Functions
Let f : R $\to$ R be a function defined by $f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$. Then $f\left( {{1 \over {100}}} \right) + f\left( {{2 \over {100}}} \right) + f\left( {{3 \over {100}}} \right) + \,\,\,.....\,\,\, + \,\,\,f\left( {{{99} \over {100}}} \right)$ is equal to ______________.
Explanation:
Given,
$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$
$\therefore$ $f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$
$ = {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$
$ = {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$
$ = {{2{e^2}} \over {e(e + {e^{2x}})}}$
$ = {{2e} \over {e + {e^{2x}}}}$
$\therefore$ $f(x) + f(1 - x) = {{2{e^{2x}}} \over {{e^{2x}} + e}} + {{2e} \over {{e^{2x}} + e}}$
$ = {{2({e^{2x}} + e)} \over {{e^{2x}} + e}}$
$ = 2$ ...... (1)
Now,
$f\left( {{1 \over {100}}} \right) + f\left( {{{99} \over {100}}} \right)$
$ = f\left( {{1 \over {100}}} \right) + f\left( {1 - {1 \over {100}}} \right)$
$ = 2$ [as $f(x) + f(1 - x) = 2$]
Similarly,
$f\left( {{2 \over {100}}} \right) + f\left( {1 - {2 \over {100}}} \right) = 2$
$ \vdots $
$f\left( {{{49} \over {100}}} \right) + f\left( {1 - {{49} \over {100}}} \right) = 2$
$\therefore$ Total sum $ = 49 \times 2$
Remaining term $ = f\left( {{{50} \over {100}}} \right) = f\left( {{1 \over 2}} \right)$
Put $x = {1 \over 2}$ in equation (1), we get
$f\left( {{1 \over 2}} \right) + f\left( {1 - {1 \over 2}} \right) = 2$
$ \Rightarrow 2f\left( {{1 \over 2}} \right) = 2$
$ \Rightarrow f\left( {{1 \over 2}} \right) = 1$
$\therefore$ Sum $ = 49 \times 2 + 1 = 99$
Let $f:R \to R$ be a function defined by
$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)(2 + {x^{25}})} \right)^{{1 \over {50}}}}$. If the function $g(x) = f(f(f(x))) + f(f(x))$, then the greatest integer less than or equal to g(1) is ____________.
Explanation:
Given,
$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)\left( {2 + {x^{25}}} \right)} \right)^{{1 \over {50}}}}$
and $g(x) = f\left( {f\left( {f\left( x \right)} \right)} \right) + f\left( {f\left( x \right)} \right)$
$\therefore$ $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$
Now, $f(1) = {\left( {2\left( {1 - {{{1^{25}}} \over 2}} \right)\left( {2 + {1^{25}}} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( {2\left( {1 - {1 \over 2}} \right)\left( {2 + 1} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( 3 \right)^{{1 \over {50}}}}$
$\therefore$ $f\left( {f\left( 1 \right)} \right) = f\left( {{3^{{1 \over {50}}}}} \right)$
$ = {\left( {2\left( {1 - {{{{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \over 2}} \right)\left( {2 + {{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( {2\left( {1 - {{{3^{{1 \over 2}}}} \over 2}} \right)\left( {2 + {3^{{1 \over 2}}}} \right)} \right)^{{1 \over {50}}}}$
$ = {\left( {2 \times \left( {{{2 - \sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)} \right)^{{1 \over {50}}}}$
$ = {\left[ {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \right]^{{1 \over {50}}}}$
$ = {\left( {4 - 3} \right)^{{1 \over {50}}}}$
$ = {1^{{1 \over {50}}}} = 1$
Now, $f\left( {f\left( {f\left( 1 \right)} \right)} \right) = f(1) = {3^{{1 \over {50}}}}$
$\therefore$ $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$
$ = {3^{{1 \over {50}}}} + 1$
Now, greatest integer less than or equal to $g(1)$
$ = \left[ {g(1)} \right]$
$ = \left[ {{3^{{1 \over {50}}}} + 1} \right]$
$ = \left[ {{3^{{1 \over {50}}}}} \right] + \left[ 1 \right]$
$ = [1.02] + 1$
$ = 1 + 1 = 2$
The number of one-one functions f : {a, b, c, d} $\to$ {0, 1, 2, ......, 10} such
that 2f(a) $-$ f(b) + 3f(c) + f(d) = 0 is ___________.
Explanation:
Given one-one function
$f:\{ a,b,c,d\} \to \{ 0,1,2,\,\,....\,\,10\} $
and $2f(a) - f(b) + 3f(c) + f(d) = 0$
$ \Rightarrow 3f(c) + 2f(a) + f(d) = f(b)$
Case I:
(1) Now let $f(c) = 0$ and $f(a) = 1$ then
$3 \times 0 + 2 \times 1 + f(d) = f(b)$
$ \Rightarrow 2 + f(d) = f(b)$
Now possible value of $f(d) = 2,3,4,5,6,7,$ and $8$.
f(d) can't be 9 and 10 as if $f(d) = 9$ or 10 then $f(b) = 2 + 9 = 11$ or $f(b) = 2 + 10 = 12$, which is not possible as here any function's maximum value can be 10.
