Let a – 2b + c = 1. If $$f(x)=\left| {\matrix{ {x + a} & {x + 2} & {x + 1} \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$, then:

R1 $$ \to $$ R1 + R3 – 2R2 f(x) = $$\left| {\matrix{ {a + c - 2b} & 0 & 0 \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$ = (a + c – 2b) ((x + 3)2 – (x + 2)(x + 4)) = x2 + 6x + 9 – x2 – 6x – 8 = 1 $$ \therefore $$ f(x) = 1 $$ \Rightarrow $$ f(50) = 1

Functions — Page 4 | JEE Mathematics

Q31

Let

\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)}

where the function ƒ satisfies ƒ(x + y) = ƒ(x)ƒ(y) for all natural numbers x, y and ƒ(1) = 2. then the natural number 'a' is

A 2

B 16

C 4

D 3

Correct Answer

Option D

Solution

Given ƒ(1) = 2 and ƒ(x + y) = ƒ(x)ƒ(y) When x = 1 and y = 1 then, ƒ(1 + 1) = ƒ(1)ƒ(1) $\Rightarrow$ f(2) = (f(1))2 = 22 Also when x = 2 and y = 1 then, ƒ(2 + 1) = ƒ(2)ƒ(1) $\Rightarrow$ f(3) = 23 $\therefore$ Similarly f(4) = 24 . . . . f(x) = 2x $\therefore$ f(a + k) = 2a + k Now given,

\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)}

$\Rightarrow$

\sum\limits_{k = 1}^{10} {{2^{a + k}}}

=

16\left( {{2^{10}} - 1} \right)

$\Rightarrow$

{2^{a + 1}} + {2^{a + 2}} + .... + {2^{a + 10}}

=

16\left( {{2^{10}} - 1} \right)

$\Rightarrow$

{2^a}\left[ {2 + {2^2} + .... + {2^{10}}} \right]

=

16\left( {{2^{10}} - 1} \right)

$\Rightarrow$

{2^a} \times {{2\left( {{2^{10}} + 1} \right)} \over {2 - 1}}

=

16\left( {{2^{10}} - 1} \right)

$\Rightarrow$

{2^a} = 8

$\Rightarrow$

a

= 3

Q32

If

f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)

,

\left| x \right| < 1

then

f\left( {{{2x} \over {1 + {x^2}}}} \right)

is equal to

A 2f(x2)

B 2f(x)

C (f(x))2

D -2f(x)

Correct Answer

Option B

Solution

Given,

f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)

f\left( {{{2x} \over {1 + {x^2}}}} \right)

=

\ln \left( {{{1 - {{2x} \over {1 + {x^2}}}} \over {1 + {{2x} \over {1 + {x^2}}}}}} \right)

=

\ln \left( {{{{x^2} - 2x + 1} \over {{x^2} + 2x + 1}}} \right)

=

\ln {\left( {{{1 - x} \over {1 + x}}} \right)^2}

=

2\ln \left( {{{1 - x} \over {1 + x}}} \right)

= 2f(x)

Q33

The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is :

A 65

\times

(15)!

B 56

\times

15

C (15)!

\times

6!

D 5!

\times

6!

Correct Answer

Option C

Solution

Given that $f(k)$ is a multiple of 3 whenever $k$ is a multiple of 4, we need to consider how to map elements from the domain {1, 2, 3, ..., 20} to the codomain {1, 2, 3, ..., 20} following this rule.

1.

We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}.

There are 5 elements in this subset.

2.

We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}.

There are 6 elements in this subset.

3.

According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3.

This can be done in ${ }^6C_5 \cdot 5! = 6!$ ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.

4.

The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4).

This can be done in $15!$ ways.

So, combining these two cases, the total number of onto functions $f$ is $6! \times 15!$ , which corresponds to option C.

