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Functions

JEE Mathematics · 125 questions · Page 7 of 13 · Click an option or "Show Solution" to reveal answer

Q61
The number of bijective functions f:{1,3,5,7,…,99}→{2,4,6,8,….100}f:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots .100\}, such that f(3)≥f(9)≥f(15)≥f(21)≥…..f(99)f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots . . f(99), is ____________.
A 50P17{ }^{50} P_{17}
B 50P33{ }^{50} P_{33}
C 33!×1733 ! \times 17!
D 50!2\dfrac{50!}{2}
Correct Answer
Option B
Solution

As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction

f(3)>f(9)>f(15) ....... >f(99)f(3) > f(9) > f(15)\,.......\, > f(99)

So number of ways

=50C17 . 1 . 33!= {}^{50}{C_{17}}\,.\,1\,.\,33!
=50P33= {}^{50}{P_{33}}
Q62
Let f(x)=x−1x+1, x∈R−{0,−1,1}f(x) = {{x - 1} \over {x + 1}},\,x \in R - \{ 0, - 1,1\} . If fn+1(x)=f(fn(x)){f^{n + 1}}(x) = f({f^n}(x)) for all n ∈\in N, then f6(6)+f7(7){f^6}(6) + {f^7}(7) is equal to :
A 76{7 \over 6}
B −32 - {3 \over 2}
C 712{7 \over {12}}
D −1112 - {{11} \over {12}}
Correct Answer
Option B
Solution

Given,

f(x)=x−1x+1f(x) = {{x - 1} \over {x + 1}}

Also given,

fn+1(x)=f(fn(x)){f^{n + 1}}(x) = f({f^n}(x))

..... (1) ∴\therefore For

n=1n = 1
f1+1(x)=f(f1(x)){f^{1 + 1}}(x) = f({f^1}(x))
⇒f2(x)=f(f(x))\Rightarrow {f^2}(x) = f(f(x))
=f(x−1x+1)= f\left( {{{x - 1} \over {x + 1}}} \right)
=x−1x+1−1x−1x+1+1= {{{{x - 1} \over {x + 1}} - 1} \over {{{x - 1} \over {x + 1}} + 1}}
=x−1−x−1x+1x−1+x+1x+1= {{{{x - 1 - x - 1} \over {x + 1}}} \over {{{x - 1 + x + 1} \over {x + 1}}}}
=−22x= {{ - 2} \over {2x}}
=−1x= - {1 \over x}

From equation (1), when n = 2

f2+1(x)=f(f2(x)){f^{2 + 1}}(x) = f({f^2}(x))
⇒f3(x)=f(f2(x))\Rightarrow {f^3}(x) = f({f^2}(x))
=f(−1x)= f\left( { - {1 \over x}} \right)
=−1x−1−1x+1= {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}
=−1−xx−1+xx= {{{{ - 1 - x} \over x}} \over {{{ - 1 + x} \over x}}}
=−1−x−1+x=−(x+1)x−1= {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}

Similarly,

f4(x)=f(f3(x)){f^4}(x) = f({f^3}(x))
=f(−x+1x−1)= f\left( {{{ - x + 1} \over {x - 1}}} \right)
=−(x+1)x−1−1−(x+1)x−1+1= {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}
=−x−1−x+1x−1−x−1+x−1x−1= {{{{ - x - 1 - x + 1} \over {x - 1}}} \over {{{ - x - 1 + x - 1} \over {x - 1}}}}
=−2x−2=x= {{ - 2x} \over { - 2}} = x

∴\therefore

f5(x)=f(f4(x)){f^5}(x) = f({f^4}(x))
=f(x)= f(x)
=x−1x+1= {{x - 1} \over {x + 1}}
f6(x)=f(f5(x)){f^6}(x) = f({f^5}(x))
=f(x−1x+1)= f\left( {{{x - 1} \over {x + 1}}} \right)
=−1x= - {1 \over x}

(Already calculated earlier)

f7(x)=f(f6(x)){f^7}(x) = f({f^6}(x))
=f(−1x)= f\left( { - {1 \over x}} \right)
=−1x−1−1x+1= {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}
=−(x+1)x−1= {{ - (x + 1)} \over {x - 1}}

∴\therefore

f6(6)=−16{f^6}(6) = - {1 \over 6}

and

f7(7)=−(7+1)7−1=−86{f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} = - {8 \over 6}

