  1. 1 , 0

  2. 3, 2

  3. 1 , 2

  4. 3 , 0


Correct Option: A

Let $\displaystyle f(x)=ax^{2}+bx+c,$ where $a,b,c$ are rational, and $f: Z\rightarrow Z,$ where $Z$ is the set of integers. Then $a+b$ is

  1. a negative integer

  2. an integer

  3. nonintegral rational number

  4. none of these


Correct Option: B
Explanation:

$f:Z \rightarrow Z$ is defined as $f(x)=ax^2+bx+c$

which implies for integer inputs, the function gives integer outputs.

$ \Rightarrow f(0)=c=Z _1$ ...(1) (where $Z _1$ is some integer)

Similarly, $f(1)=a+b+c=Z _2$ ...(2) (where $Z _2$ is some integer)

(2) - (1) gives $a+b =Z _2-Z _1$, which is also an integer.

The positive integers $x$ for which $f(x)=x^{3}-8x^{2}+20x-13$ is a prime is

  1. $2$

  2. $3$

  3. $4$

  4. $5$


Correct Option: A

If $f\quad \left( x \right) ={ x }^{ 2 }+2bx+{ 2c }^{ 2 }\quad and\quad g\quad (x)\quad ={ -x }^{ 2 }\quad -2cx+{ b }^{ 2 }\quad are\quad such\quad that\quad min\quad f\quad (x)\quad >\quad max\quad g\quad (x),\quad then$ relation between b and c, is

  1. none relation

  2. 0 < c < b/2

  3. $\left| c \right| <\frac { \left| b \right| }{ \sqrt { 2 } } $

  4. $\left| c \right| >\sqrt { 2 } \left| b \right| $


Correct Option: A

If $f(x)$ is a polynomial function satisfying $f(x)f\left(\dfrac{1}{x}\right)=f(x)+\left(\dfrac{1}{x}\right)$ and $f(3)=28$, then $f(4)=$

  1. $63$

  2. $65$

  3. $66$

  4. $27$


Correct Option: A

If $f\left(x\right)$ is a polynomial such that $ f\left(a\right) f\left(b\right)<0$, then number of zeros lieing between $a$ and $b$ is 

  1. $one$

  2. $at least one$

  3. $two$

  4. $at most 2$


Correct Option: A

If $ P ( X ) = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 5 $ and P ( a ) = P ( b ) = P ( c ) = 0 then the value of ( 2 - a ) ( 2 - b ) ( 2 - c ) is

  1. 3

  2. 5

  3. 7

  4. 9


Correct Option: A

If f : R $\rightarrow$ R, g : R $\rightarrow$ R and h : R $\rightarrow$ R is such that $f(x) = x^2, g(x) = tan  x$ and $h(x) = log  x$, then the value of [ho(gof)], if $x = \displaystyle \dfrac{\pi}{2}$ will be

  1. 0

  2. 1

  3. -1

  4. 10


Correct Option: A
Explanation:

$ho(gof) =(hof)(f(x))$
$=(hog)(x^2)=(hof) (\dfrac{\pi}{4}) = h(g(\dfrac{\pi}{4}))$
$= h(tan \dfrac{ \pi}{4}) = h(1) = log 1 =0$

If f is a constant function and f(100)=100  then f(2007)=_____

  1. 2007

  2. 100

  3. 0

  4. None of these


Correct Option: B
Explanation:

A constant function $ f(x) $ will have the result as a constant for any value of $ x $

So, if $ f(100) = 100 $
then $ f(2007 ) $ is also $ 100 $

The number of elements of an identity function defined on a set containing four elements is______

  1. $\displaystyle 2^{2}$

  2. $\displaystyle 2^{4}$

  3. $\displaystyle 2^{8}$

  4. $\displaystyle 2^{16}$


Correct Option: A
Explanation:

If an element is related to itself, it is called an identity function. That is $ f(x) = x $

So, if  the set has $ 4 $ elements, then the function will also have $ 4 = 2^2 $ elements.

On differentiating an identity function, we get?

  1. Signum function

  2. Sinc function

  3. Constant function

  4. None


Correct Option: C
Explanation:

Derivative of an Identity function gives a Constant function



$\dfrac{d(cx)}{dx} = c$ 
Where $c$ is a constant.

