Functions

$f(x) = \sqrt {3 - x} + \sqrt {x + 2} $

$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$

$ \Rightarrow \sqrt x + 2 = \sqrt 3 - x$

$ \Rightarrow x = {1 \over 2}$

$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10} $

${y_{\min }}$ at $x = - 2$ or $x = 3$, i.e., $\sqrt 5 $

$\therefore$ $f(x) \in [\sqrt 5 ,\sqrt {10} ]$

$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$ ..... (i)

$n \to n + 1$

$f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)$ ...... (ii)

(i) and (ii) gives

$3f(3) - 2f(2) = 0$

$4f(4) - 3f(3) = 0$

$ \vdots $

$(n + 1)f(n + 1) - nf(n) = 0$

$ \Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}$

$f(n) = {1 \over {2n}}$

${1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100$

$f(x) = {{{x^2} + 2x + 1} \over {({x^2} + 1)}}$

$ \Rightarrow f'(x) = {{({x^2} + 1)(2x + 2) - ({x^2} + 2x + 1)(2x)} \over {{{({x^2} + 1)}^2}}}$

$ \Rightarrow f'(x) = {{2 - 2{x^2}} \over {{{({x^2} + 1)}^2}}}$

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

$ \therefore $ $f'(x) = 0$ at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima.

Option A : $f(x)$ is many-one in $( - \infty , - 1)$.

As x = -1 is a point of maxima/minima. So it is a boundary point in the range $( - \infty , - 1)$. So, $f(x)$ is one-one in $( - \infty , - 1)$.

$ \therefore $ Option A is incorrect.

Option B : $f(x)$ is one-one in $( - \infty ,\infty )$.

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $( - \infty ,\infty )$. So, $f(x)$ is many-one function in $( - \infty ,\infty )$.

$ \therefore $ Option B is incorrect.

Option C : $f(x)$ is one-one in $[1,\infty )$ but not in $( - \infty ,\infty )$.

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $[1,\infty )$. So, $f(x)$ is one-one in $[1,\infty )$.

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $( - \infty ,\infty )$. So, $f(x)$ is many-one function in $( - \infty ,\infty )$ .

$ \therefore $ Option C is correct.

Option D : $f(x)$ is many-one in $(1,\infty )$.

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $[1,\infty )$. So, $f(x)$ is one-one in $[1,\infty )$.

$ \therefore $ Option D is incorrect.

Note :

Methods to Check One-One Function :

(i) If a function is one-one, any line parallel to $x$-axis cuts the graph of the function maximum at one point.

(ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.

$f(1) = a + b + c = 3$ ..... (i)

$f(3) = 9a + 3b + c = 4$ .... (ii)

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

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

OR $4a - 2b + 2c = 7$ ..... (iii)

From (i) and (ii) $8a + 2b = 1$ ..... (iv)

From (iii) $ - (2) \times $ (i)

$ \Rightarrow 2a - 4b = 1$ ..... (v)

From (iv) and (v) $a = {1 \over 6},\,b = {{ - 1} \over 6}$ and $c = 3$

$f( - 2) = 4a - 2b + c$

$ = {4 \over 6} + {2 \over 6} + 3 = 4$

$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)$

${{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0$

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

$f(x) = \alpha {x^5} + \beta {x^3} + \gamma x$

$\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)} $

$ = 0\alpha + 0\beta + 0\gamma $

$ = 0$

$\therefore$ $f\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} $

$f,g:N - \{ 1\} \to N$ defined as

$f(a) = \alpha $, where $\alpha$ is the maximum power of those primes p such that p^$\alpha$ divides a.

$g(a) = a + 1$,

Now,

$\matrix{ {f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \cr {f(3) = 1,} & {g(3) = 4} & \Rightarrow & {(f + g)\,(3) = 5} \cr {f(4) = 2,} & {g(4) = 5} & \Rightarrow & {(f + g)\,(4) = 7} \cr {f(5) = 1,} & {g(5) = 6} & \Rightarrow & {(f + g)\,(5) = 7} \cr } $

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

$\therefore$ $f + g$ is not one-one

Now, $\because$ ${f_{\min }} = 1,\,{g_{\min }} = 3$

So, there does not exist any $x \in N - \{ 1\} $ such that $(f + g)(x) = 1,2,3$

$\therefore$ $f + g$ is not onto

As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction $f(3) > f(9) > f(15)\,.......\, > f(99)$

So number of ways $ = {}^{50}{C_{17}}\,.\,1\,.\,33!$

$ = {}^{50}{P_{33}}$

Given, $f(1) + f(2) = f(3)$

It means $f(1),f(2)$ and $f(3)$ are dependent on each other. But there is no condition on $f(4)$, so $f(4)$ can be $f(4) = 1,2,3,4,5,6$.

For $f(1),f(2)$ and we have to find how many functions possible which will satisfy the condition $f(1) + f(2) = f(3)$

Case 1 :

When $f(3) = 2$ then possible values of $f(1)$ and $f(2)$ which satisfy $f(1) + f(2) = f(3)$ is $f(1) = 1$ and $f(2) = 1$.