So,

f6(6)+f7(7){f^6}(6) + {f^7}(7)
=−16−86= - {1 \over 6} - {8 \over 6}
=−32= - {3 \over 2}
Q63
Let f : R →\to R be defined as f (x) = x −- 1 and g : R −- {1, −-1} →\to R be defined as g(x)=x2x2−1g(x) = {{{x^2}} \over {{x^2} - 1}}. Then the function fog is :
A one-one but not onto
B onto but not one-one
C both one-one and onto
D neither one-one nor onto
Correct Answer
Option D
Solution
f:R→Rf:R \to R

defined as

f(x)=x−1f(x) = x - 1

and

g:R→{1,−1}→R, g(x)=x2x2−1g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}

Now

fog(x)=x2x2−1−1=1x2−1fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}

∴\therefore Domain of

fog(x)=R−{−1,1}fog(x) = R - \{ - 1,1\}

And range of

fog(x)=(−∞,−1]∪(0,∞)fog(x) = ( - \infty , - 1] \cup (0,\infty )

Now,

ddx(fog(x))=−1x2−1 . 2x=2x1−x2{d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}

∴\therefore

ddx(fog(x))>0{d \over {dx}}(fog(x)) > 0

for

2x(1−x)(1+x)>0{{2x} \over {(1 - x)(1 + x)}} > 0
⇒x(x−1)(x+1)\Rightarrow {x \over {(x - 1)(x + 1)}}

\therefore

x \in ( - \infty , - 1) \cup (0,1)

andand

{d \over {dx}}(fog(x)) ∴\therefore

fog(x)fog(x)

is neither one-one nor onto.

Q64
Let f : N →\to R be a function such that f(x+y)=2f(x)f(y)f(x + y) = 2f(x)f(y) for natural numbers x and y. If f(1) = 2, then the value of α\alpha for which ∑k=110f(α+k)=5123(220−1)\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} holds, is :
A 2
B 3
C 4
D 6
Correct Answer
Option C
Solution

Given,

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

and

f(1)=2f(1) = 2

For x = 1 and y = 1,

f(1+1)=2f(1)f(1)f(1 + 1) = 2f(1)f(1)
⇒f(2)=2(f(1))2=2(2)2=23\Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}

For x = 1, y = 2,

f(1+2)=2f(1)y(2)f(1 + 2) = 2f(1)y(2)
⇒f(3)=2 . 2 . 23=25\Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}

For x = 1, y = 3,

f(1+3)=2f(1)f(3)f(1 + 3) = 2f(1)f(3)
⇒f(4)=2 . 2 . 25=27\Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}

For x = 1, y = 4,

f(1+4)=2f(1)f(4)f(1 + 4) = 2f(1)f(4)
⇒f(5)=2 . 2 . 27=29\Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}

..... (1) Also given

∑k=110f(α+k)=5123(220−1)\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)}
⇒f(α+1)+f(α+2)+f(α+3) +  ...  +  f(α+10)=5123(220−1)\Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 1)
⇒f(α+1)+f(α+2)+f(α+3) +  ....  +f(α+10)=29((22)10−1)22−1\Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,....\,\, + f(\alpha + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}

This represent a G.P with first term = 29 and common ratio = 22 ∴\therefore First term

=f(α+1)=29= f(\alpha + 1) = {2^9}

..... (2) From equation (1),

f(5)=29f(5) = {2^9}

∴\therefore From (1) and (2), we get

f(α+1)=29=f(5)f(\alpha + 1) = {2^9} = f(5)
⇒f(α+1)=f(5)\Rightarrow f(\alpha + 1) = f(5)
⇒f(α+1)=f(4+1)\Rightarrow f(\alpha + 1) = f(4 + 1)

Comparing both sides we get,

α=4\alpha = 4
Q65
Let f:R→Rf:R \to R and g:R→Rg:R \to R be two functions defined by f(x)=log⁡e(x2+1)−e−x+1f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1 and g(x)=1−2e2xexg(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}. Then, for which of the following range of α\alpha, the inequality f(g((α−1)23))>f(g(α−53))f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right) holds ?
A (2, 3)
B (−-2, −-1)
C (1, 2)
D (−-1, 1)
Correct Answer
Option A
Solution
f(x)=log⁡e(x2+1)−e−x+1f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1
f′(x)=2xx2+1+e−xf'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}
=2x+1x+e−x>0  ∀x∈R= {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R
g(x)=e−x−2exg(x) = {e^{ - x}} - 2{e^x}
g′(x)−−e−x−2exg'(x) - - {e^{ - x}} - 2{e^x}

\Rightarrow

f(x)isincreasingandg(x)isdecreasingfunction.f(x) is increasing and g(x) is decreasing function.

f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha - {5 \over 3}} \right)} \right)

\Rightarrow {{{{(\alpha - 1)}^2}} \over 3}

=α2−5α+6= {\alpha ^2} - 5\alpha + 6

= (\alpha - 2)(\alpha - 3)

=α∈(2, 3)= \alpha \in (2,\,3)
Q66
The total number of functions, f:{1,2,3,4}→{1,2,3,4,5,6} f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\} such that f(1)+f(2)=f(3)f(1)+f(2)=f(3), is equal to :
A 60
B 90
C 108
D 126
Correct Answer
Option B
Solution