If $f,g,h$ are three functions from a set of positive real numbers into itself satisfying the condition,
$f(x) \cdot g(x)=h \sqrt{x^2 + y^2}$ such that $x,y \epsilon (0,\infty)$.then, $\dfrac{f(x)}{g(x)}$ is a?

  1. Constant function

  2. Identity function

  3. Zero function

  4. Signum function


Correct Option: A

A constant function is a periodic function.

  1. True

  2. False


Correct Option: A
Explanation:

$f(x)=f(x+a)$
A constant function is a period with any value of $'a'$ as a period

Let $f(-2, 2)\rightarrow(-2, 2)$ be a continuous function given $f(x)=f{(x}^{2})$. Given $f(0)=\dfrac{1}{2}$ then the $4f(\dfrac{1}{2})$

  1. $4$

  2. $2$

  3. $-2$

  4. $1$


Correct Option: A

If $f\left( x \right)$ is a function satisfying  $f\left( x \right).f\left( {\frac{1}{x}} \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right)$ and $f\left( 4 \right) = 65$ then find $f\left( 6 \right)$

  1. $217$

  2. $215$

  3. $-216$

  4. $-217$


Correct Option: D

Let $f\left( x \right) = p{x^2} + qx - \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right),\,\left( {p,q,a,b,c \in R} \right)(a,b,c$ are distinct). If both roots of $f(x)=0$ are non-real, then 

  1. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$

  2. $2\left( {p + q} \right) - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$

  3. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] < 0$

  4. $p - 2q - 2 - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] > 0$


Correct Option: C

If $f(x)$ is a polynomial function satisfying the condition $f(x) \times f\left(\dfrac{1}{x}\right)=f(x)+f\left(\dfrac{1}{x}\right)$ and $f(2)=9$ then

  1. $2f(4) =3 f(6)$

  2. $14f(1) = f(3)$

  3. $ 9f(3) = 2f(5)$

  4. $f(10) = f(11)$


Correct Option: B,C
Explanation:

The polynomial which satisfies $f(x)f(1/x)=f(x)+f(1/x)$ is $ \pm x^n+1$ (standard result)
Given that $f(2) = 9 \ \Rightarrow \pm 2^n + 1 = 9 \ \Rightarrow 2^n = 8 $
(-ve sign not possible here)
$ \Rightarrow n=3$
Hence the function is $ f(x)=x^3+1$
$ \Rightarrow f(1) = 2, \; f(3)=28 , \; f(5)=126$
$ f(4) = 65, \; f(6) = 217$
Using these, we see only option B and C are correct. 

If $\displaystyle f(x)=27x^{3}+\frac{1}{x^{3}}$ and $\alpha,\beta$ are the roots of $\displaystyle 3x+\frac{1}{x}=2$ is

  1. $f(\alpha)=f(\beta)$

  2. $f(\alpha)=10$

  3. $f(\beta)=-10$

  4. none of these


Correct Option: A,C
Explanation:
$3x+\dfrac{1}{x}=2$
Cubing the above equation, we get
$ \left(3x+\dfrac{1}{x}\right)^3=2^3$
$\therefore 27x^3+\dfrac{1}{x^3}+3(3x)\left(\dfrac{1}{x}\right)\left(3x+\dfrac{1}{x}\right)=8$

$\therefore 27x^3+\dfrac{1}{x^3}+9\left(3x+\dfrac{1}{x}\right)=8$

$\therefore 27x^3+\dfrac{1}{x^3}+9(2)=8$

$\therefore 27x^3+\dfrac{1}{x^3}=-10$

$\alpha$ and $\beta$ are roots of above equation.
$\therefore 27\alpha^3+\dfrac{1}{\alpha^3}=-10$ ...(1)
and $27\beta^3+\dfrac{1}{\beta^3}=-10$         ...(2)

$f(x)=27x^3+\dfrac{1}{x^3}$

$f(\alpha)=27\alpha^3+\dfrac{1}{\alpha^3}$
$\implies f(\alpha) =-10$   ...(from 1)

Similarly, $f(\beta)=27\beta^3+\dfrac{1}{\beta^3}$
$\implies f(\beta)=-10$   ...(from 2)

$\therefore f(\alpha)=f(\beta)=-10$


So, the correct options are option (A) and (C)

If a function satisfies $(x-y)f(x+y)-(x+y)f(x-y)=2(x^{2}y-y^{3}),\forall x,y\in R$ and $ f(1)=2,$ then