And $f(4)$ can be = 1, 2, 3, 4, 5, 6

$\therefore$ Total possible functions $=1\times6=6$

Case 2 :

When $f(3) = 3$ then possible values

(1) $f(1) = 1$ and $f(2) = 2$

(2) $f(1) = 2$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5, 6.

$\therefore$ Total functions $ = 2 \times 6 = 12$

Case 3 :

When $f(3) = 4$ then

(1) $f(1) = 1$ and $f(2) = 3$

(2) $f(1) = 2$ and $f(2) = 2$

(3) $f(1) = 3$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5, 6

$\therefore$ Total functions $ = 3 \times 6 = 18$

Case 4 :

When $f(3) = 5$ then

(1) $f(1) = 1$ and $f(4) = 4$

(2) $f(1) = 2$ and $f(4) = 3$

(3) $f(1) = 3$ and $f(4) = 2$

(4) $f(1) = 4$ and $f(4) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5 and 6

$\therefore$ Total functions $ = 4 \times 6 = 24$

Case 5 :

When $f(3) = 6$ then

(1) $f(1) = 1$ and $f(2) = 5$

(2) $f(1) = 2$ and $f(2) = 4$

(3) $f(1) = 3$ and $f(2) = 3$

(4) $f(1) = 4$ and $f(2) = 2$

(5) $f(1) = 5$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5 and 6

$\therefore$ Total possible functions $ = 5 \times 6 = 30$

$\therefore$ Total functions from those 5 cases we get

$ = 6 + 12 + 18 + 24 + 30$

$ = 90$

When n = 1, 5, 9, 13 then ${{n + 1} \over 2}$ will give all odd numbers.

When n = 3, 7, 11, 15 .....

n $-$ 1 will be even but not divisible by 4

When n = 2, 4, 6, 8 .....

Then 2n will give all multiples of 4

So range will be N.

And no two values of n give same y, so function is one-one and onto.

$f:R \to R$ defined as

$f(x) = x - 1$ and $g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}$

Now $fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}$

$\therefore$ Domain of $fog(x) = R - \{ - 1,1\} $

And range of $fog(x) = ( - \infty , - 1] \cup (0,\infty )$

Now, ${d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}$

$\therefore$ ${d \over {dx}}(fog(x)) > 0$ for ${{2x} \over {(1 - x)(1 + x)}} > 0$

$ \Rightarrow {x \over {(x - 1)(x + 1)}} < 0$

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

and ${d \over {dx}}(fog(x)) < 0$ for $x \in ( - 1,0) \cup (1,\infty )$

$\therefore$ $fog(x)$ is neither one-one nor onto.

Given,

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

Also given,

${f^{n + 1}}(x) = f({f^n}(x))$ ..... (1)

$\therefore$ For $n = 1$

${f^{1 + 1}}(x) = f({f^1}(x))$

$ \Rightarrow {f^2}(x) = f(f(x))$

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

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

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

$ = {{ - 2} \over {2x}}$

$ = - {1 \over x}$

From equation (1), when n = 2

${f^{2 + 1}}(x) = f({f^2}(x))$

$ \Rightarrow {f^3}(x) = f({f^2}(x))$

$ = f\left( { - {1 \over x}} \right)$

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

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

$ = {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}$

Similarly,

${f^4}(x) = f({f^3}(x))$

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

$ = {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}$

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

$ = {{ - 2x} \over { - 2}} = x$

$\therefore$ ${f^5}(x) = f({f^4}(x))$

$ = f(x)$

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

${f^6}(x) = f({f^5}(x))$

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

$ = - {1 \over x}$ (Already calculated earlier)

${f^7}(x) = f({f^6}(x))$

$ = f\left( { - {1 \over x}} \right)$

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

$ = {{ - (x + 1)} \over {x - 1}}$

$\therefore$ ${f^6}(6) = - {1 \over 6}$

and ${f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} = - {8 \over 6}$

So, ${f^6}(6) + {f^7}(7)$

$ = - {1 \over 6} - {8 \over 6}$

$ = - {3 \over 2}$

Given,

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

and $f(1) = 2$

For x = 1 and y = 1,

$f(1 + 1) = 2f(1)f(1)$

$ \Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$

For x = 1, y = 2,

$f(1 + 2) = 2f(1)y(2)$

$ \Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$

For x = 1, y = 3,

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

$ \Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$

For x = 1, y = 4,

$f(1 + 4) = 2f(1)f(4)$

$ \Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$ ..... (1)

Also given

$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $

$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 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(\alpha + 1) = {2^9}$ ..... (2)

From equation (1), $f(5) = {2^9}$

$\therefore$ From (1) and (2), we get

$f(\alpha + 1) = {2^9} = f(5)$

$ \Rightarrow f(\alpha + 1) = f(5)$

$ \Rightarrow f(\alpha + 1) = f(4 + 1)$

Comparing both sides we get,

$\alpha = 4$

$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$

$f'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}$

$ = {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R$

$g(x) = {e^{ - x}} - 2{e^x}$

$g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R$

$\Rightarrow$ 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} < \alpha - {5 \over 3}$

$ = {\alpha ^2} - 5\alpha + 6 < 0$

$ = (\alpha - 2)(\alpha - 3) < 0$

$ = \alpha \in (2,\,3)$