Q34

Let fk(x) =

{1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)

for k = 1, 2, 3, ... Then for all x

\in

R, the value of f4(x)

-

f6(x) is equal to

A

{1 \over 4}

B

{5 \over {12}}

C

{{ - 1} \over {12}}

D

{1 \over {12}}

Correct Answer

Option D

Solution

f4(x) $-$ f6(x) =

{1 \over 4}

(sin4 x + cos4 x) $-$

{1 \over 6}

(sin6 x + cos6 x) =

{1 \over 4}

(1 $-$

{1 \over 2}

sin2 2x) $-$

{1 \over 6}

(1 $-$

{3 \over 4}

sin2 2x) =

{1 \over 12}

Q35

Let ƒ(x) = ax (a > 0) be written as ƒ(x) = ƒ1 (x) + ƒ2 (x), where ƒ1 (x) is an even function of ƒ2 (x) is an odd function. Then ƒ1 (x + y) + ƒ1 (x – y) equals

A 2ƒ1 (x)ƒ1 (y)

B 2ƒ1 (x + y)ƒ1 (x – y)

C 2ƒ1 (x)ƒ2 (y)

D 2ƒ1 (x + y)ƒ2 (x – y)

Correct Answer

Option A

Solution

f(x) = ax As f1(x) is even function then f1(x) =

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

=

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

As f2(x) is odd function then f2(x) =

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

=

{{{a^x} - {a^{ - x}}} \over 2}

Now, ƒ1 (x + y) + ƒ1 (x – y) =

{{{a^{x + y}} + {a^{ - \left( {x + y} \right)}} + {a^{x + y}} - {a^{ - \left( {x - y} \right)}}} \over 2}

Also ƒ1 (x)ƒ1 (y) =

\left( {{{{a^x} + {a^{ - x}}} \over 2}} \right)\left( {{{{a^y} + {a^{ - y}}} \over 2}} \right)

=

{{{a^{x + y}} + {a^{x - y}} + {a^{ - x + y}} + {a^{ - x - y}}} \over 4}

=

{{{{f_1}\left( {x + y} \right) + {f_2}\left( {x - y} \right)} \over 2}}

$\therefore$ ƒ1 (x + y) + ƒ1 (x – y) = 2ƒ1 (x)ƒ1 (y)

Q36

Let a – 2b + c = 1. If

f(x)=\left| \begin{array}{lll}{x + a} & {x + 2} & {x + 1} \\ {x + b} & {x + 3} & {x + 2} \\ {x + c} & {x + 4} & {x + 3} \end{array} \right|

, then:

A ƒ(50) = 1

B ƒ(–50) = –1

C ƒ(50) = –501

D ƒ(–50) = 501

Correct Answer

Option A

Solution

R1 $\to$ R1 + R3 – 2R2 f(x) =

\left| \begin{array}{lll}{a + c - 2b} & 0 & 0 \\ {x + b} & {x + 3} & {x + 2} \\ {x + c} & {x + 4} & {x + 3} \end{array} \right|

= (a + c – 2b) ((x + 3)2 – (x + 2)(x + 4)) = x2 + 6x + 9 – x2 – 6x – 8 = 1 $\therefore$ f(x) = 1 $\Rightarrow$ f(50) = 1

Q37

For a suitably chosen real constant a, let a function,

f:R - \left\{ { - a} \right\} \to R

be defined by

f(x) = {{a - x} \over {a + x}}

. Further suppose that for any real number

x \ne - a

and

f(x) \ne - a

, (fof)(x) = x. Then

f\left( { - {1 \over 2}} \right)

is equal to :

A

{1 \over 3}

B –3

C

- {1 \over 3}

D 3

Correct Answer

Option D

Solution

Given,

f(x) = {{a - x} \over {a + x}}

and (fof)(x) = x $\Rightarrow$ f(f(x)) =

{{a - f\left( x \right)} \over {a + f\left( x \right)}}

= x $\Rightarrow$

{{a - \left( {{{a - x} \over {a + x}}} \right)} \over {a + \left( {{{a - x} \over {a + x}}} \right)}}