  1. $f(x)$ must be polynomial function

  2. $f(3)=12$

  3. $f(0)=0$

  4. $f(x)$ may not be differentiable


Correct Option: A,B,C
Explanation:

$(x-y)f(x+y)-(x+y)f(x-y)$ $=2y((x-y)(x+y))$


Let $x-y=u, x+y=v$

$uf(v)-vf(u)=uv(v-u)$

$\displaystyle \frac{f(v)}{v}-\frac{f(u)}{u}=v-u$

$\Rightarrow  \displaystyle \left ( \frac{f(v)}{v}-v \right )=\left ( \frac{f(u)}{u}-u \right )=constant$

Let $\displaystyle \frac{f(x)}{x}-x=\lambda $

$\Rightarrow f(x)=(\lambda x+x^{2})$

$f(1)=2$

$\lambda +1=2\Rightarrow \lambda =1$

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

If $g(x)$ is a polynomial satisfying $g(x) g(y) = g(x) + g(y) + g(xy) - 2$ for all real $x$ and $y$ and $g(2) = 5$ then $g(3)$ is equal  to -

  1. $10$

  2. $24$

  3. $21$

  4. none of these


Correct Option: A
Explanation:

$g(x) g(y) = g (x) + g (y) + g (xy) - 2$


Substitute $x = 2$ & $y = 1$

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

$\Rightarrow 4g (1) = 8 $

$\Rightarrow g (1) = 2$

$g (x) g(y) = g (x) + g (y) + g(xy)-2$, now substitute $y\, =\, \displaystyle \frac{1}{x}$

Now $g (x)\, g \left (\displaystyle \frac{1}{x} \right )\, =\, g (x)\, +\, g \left (\displaystyle \frac{1}{x} \right )$

$g(x) = 1\, \pm\, x^n$

$\therefore\, 5\, =\, 1\, \pm\, 2^n\, (\because\, g (2)\, =\, 5)$

$2^n=4$

So, $n = 2$

Now $g (3) = 1$ + $3^2=10$

Write a rational function $f$ that has vertical asymptote at $x=4$, a horizontal asymptote at $y=5$ and a zero at $x=-7$.

  1. $f(x)=\dfrac{5(x-7)}{(x-4)}$

  2. $f(x)=\dfrac{5(x+7)}{(x-4)}$

  3. $f(x)=\dfrac{5(x-7)}{(x+4)}$

  4. $f(x)=\dfrac{(x+7)}{(x+4)}$


Correct Option: B
Explanation:

Since the rational function $f$ has the vertical asymptote at $x=4$, then the denominator of $f$ contains the term $(x-4)$.

Thus function $f(x)$ is of the form $f=\dfrac{g(x)}{x-4}$.
Since the horizontal asymptote exists $y=5$, the numerator $g(x)$ of $f(x)$ has to be of the same degree as the denominator with a leading coefficient equal to $5$.
Also $g(x)$ must contain the term $(x+7)$ since $f$ has zero at $x=-7$.
Hence, $f(x)=\dfrac{5(x+7)}{(x-4)}$

A large mixing tank currently contains $200$ gallons of water into which $10$ pounds of sugar have been mixed. A tap will open pouring $20$ gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of $2$ pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after $14$ minutes. Then 

  1. the concentration is greater than at the beginning?

  2. the concentration lesser than at the beginning?

  3. the concentration equal to the concentration at the beginning?

  4. None of the above


Correct Option: A
Explanation:

Let $t$ be the number of minutes since the tap opened. 

Since the water increases at $20$ gallons per minute, and the sugar increases at $2$ pound per minute, these are constant rates of change. 

This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. 

We can write an equation independently for each:

$\text{Water}: \: W(t)=200+20t$ in gallons

$\text{Sugar}: \: S(t)=10+2t$ in pounds

The concentration $C$ will be the ratio of pounds of sugar to gallons of water.

$C(t)=\dfrac{10+2t}{200+20t}$

The concentration after $14$ minutes is given by evaluating $C(t)$ at $t=14$

$\therefore C(14)=\dfrac{10+28}{200+280}=\dfrac{19}{240}$

This means the concentration is $19$ pounds of sugar to $240$ gallons of water.

At the beginning, the concentration is $C(0)=\dfrac{10+0}{200+0}=\dfrac{1}{20}$.

Since $\dfrac{19}{240}\approx 0.08 > \dfrac{1}{20}=0.05$, the concentration is greater after $14$ minutes than at the beginning.

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