Given,

f(1)+f(2)=f(3)f(1) + f(2) = f(3)

It means

f(1),f(2)f(1),f(2)

and

f(3)f(3)

are dependent on each other. But there is no condition on

f(4)f(4)

, so

f(4)f(4)

can be

f(4)=1,2,3,4,5,6f(4) = 1,2,3,4,5,6

. For

f(1),f(2)f(1),f(2)

and we have to find how many functions possible which will satisfy the condition

f(1)+f(2)=f(3)f(1) + f(2) = f(3)

Case 1 : When

f(3)=2f(3) = 2

then possible values of

f(1)f(1)

and

f(2)f(2)

which satisfy

f(1)+f(2)=f(3)f(1) + f(2) = f(3)

is

f(1)=1f(1) = 1

and

f(2)=1f(2) = 1

. And

f(4)f(4)

can be = 1, 2, 3, 4, 5, 6 ∴\therefore Total possible functions

=1×6=6=1\times6=6

Case 2 : When

f(3)=3f(3) = 3

then possible values (1)

f(1)=1f(1) = 1

and

f(2)=2f(2) = 2

(2)

f(1)=2f(1) = 2

and

f(2)=1f(2) = 1

And

f(4)f(4)

can be = 1, 2, 3, 4, 5, 6. ∴\therefore Total functions

=2×6=12= 2 \times 6 = 12

Case 3 : When

f(3)=4f(3) = 4

then (1)

f(1)=1f(1) = 1

and

f(2)=3f(2) = 3

(2)

f(1)=2f(1) = 2

and

f(2)=2f(2) = 2

(3)

f(1)=3f(1) = 3

and

f(2)=1f(2) = 1

And

f(4)f(4)

can be = 1, 2, 3, 4, 5, 6 ∴\therefore Total functions

=3×6=18= 3 \times 6 = 18

Case 4 : When

f(3)=5f(3) = 5

then (1)

f(1)=1f(1) = 1

and

f(4)=4f(4) = 4

(2)

f(1)=2f(1) = 2

and

f(4)=3f(4) = 3

(3)

f(1)=3f(1) = 3

and

f(4)=2f(4) = 2

(4)

f(1)=4f(1) = 4

and

f(4)=1f(4) = 1

And

f(4)f(4)

can be = 1, 2, 3, 4, 5 and 6 ∴\therefore Total functions

=4×6=24= 4 \times 6 = 24

Case 5 : When

f(3)=6f(3) = 6

then (1)

f(1)=1f(1) = 1

and

f(2)=5f(2) = 5

(2)

f(1)=2f(1) = 2

and

f(2)=4f(2) = 4

(3)

f(1)=3f(1) = 3

and

f(2)=3f(2) = 3

(4)

f(1)=4f(1) = 4

and

f(2)=2f(2) = 2

(5)

f(1)=5f(1) = 5

and

f(2)=1f(2) = 1

And

f(4)f(4)

can be = 1, 2, 3, 4, 5 and 6 ∴\therefore Total possible functions

=5×6=30= 5 \times 6 = 30

∴\therefore Total functions from those 5 cases we get

=6+12+18+24+30= 6 + 12 + 18 + 24 + 30
=90= 90
Q67
Let f,g:N−{1}→Nf, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N} be functions defined by f(a)=αf(a)=\alpha, where α\alpha is the maximum of the powers of those primes pp such that pαp^{\alpha} divides aa, and g(a)=a+1g(a)=a+1, for all a∈N−{1}a \in \mathbb{N}-\{1\}. Then, the function f+gf+g is
A one-one but not onto
B onto but not one-one
C both one-one and onto
D neither one-one nor onto
Correct Answer
Option D
Solution
f,g:N−{1}→Nf,g:N - \{ 1\} \to N

defined as

f(a)=αf(a) = \alpha

, where α\alpha is the maximum power of those primes p such that pα\alpha divides a.

g(a)=a+1g(a) = a + 1

, Now,

f(2)=1,g(2)=3⇒(f+g) (2)=4f(3)=1,g(3)=4⇒(f+g) (3)=5f(4)=2,g(4)=5⇒(f+g) (4)=7f(5)=1,g(5)=6⇒(f+g) (5)=7\begin{array}{lll}{f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \\ {f(3) = 1,} & {g(3) = 4} & \Rightarrow & {(f + g)\,(3) = 5} \\ {f(4) = 2,} & {g(4) = 5} & \Rightarrow & {(f + g)\,(4) = 7} \\ {f(5) = 1,} & {g(5) = 6} & \Rightarrow & {(f + g)\,(5) = 7} \end{array}

∵\because

(f+g) (5)=(f+g) (4)(f + g)\,(5) = (f + g)\,(4)