= x $\Rightarrow$

{{{a^2} + ax - a + x} \over {{a^2} + ax + a + x}}

= x $\Rightarrow$ (a2 – a) + x(a + 1) = (a2 + a)x + x2(a – 1) $\Rightarrow$ a(a – 1) + x(1 – a2) – x2(a – 1) = 0 $\Rightarrow$ a = 1 $\therefore$ f(x) =

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

So,

f\left( { - {1 \over 2}} \right)

=

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

= 3

Q38

If f(x + y) = f(x)f(y) and

\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2

, x, y

\in

N, where N is the set of all natural number, then the value of

{{f\left( 4 \right)} \over {f\left( 2 \right)}}

is :

A

{2 \over 3}

B

{1 \over 9}

C

{1 \over 3}

D

{4 \over 9}

Correct Answer

Option D

Solution

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

\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2

$\Rightarrow$ f(1) + f(2) + f(3) + ........

$\infty$ = 2 ....(

1) On f(x + y) = f(x) f(y) * Put x = 1, y = 1 f(2) = (f(1))2 * Put x = 2, y = 1 f(3) = f(2). f(1) = f((1))3 * Put x = 2, y = 2 f(4) = f((2))2 = f((1))4 Now put these values in equation (1) f(1) + f((1))2 + f((1))3 + .......

= 2 $\Rightarrow$

{{f\left( 1 \right)} \over {1 - f\left( 1 \right)}}

= 2 $\Rightarrow$ f(1) =

{2 \over 3}

Now f(2) =

{\left( {{2 \over 3}} \right)^2}

and f(4) =

{\left( {{2 \over 3}} \right)^4}

$\therefore$

{{f\left( 4 \right)} \over {f\left( 2 \right)}}

=

{{{{\left( {{2 \over 3}} \right)}^4}} \over {{{\left( {{2 \over 3}} \right)}^2}}}

=

{4 \over 9}

Q39

Let f : R

\to

R be a function which satisfies f(x + y) = f(x) + f(y)

\forall

x, y

\in

R. If f(1) = 2 and g(n) =

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

, n

\in

N then the value of n, for which g(n) = 20, is :

A 20

B 9

C 5

D 4

Correct Answer

Option C

Solution

Given f(1) = 2 ; f(x + y) = f(x) + f(y) When x = y = 1 $\Rightarrow$ f(2) = 2 + 2 = 4 When x = 2, y = 1 $\Rightarrow$ f(3) = 4 + 2 = 6 g(n) =

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

= f(1) + f(2) +.........+ f(n - 1) = 2 + 4 + 6 + ......+ 2(n - 1) = 2 $\times$

\frac{\left( n-1\right) \left( n\right) }{2}

= n2 - n Given g(n) = 20 $\Rightarrow$ n2– n = 20 $\Rightarrow$ n2 – n – 20 = 0 $\Rightarrow$ n = 5

Q40

Let ƒ : (1, 3)

\to

R be a function defined by

f(x) = {{x\left[ x \right]} \over {1 + {x^2}}}

, where [x] denotes the greatest integer

\le

x. Then the range of ƒ is

A

\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 \over 5}} \right]

B

\left( {{3 \over 5},{4 \over 5}} \right)

C

\left( {{2 \over 5},{4 \over 5}} \right]

D

\left( {{2 \over 5},{3 \over 5}} \right] \cup \left( {{3 \over 4},{4 \over 5}} \right)

Correct Answer

Option A

Solution

f(x) =

\left\{ \begin{array}{ll}{{x \over {{x^2} + 1}},} & {1 < x < 2} \\ {{{2x} \over {{x^2} + 1}},} & {2 \le x < 3} \end{array} \right.

$\therefore$ f(x) is decreasing function $\therefore$ Range is

\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 \over 5}} \right]

.