$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1$ are = 7
(2) When $f(c) = 0$ and $f(a) = 2$ then
$3 \times 0 + 2 \times 2 + f(d) = f(b)$
$ \Rightarrow 4 + f(d) = f(b)$
$\therefore$ possible value of $f(d) = 1,3,4,5,6$
$\therefore$ Total possible functions in this case = 5
(3) When $f(c) = 0$ and $f(a) = 3$ then
$3 \times 0 + 2 \times 3 + f(d) = f(b)$
$ \Rightarrow 6 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 1,2,4$
$\therefore$ Total possible functions in this case = 3
(4) When $f(c) = 0$ and $f(a) = 4$ then
$3 \times 0 + 2 \times 4 + f(d) = f(b)$
$ \Rightarrow 8 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 1,2$
$\therefore$ Total possible functions in this case = 2
(5) When $f(c) = 0$ and $f(a) = 5$ then
$3 \times 0 + 2 \times 5 + f(d) = f(b)$
$ \Rightarrow 10 + f(d) = f(b)$
Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.
$\therefore$ No function is possible in this case.
$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1,2,3$ and $4$ are $ = 7 + 5 + 3 + 2 = 17$
Case II:
(1) When $f(c) = 1$ and $f(a) = 0$ then
$3 \times 1 + 2 \times 0 + f(d) = f(b)$
$ \Rightarrow 3 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 2,3,4,5,6,7$
$\therefore$ Total possible functions in this case = 6
(2) When $f(c) = 1$ and $f(a) = 2$ then
$3 \times 1 + 2 \times 2 + f(d) = f(b)$
$ \Rightarrow 7 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 0,3$
$\therefore$ Total possible functions in this case = 2
(3) When $f(c) = 1$ and $f(a) = 3$ then
$3 \times 1 + 2 \times 3 + f(d) = f(b)$
$ \Rightarrow 9 + f(d) = f(b)$
$\therefore$ Possible value of $f(d) = 0$
$\therefore$ Total possible functions in this case = 1
$\therefore$ Total possible functions when $f(c) = 1$ and $f(a) = 0,2$ and $3$ are
$ = 6 + 2 + 1 = 9$
Case III:
(1) When $f(c) = 2$ and $f(a) = 0$ then
$3 \times 2 + 2 \times 0 + f(d) = f(b)$
$ \Rightarrow 6 + f(d) = f(b)$
$\therefore$ Possible values of $f(d) = 1,3,4$
$\therefore$ Total possible functions in this case = 3
(2) When $f(c) = 2$ and $f(a) = 1$ then,
$3 \times 2 + 2 \times 1 + f(d) = f(b)$
$ \Rightarrow 8 + f(d) = f(b)$
$\therefore$ Possible values of $f(d) = 0$
$\therefore$ Total possible function in this case = 1
$\therefore$ Total possible functions when $f(c) = 2$ and $f(a) = 0,1$ are
$ = 3 + 1 = 4$
Case IV:
(1) When $f(c) = 3$ and $f(a) = 0$ then
$3 \times 3 + 2 \times 0 + f(d) = f(b)$
$ \Rightarrow 9 + f(d) = f(b)$
$\therefore$ Possible values of $f(d) = 1$
$\therefore$ Total one-one functions from four cases
$ = 17 + 9 + 4 + 1 = 31$
$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$ is :
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n $\ge$ 0.
Then which of the following statements is true?
Let g : R $ \to $ R be given as g(x) = 2x $-$ 3. Then, the sum of all the values of x for which f$-$1(x) + g$-$1(x) = ${{13} \over 2}$ is equal to :
$f(x) = {{\cos e{c^{ - 1}}x} \over {\sqrt {x - [x]} }}$, where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to :
f + g, f $-$ g, f/g, g/f, g $-$ f where $(f \pm g)(x) = f(x) \pm g(x),(f/g)x = {{f(x)} \over {g(x)}}$
function f(x) = (4a $-$ 3)(x + loge 5) + 2(a $-$ 7) cot$\left( {{x \over 2}} \right)$ sin2$\left( {{x \over 2}} \right)$, x $\ne$ 2n$\pi$, n$\in$N has critical points, is :
$f(k) = \left\{ {\matrix{ {k + 1} & {if\,k\,is\,odd} \cr k & {if\,k\,is\,even} \cr } } \right.$
Then the number of possible functions $g:A \to A$ such that $gof = f$ is :
such that f(m . n) = f(m) . f(n) for every m, n $\in$ S and m . n $\in$ S is equal to _____________.
