Functions — Page 2 | JEE Mathematics

Q11

The period of

{\sin ^2}\theta

is

A

{\pi ^2}

B

\pi

C

2\pi

D

\pi /2

Correct Answer

Option B

Solution

The period of

{\sin ^2}\theta

is = $\pi$ Note : (1) When

n

is odd then the period of

{\sin ^n}\theta

,

{\cos ^n}\theta

,

{\csc ^n}\theta

,

{\sec ^n}\theta

=

2\pi

(2) When

n

is even then the period of

{\sin ^n}\theta

,

{\cos ^n}\theta

,

{\csc ^n}\theta

,

{\sec ^n}\theta

= $\pi$ (3) When

n

is even/odd then the period of

{\tan ^n}\theta

,

{\cot ^n}\theta

= $\pi$ (3) When

n

is even/odd then the period of

\left| {{{\sin }^n}\theta } \right|

,

\left| {{{\cos }^n}\theta } \right|

,

\left| {{{\csc }^n}\theta } \right|

,

\left| {{{\sec }^n}\theta } \right|

,

\left| {{{\tan }^n}\theta } \right|

,

\left| {{{\cot }^n}\theta } \right|

= $\pi$

Q12

A function

f

from the set of natural numbers to integers defined by

f\left( n \right) = \left\{ \begin{array}{ll}{{{n - 1} \over 2},\,when\,n\,is\,odd} \\ { - {n \over 2},\,when\,n\,is\,even} \end{array} \right.

$ is

A neither one -one nor onto

B one-one but not onto

C onto but not one-one

D one-one and onto both

Correct Answer

Option D

Solution

We have

f:N \to I

If

x

and

y

are two even natural numbers, then

f\left( x \right) = f\left( y \right) \Rightarrow {{ - x} \over 2} = {{ - y} \over 2} \Rightarrow x = y

Again if

x

and

y

are two odd natural numbers then

f\left( x \right) = f\left( y \right) \Rightarrow {{x - 1} \over 2} = {{y - 1} \over 2} \Rightarrow x = y

$\therefore$

f

is onto.

Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.

$\therefore$

f

is onto. Hence

f

is one one and onto both.

Q13

If

f:R \to R

satisfies

f

(x + y) =

f

(x) +

f

(y), for all x, y

\in

R and

f

(1) = 7, then

\sum\limits_{r = 1}^n {f\left( r \right)}

is

A

{{7n\left( {n + 1} \right)} \over 2}

B

{{7n} \over 2}

C

{{7\left( {n + 1} \right)} \over 2}

D

7n + \left( {n + 1} \right)

Correct Answer

Option A

Solution

f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).

Function should be

f(x)=mx

f\left( 1 \right) = 7;

$\therefore$

m=7,

f\left( x \right) = 7x

\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}}

Q14

Domain of definition of the function f(x) =

{3 \over {4 - {x^2}}}

+

{\log _{10}}\left( {{x^3} - x} \right)

, is

A (-1, 0)

\cup

(1, 2)

\cup

(2,

\infty

)

B (1, 2)

C (-1, 0)

\cup

(1, 2)

D (1, 2)

\cup

(2,

\infty

)

Correct Answer

Option A

Solution

f\left( x \right) = {3 \over {4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)

4 - {x^2} \ne 0;\,\,\,{x^3} - x > 0;

x \ne \pm \sqrt 4

and

- 1 < x < 0

or

\,\,\,1 < x < \infty

$\therefore$

D = \left( { - 1,0} \right) \cup \left( {1,\infty } \right) - \left\{ {\sqrt 4 } \right\}

D = \left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right).

Q15

The domain of the function

f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}

A [1, 2]

B [2, 3)

C [1, 2)

D [2, 3]

Correct Answer

Option B

Solution

f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}

is defined if

(i)

\,\,\, - 1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4

and

(ii)

9 - {x^2} > 0 \Rightarrow - 3 < x < 3

Taking common solution of

\left( i \right)

and

\left( {ii} \right),

we get

2 \le x < 3

$\therefore$ Domain

= \left[ {2,\left. 3 \right)} \right.

