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Functions

JEE Mathematics · 125 questions · Page 1 of 13 · Click an option or "Show Solution" to reveal answer

Q1
The range of f(x)=4sin⁡−1(x2x2+1)f(x)=4 \sin ^{-1}\left(\dfrac{x^{2}}{x^{2}+1}\right) is
A [0,2π][0,2 \pi]
B [0,2π)[0,2 \pi)
C [0,π)[0, \pi)
D [0,π][0, \pi]
Correct Answer
Option B
Solution
x21+x2=1−11+x2<1∴0≤x21+x2<1⇒0≤sin⁡−1(x21+x2)<π2⇒0≤4sin⁡−1(x21+x2)<2π\begin{aligned} & \frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}<1 \\\\ \therefore & 0 \leq \frac{x^2}{1+x^2}<1 \\\\ \Rightarrow & 0 \leq \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<\frac{\pi}{2} \\\\ \Rightarrow & 0 \leq 4 \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<2 \pi \end{aligned}
Q2
If $$f(x)=\left\{\begin{array}{cc}2+2 x, & -1 \leq x
A [0,1)[0,1)
B [0,3)[0,3)
C (0,1](0,1]
D [0,1][0,1]
Correct Answer
Option D
Solution
f(g(x))={2+2g(x),−1≤g(x)<0.....(1)1−g(x)3,0≤g(x)≤3.....(2)f(g(x)) = \left\{ \begin{array}{lll}{2 + 2g(x),} & { - 1 \le g(x) < 0} & {.....(1)} \\ {1 - {{g(x)} \over 3},} & {0 \le g(x) \le 3} & {.....(2)} \end{array} \right.
 By (1) x∈ϕ\text{ By (1) } x \in \phi

And by (2)

x∈[−3,0]x \in[-3,0]

and

x∈[0,1]x \in[0,1]

Range of

f(g(x))\mathrm{f(g(x))}

is

[0,1][0,1]
Q3
The function f : N → \to N defined by f (x) = x −- 5 [x5],\left[ {{x \over 5}} \right], Where N is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is :
A one-one and onto
B one-one but not onto.
C onto but not one-one.
D neither one-one nor onto.
Correct Answer
Option D
Solution

f(1) = 1 - 5

[15]\left[ {{1 \over 5}} \right]

= 1 f(6) = 6 - 5

[65]\left[ {{6 \over 5}} \right]

= 1 So, this function is many to one. f(10) = 10 - 5

[105]\left[ {{10 \over 5}} \right]

= 0 which is not present in the set of natural numbers. So this function is neither one-one nor onto.

Q4
The function f(x)f\left( x \right) =log⁡(x+x2+1) = \log \left( {x + \sqrt {{x^2} + 1} } \right), is
A neither an even nor an odd function
B an even function
C an odd function
D a periodic function
Correct Answer
Option C
Solution
f(x)=log⁡(x+x2+1)f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)
f(−x)=log⁡{−x+x2+1}f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\}
=log⁡{−x2+x2+1x+x2+1}= \log \left\{ {{{ - {x^2} + {x^2} + 1} \over {x + \sqrt {{x^2} + 1} }}} \right\}
=−log⁡(x+x2+1)=−f(x)= - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\left( x \right)
⇒f(x)    \Rightarrow f\left( x \right)\,\,\,\,

is an odd function.

Q5
A real valued function f(x) satisfies the functional equation f(x - y) = f(x)f(y) - f(a - x)f(a + y) where a is given constant and f(0) = 1, f(2a - x) is equal to
A - f(x)
B f(x)
C f(a) + f(a - x)
D f(- x)
Correct Answer
Option A
Solution
f(2a−x)=f(a−(x−a))f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right)
=f(a)f(x−a)−f(0)f(x)= f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right)
=f(a)f(x−a)−f(x)= f\left( a \right)f\left( {x - a} \right) - f\left( x \right)
=−f(x)= - f\left( x \right)
[\left[ {} \right.

as

x=0,y=0,f(0)=f2(0)−f2(a)x = 0,y = 0,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right)
        ⇒f2(a)=0⇒f(a)=0]\,\,\,\,\,\,\,\, \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( a \right) = \left. 0 \right]
⇒f(2a−x)=−f(x)\Rightarrow f\left( {2a - x} \right) = - f\left( x \right)
Q6
Let f : R → \to R be defined by f(x) = x1+x2,x∈R{x \over {1 + {x^2}}},x \in R. Then the range of f is :
A [−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right]
B R−[−12,12]R - \left[ { - {1 \over 2},{1 \over 2}} \right]
C (−- 1, 1) −- {0}
D R −- [−-1, 1]
Correct Answer
Option A
Solution

f(0) = 0 & f(x) is odd Further, if x > 0 then f(x) =

f(x)=1x+1x∈(0,12]f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]