Explanation:
Put m = 1 f(n) = f(1) . f(n) $\Rightarrow$ f(1) = 1
Put m = n = 2
$f(4) = f(2).f(2)\left\{ \matrix{ f(2) = 1 \Rightarrow f(4) = 1 \hfill \cr or \hfill \cr f(2) = 2 \Rightarrow f(4) = 4 \hfill \cr} \right.$
Put m = 2, n = 3
$f(6) = f(2).f(3)\left\{ \matrix{ when\,f(2) = 1 \hfill \cr f(3) = 1\,to\,7 \hfill \cr \hfill \cr f(2) = 2 \hfill \cr f(3) = 1\,or\,2\,or\,3 \hfill \cr} \right.$
f(5), f(7) can take any value
Total = (1 $\times$ 1 $\times$ 7 $\times$ 1 $\times$ 7 $\times$ 1 $\times$ 7) + (1 $\times$ 1 $\times$ 3 $\times$ 1 $\times$ 7 $\times$ 1 $\times$ 7)
= 490
Explanation:
$\Rightarrow$ f(1) + f(2) = 3 + f(3) = 3
The only possibility is : 0 + 1 + 2 = 3
$\Rightarrow$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.
$\left| \!{\underline {\, 3 \,}} \right. $ $\times$ $\left| \!{\underline {\, 5 \,}} \right. $ = 720
Explanation:
We know, x2 + x + 1 = (x $-$ $\omega$) (x $-$ $\omega$2)
Given, p(x) is divisible by x2 + x + 1. So, roots of p(x) is $\omega$ and $\omega$2.
As root satisfy the equation,
So, put x = $\omega$
p($\omega$) = f($\omega$3) + $\omega$g($\omega$3) = 0
= f(1) + $\omega$g(1) = 0 [$\omega$3 = 1]
= f(1) + $\left( { - {1 \over 2} + {{i\sqrt 3 } \over 2}} \right)$ g(1) = 0
$ \Rightarrow $ f(1) $-$ ${{g(1)} \over 2} + i\left( {{{\sqrt 3 g(1)} \over 2}} \right)$ = 0 + i0
Comparing both sides, we get
f(1) $-$ ${{g(1)} \over 2}$ = 0
and ${{{\sqrt 3 } \over 2}g(1) = 0}$ $ \Rightarrow $ g(1) = 0
So, f(1) = 0
Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0
Explanation:
Replace $x $ with $ {1 \over x}$
$af\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x$ ..... (ii)
(i) + (ii)
$(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )$
${{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2$
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 :
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 :
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:
$f(x) = {{x\left[ x \right]} \over {1 + {x^2}}}$ , where [x] denotes the greatest integer $ \le $ x. Then the range of ƒ is
f(x) = ${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$, x $ \in $ (-1, 1), is :
(goƒ) (x) = 4x2 - 10x + 5, then ƒ$\left( {{5 \over 4}} \right)$ is equal to:
f(x + y) = f(x)f(y) for all x, y $ \in $ R and f(1) = 3.
If $\sum\limits_{i = 1}^n {f(i)} = 363$ then n is equal to ________ .
Explanation:
put x = y = 1
$ \therefore $ f(2) = (ƒ(1))2 = 32
put x = 2, y = 1
$ \therefore $ f(3) = (ƒ(1))3 = 33
Similarly f(x) = 3x
$ \Rightarrow $ f(i) = 3i
Given, $\sum\limits_{i = 1}^n {f(i)} = 363$
$ \Rightarrow $ 3 + 32 + 33 +.... + 3n = 363
$ \Rightarrow $ ${{3\left( {{3^n} - 1} \right)} \over {3 - 1}}$ = 363
$ \Rightarrow $ 3n - 1 = ${{363 \times 2} \over 3}$ = 242
$ \Rightarrow $ 3n = 243 = 35
$ \Rightarrow $ n = 5
C = {f : A $ \to $ B | 2 $ \in $ f(A) and f is not one-one} is ______.
Explanation:
Case 1 : When 2 is the image of all element of set A.
Number of ways this is possible = 1
Case 2 : When one image is 2 and other one image is one of {1, 3, 4}.
Number of ways we can choose one of {1, 3, 4} is = 3C1.
Now divide 3 elements {a, b, c} of set A into two parts.
We can do this ${{3!} \over {2!1!}}$ ways.
Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.
We can do that 2! ways.
So number of functions in this case
= 3C1 $ \times $ ${{3!} \over {2!1!}}$ $ \times $ 2! = 18
$ \therefore $ Total number of functions = 1 + 18 = 19
$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$ is
ƒ(x + y) = ƒ(x)ƒ(y) for all natural numbers x, y and ƒ(1) = 2. then the natural number 'a' is
ƒ(x) = ${{{x^2}} \over {1 - {x^2}}}$ , is surjective, then A is equal to
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
f(n) = $\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$;
and g(n) = n $-$($-$ 1)n.
Then fog is -
Define a function $f$ : A $ \to $ R as $f(x)$ = ${{2x} \over {x - 1}}$,
then $f$ is :
and f3 (x) = $1 \over {1 - x}$ be three given
functions. If a function, J(x) satisfies
(f2 o J o f1) (x) = f3 (x) then J(x) is equal to :