Q16

The graph of the function y = f(x) is symmetrical about the line x = 2, then

A

f\left( x \right) = - f\left( { - x} \right)

B

f\left( {2 + x} \right) = f\left( {2 - x} \right)

C

f\left( x \right) = f\left( { - x} \right)

D

f\left( {x + 2} \right) = f\left( {x - 2} \right)

Correct Answer

Option B

Solution

Let us consider a graph symm. with respect to line

x=2

as shown in the figure. From the figure

f\left( {{x_1}} \right) = f\left( {{x_2}} \right),

where

{x_1} = 2 - x

and

{x_2} = 2 + x

$\therefore$

\,\,\,\,f\left( {2 - x} \right) = f\left( {2 + x} \right)

Q17

If

f:R \to S

, defined by

f\left( x \right) = \sin x - \sqrt 3 \cos x + 1

, is onto, then the interval of

S

is

A [-1, 3]

B [-1, 1]

C [0, 1]

D [0, 3]

Correct Answer

Option A

Solution

f\left( x \right)

is onto $\therefore$

S=

range of

f(x)

Now

f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1

= 2\sin \left( {x - {\pi \over 3}} \right) + 1

As

1 - \le \sin \left( {x - {\pi \over 3}} \right) \le 1

- 1 \le 2\sin \left( {x - {\pi \over 3}} \right) + 1 \le 3

$\therefore$

f\left( x \right) \in \left[ { - 1,3} \right] = S

Q18

The range of the function f(x) =

{}^{7 - x}{P_{x - 3}}

is

A {1, 2, 3, 4, 5}

B {1, 2, 3, 4, 5, 6}

C {1, 2, 3, 4}

D {1, 2, 3}

Correct Answer

Option D

Solution

The range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.

The domain of $f(x)$ is the set of all real numbers such that (i) $x \geq 3$ (since the permutation function is only defined for non-negative integers) (ii) 7 - x > 0 $\Rightarrow$ x < 7 (iii) x - 3 $\le$ 7 - x $\Rightarrow$ 2x $\le$ 10 $\Rightarrow$ x $\le$ 5.

To find the range of $f(x)$ , we need to consider what values the expression ${}^{7-x}P_{x-3}$ can take as $x$ varies over its domain.

For $x=3$ , we have ${}^{7-x}P_{x-3} = {}^{4}P_{0} = 1$ .

For $x=4$ , we have ${}^{7-x}P_{x-3} = {}^{3}P_{1} = 3$ .

For $x=5$ , we have ${}^{7-x}P_{x-3} = {}^{2}P_{2} = 2$ .

Therefore, the range of $f(x)$ is the set of all non-negative integers less than or equal to 3, i.e., ${1, 2, 3}$ .

Q19

For x

\in

(0, 3/2), let f(x) =

\sqrt x

, g(x) = tan x and h(x) =

{{1 - {x^2}} \over {1 + {x^2}}}

. If

\phi

(x) = ((hof)og)(x), then

\phi \left( {{\pi \over 3}} \right)

is equal to :

A

\tan {{7\pi } \over {12}}

B

\tan {{11\pi } \over {12}}

C

\tan {\pi \over {12}}

D

\tan {{5\pi } \over {12}}

Correct Answer

Option B

Solution

\phi \left( x \right) = \left( {\left( {hof} \right)og} \right)(x) = h\left( {\sqrt {\tan x} } \right)

\Rightarrow \phi (x) = {{1 - \tan x} \over {1 + \tan x}} = \tan \left( {{\pi \over 4} - 4} \right)

$\therefore$

\phi \left( {{\pi \over 3}} \right) = \tan \left( {{\pi \over 4} - {\pi \over 3}} \right)

\Rightarrow \tan \left( { - {\pi \over {12}}} \right) = \tan {{11\pi } \over {12}}

Q20

Let

f:( - 1,1) \to B

, be a function defined by

f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}

, then

f

is both one-one and onto when B is the interval

A

\left( {0,{\pi \over 2}} \right)

B

\left[ {0,{\pi \over 2}} \right)

C

\left[ { - {\pi \over 2},{\pi \over 2}} \right]

D

\left( { - {\pi \over 2},{\pi \over 2}} \right)

Correct Answer

Option D

Solution

Given

\,\,f\left( x \right) = {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x

for

x \in \left( { - 1,1} \right)

If

\,\,x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 4},{\pi \over 4}} \right)

\Rightarrow 2{\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)

Clearly, range of

f\left( x \right) = \left( { - {\pi \over 2},{\pi \over 2}} \right)

For

f

to be onto, co-domain $=$ range $\therefore$ Co-domain of function

= B = \left( { - {\pi \over 2},{\pi \over 2}} \right).