Hence,

f(x)∈[−12,12]f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]
Q7
The real valued function f(x)=cos⁡ec−1xx−[x]f(x) = {{\cos e{c^{ - 1}}x} \over {\sqrt {x - [x]} }}, where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to :
A all real except integers
B all non-integers except the interval [ −-1, 1 ]
C all integers except 0, −-1, 1
D all real except the interval [ −-1, 1 ]
Correct Answer
Option B
Solution

Domain of

cos⁡ec−1x\cos e{c^{ - 1}}x

:

x∈(−∞,−1]∪[1,∞)x \in ( - \infty , - 1] \cup [1,\infty )

and,

x−[x]>0x - [x] > 0
⇒{x}>0\Rightarrow \{ x\} > 0
⇒x≠I\Rightarrow x \ne I

∴\therefore Required domain =

(−∞,−1]∪[1,∞)−I( - \infty , - 1] \cup [1,\infty ) - I
Q8
The domain of sin⁡−1[log⁡3(x3)]{\sin ^{ - 1}}\left[ {{{\log }_3}\left( {{x \over 3}} \right)} \right] is
A [1, 9]
B [-1, 9]
C [9, 1]
D [-9, -1]
Correct Answer
Option A
Solution
f(x)=sin⁡−1(log⁡3(x3))f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {{x \over 3}} \right)} \right)

exists if

    −1≤log⁡3(x3)≤1\,\,\,\, - 1 \le {\log _3}\left( {{x \over 3}} \right) \le 1
⇔3−1≤x3≤31\Leftrightarrow {3^{ - 1}} \le {x \over 3} \le {3^1}
⇔1≤x≤9\Leftrightarrow 1 \le x \le 9

or

    x∈[1,9]\,\,\,\,x \in \left[ {1,9} \right]
Q9
The function f(x)=x2+2x−15x2−4x+9,x∈Rf(x)=\dfrac{x^2+2 x-15}{x^2-4 x+9}, x \in \mathbb{R} is
A both one-one and onto.
B onto but not one-one.
C neither one-one nor onto.
D one-one but not onto.
Correct Answer
Option C
Solution

The function f(x)=x2+2x−15x2−4x+9,x∈R f(x)=\dfrac{x^2+2x-15}{x^2-4x+9}, x \in \mathbb{R} can be simplified to f(x)=(x−3)(x+5)x2−4x+9 f(x)=\dfrac{(x-3)(x+5)}{x^2-4x+9} .

For x=3 x=3 and x=−5 x=-5 , f(x) f(x) equals 0.

Therefore, f(x) f(x) is not one-one as it yields the same output for different input values.

The range of f(x) f(x) is [−2,1.6] [-2, 1.6] , indicating that f(x) f(x) does not cover all possible real values.

Consequently, f(x) f(x) is not onto.

Thus, the function is neither one-one nor onto.

Q10
Which one is not periodic?
A ∣sin⁡3x∣+sin⁡2x\left| {\sin 3x} \right| + {\sin ^2}x
B cos⁡x+cos⁡2x\cos \sqrt x + {\cos ^2}x
C cos⁡ 4x+tan⁡2x\cos \,4x + {\tan ^2}x
D cos 2x+sin⁡xcos\,2x + \sin x
Correct Answer
Option B
Solution
x\sqrt x

is non periodic function and

cos⁡(something)\cos \left( {something} \right)

is a periodic function so here in

cos⁡x\cos \sqrt x

→\to inside periodic function there is non periodic function which always produce non periodic function.

cos⁡2x{{{\cos }^2}x}

is a periodic function with period π\pi Note : (1) When

nn

is odd then the period of

sin⁡nθ{\sin ^n}\theta

,

cos⁡nθ{\cos ^n}\theta

,

csc⁡nθ{\csc ^n}\theta

,

sec⁡nθ{\sec ^n}\theta

=

2π2\pi

(2) When

nn

is even then the period of

sin⁡nθ{\sin ^n}\theta

,

cos⁡nθ{\cos ^n}\theta

,

csc⁡nθ{\csc ^n}\theta

,

sec⁡nθ{\sec ^n}\theta

= π\pi (3) When

nn

is even/odd then the period of

tan⁡nθ{\tan ^n}\theta

,

cot⁡nθ{\cot ^n}\theta

= π\pi (3) When

nn

is even/odd then the period of

∣sin⁡nθ∣\left| {{{\sin }^n}\theta } \right|

,

∣cos⁡nθ∣\left| {{{\cos }^n}\theta } \right|

,

∣csc⁡nθ∣\left| {{{\csc }^n}\theta } \right|

,

∣sec⁡nθ∣\left| {{{\sec }^n}\theta } \right|

,

∣tan⁡nθ∣\left| {{{\tan }^n}\theta } \right|

,

∣cot⁡nθ∣\left| {{{\cot }^n}\theta } \right|

= π\pi

cos⁡x+cos⁡2x\cos \sqrt x + {\cos ^2}x

= non periodic function + periodic function = non periodic function

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