∴\therefore

f+gf + g

is not one-one Now, ∵\because

fmin⁡=1, gmin⁡=3{f_{\min }} = 1,\,{g_{\min }} = 3

So, there does not exist any

x∈N−{1}x \in N - \{ 1\}

such that

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

∴\therefore

f+gf + g

is not onto

Q68
Let α,β\alpha, \beta and γ\gamma be three positive real numbers. Let f(x)=αx5+βx3+γx,x∈Rf(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R} and g:R→Rg: \mathbf{R} \rightarrow \mathbf{R} be such that g(f(x))=xg(f(x))=x for all x∈Rx \in \mathbf{R}. If a1,a2,a3,…,an\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}} be in arithmetic progression with mean zero, then the value of f(g(1n∑i=1nf(ai)))f\left(g\left(\dfrac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right) is equal to :
A 0
B 3
C 9
D 27
Correct Answer
Option A
Solution
f(g(1n∑i=1nf(ai)))f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)
a1+a2+a3 + ...... + ann=0{{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0

∴\therefore First and last term, second and second last and so on are equal in magnitude but opposite in sign.

f(x)=αx5+βx3+γxf(x) = \alpha {x^5} + \beta {x^3} + \gamma x
∑i=1nf(ai)=α(a15+a25+a35 + ..... + an5)+β(a13+a23 + .... + an3)+γ(a1+a2 + .... + an)\sum\limits_{i = 1}^n {f({a_i}) = \alpha \left( {a_1^5 + a_2^5 + a_3^5\, + \,.....\, + \,a_n^5} \right) + \beta \left( {a_1^3 + a_2^3\, + \,....\, + \,a_n^3} \right) + \gamma \left( {{a_1} + {a_2}\, + \,....\, + \,{a_n}} \right)}
=0α+0β+0γ= 0\alpha + 0\beta + 0\gamma
=0= 0

∴\therefore

f(g(1n∑i=1nf(ai)))=1n∑i=1nf(ai)=0f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right) = {1 \over n}\sum\limits_{i = 1}^n {f({a_i}) = 0}
Q69
 Let f(x)=ax2+bx+c be such that f(1)=3,f(−2)=λ and  \text{ Let } f(x)=a x^{2}+b x+c \text{ be such that } f(1)=3, f(-2)=\lambda \text{ and } f(3)=4f(3)=4. If f(0)+f(1)+f(−2)+f(3)=14f(0)+f(1)+f(-2)+f(3)=14, then λ\lambda is equal to :
A −-4
B 132\dfrac{13}{2}
C 232\dfrac{23}{2}
D 4
Correct Answer
Option D
Solution
f(1)=a+b+c=3f(1) = a + b + c = 3

..... (i)

f(3)=9a+3b+c=4f(3) = 9a + 3b + c = 4

.... (ii)

f(0)+f(1)+f(−2)+f(3)=14f(0) + f(1) + f( - 2) + f(3) = 14

OR

c+3+(4a−2b+c)+4=14c + 3 + (4a - 2b + c) + 4 = 14

OR

4a−2b+2c=74a - 2b + 2c = 7

..... (iii) From (i) and (ii)

8a+2b=18a + 2b = 1

..... (iv) From (iii)

−(2)×- (2) \times

(i)

⇒2a−4b=1\Rightarrow 2a - 4b = 1

..... (v) From (iv) and (v)

a=16, b=−16a = {1 \over 6},\,b = {{ - 1} \over 6}

and

c=3c = 3
f(−2)=4a−2b+cf( - 2) = 4a - 2b + c
=46+26+3=4= {4 \over 6} + {2 \over 6} + 3 = 4
Q70
Let f:R−0,1→Rf:\mathbb{R}-{0,1}\to \mathbb{R} be a function such that f(x)+f(11−x)=1+xf(x)+f\left(\dfrac{1}{1-x}\right)=1+x. Then f(2)f(2) is equal to
A 94\dfrac{9}{4}
B 74\dfrac{7}{4}
C 73\dfrac{7}{3}
D 92\dfrac{9}{2}
Correct Answer
Option A
Solution

f(x)+f(11−x)=1+xx=2⇒f(2)+f(−1)=3........(1)x=−1⇒f(−1)+f(12)=0.........(2)x=12⇒f(12)+f(2)=32........(3)(1)+(3)−(2)⇒2f(2)=92∴f(2)=94\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\dfrac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\dfrac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\dfrac{1}{2} \Rightarrow \mathrm{f}\left(\dfrac{1}{2}\right)+\mathrm{f}(2)=\dfrac{3}{2} ........(3) \\\\ & (1)+(3)-(2) \Rightarrow 2 \mathrm{f}(2)=\dfrac{9}{2} \\\\ & \therefore \mathrm{f}(2)=\dfrac{9}{4}\end{aligned}

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