Functions
Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{4^x}{4^x+2}$, what is the value of $f\left(\frac{1}{4}\right)+2 f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)$ is equal to
Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=2 x+1$ and $g(x)=x^2-2$ determine $(g \circ f)(x)$ is equal to
Given, the function $f(x)=\frac{a^x+a^{-x}}{2},(a>2)$, then $f(x+y)+f(x-y)$ is equal to
If $f$ is a function defined on $(0,1)$ by $f(x)=\min \{x-[x],-x-[x]\}$, then $(f \circ f o f o f)(x)$ is equal to $\rightarrow([\cdot]$ greatest integer function)
If ${({x^2} + 5x + 5)^{x + 5}} = 1$, then the number of integers satisfying this equation is
If $\frac{x^4}{(x-1)(x-2)}=f(x)+\frac{A}{x-1}+\frac{B}{x-2}$, then
Which statement among the following is true?
(i) the function $f(x)=x|x|$ is strictly increasing on $R-\{0\}$.
(ii) the function $f(x)=\log _{(1 / 4)} x$ is strictly increasing on $(0, \infty)$.
(iii) a one-one function is always an increasing function.
(iv) $f(x)=x^{1 / 3}$ is strictly decreasing on $R$
If f(x) = 4x $-$ x2, x$\in$R, and f(a + 1) $-$ f(a $-$ 1) = 0, then a is equal to
The maximum value of the function y = x(x $-$ 1)2, is
Find the area enclosed by the loop in the curve 4y2 = 4x2 $-$ x3.
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 Æ’ is
f(x) = ${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$, x $ \in $ (-1, 1), is :
(goÆ’) (x) = 4x2 - 10x + 5, then Æ’$\left( {{5 \over 4}} \right)$ 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) = (Æ’(1))2 = 32
put x = 2, y = 1
$ \therefore $ f(3) = (Æ’(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
$f(x) = {{{4^x}} \over {{4^x} + 2}}$
Then the value of $f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$ is ..........
Explanation:
$f(x) = {{{4^x}} \over {{4^x} + 2}}$
$ \because $ $f(1 - x) = {{{4^{1 - x}}} \over {{4^{1 - x}} + 2}} = {2 \over {2 + {4^x}}}$
$ \therefore $ $f(x) + f(1 - x) = {{{4^x}} \over {{4^x} + 2}} + {2 \over {2 + {4^x}}}$
$ = {{{4^x} + 2} \over {{4^x} + 2}}$
So, f(x) + f(1 $-$ x) = 1 .....(i)
$ \therefore $ $f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$
$ = \left[ {f\left( {{1 \over {40}}} \right) + f\left( {{{39} \over {40}}} \right)} \right] + \left[ {f\left( {{2 \over {40}}} \right) + f\left( {{{38} \over {40}}} \right)} \right] + ... + \left[ {f\left( {{{18} \over {40}}} \right) + f\left( {{{22} \over {40}}} \right)} \right] + \left[ {f\left( {{{19} \over {40}}} \right) + f\left( {{{21} \over {40}}} \right)} \right] + \left[ {f\left( {{{20} \over {40}}} \right) - f\left( {{1 \over 2}} \right)} \right]$
$ = \{ 1 + 1 + ... + 1 + 1\} + f\left( {{1 \over 2}} \right) - f\left( {{1 \over 2}} \right)$
$ = \{ 1 + 1 + ... + 1 + 1\}$(19 times) {from Eq. (i)}
= 19.
Suppose the function f has a local minimum at $\theta $ precisely when $\theta \in \{ {\lambda _1}\pi ,....,{\lambda _r}\pi \} $, where $0 < {\lambda _1} < ...{\lambda _r} < 1$. Then the value of ${\lambda _1} + ... + {\lambda _r}$ is .............
Explanation:
$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$
$ = 1 + \sin 2\theta + {(1 - \sin 2\theta )^2}$
$ = 1 + \sin 2\theta + 1 + {\sin ^2}2\theta - 2\sin 2\theta $
$ = {\sin ^2}2\theta - \sin 2\theta + 2$
$ = {\left( {\sin 2\theta - {1 \over 2}} \right)^2} + {7 \over 4}$
The local minimum of function 'f' occurs when
$\sin 2\theta = {1 \over 2}$
$ \Rightarrow 2\theta = {\pi \over 6},\,{{5\pi } \over 6},\,{{13\pi } \over 6},\,...$
$ \Rightarrow \theta = {\pi \over {12}},\,{{5\pi } \over {12}},\,{{13\pi } \over {12}},\,...$
but $\theta \in \{ {\lambda _1}\pi ,\,{\lambda _2}\pi ,\,...,\,{\lambda _r}\pi \} $,
where $0 < {\lambda _1} < .... < {\lambda _r} < 1$.
$ \therefore $ $\theta = {\pi \over {12}},\,{{5\pi } \over {12}}$
So, ${\lambda _1} + ... + {\lambda _r} = {1 \over {12}} + {5 \over {12}} = 0.50$
$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$
If $\alpha ,\,\beta \in [0,2]$ are such that $\{ x \in [0,2]:f(x) \ge 0\} = [\alpha ,\beta ]$, then the value of $\beta - \alpha $ is ..........
Explanation:
$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$
$ = (3 - \sin (2\pi x))\left[ {{{\sin \pi x} \over {\sqrt 2 }} - {{\cos \pi x} \over {\sqrt 2 }}} \right] - \left\{ {{{\sin 3\pi x} \over {\sqrt 2 }} + {{\cos (3\pi x)} \over {\sqrt 2 }}} \right\}$
$ = (3 - \sin (2\pi x)){{[\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}[3\sin (\pi x) - 4{\sin ^3}(\pi x) + 4{\cos ^3}(\pi x) - 3\cos (\pi x)]$
$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[3 - \sin (2\pi x) - 3 + 4\{ {\sin ^2}(\pi x) + {\cos ^2}(\pi x) + \sin (\pi x)\cos (\pi x)\} ]$
$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[4 + \sin (2\pi x)]$
As, $f(x) \ge 0\forall \in [\alpha ,\beta ]$, where $\alpha ,\beta \in [0,2]$, so
$\sin (\pi x) - \cos (\pi x) \ge 0$
as $4 + \sin (2\pi x) > 0\,\forall x \in R$.
$ \Rightarrow \pi x \in \left[ {{\pi \over 4},{{5\pi } \over 4}} \right] \Rightarrow x \in \left[ {{1 \over 4},{5 \over 4}} \right]$
$ \therefore $ $\alpha = {1 \over 4}$ and $\beta = {5 \over 4}$
Therefore the value of $(\beta - \alpha ) = 1$
$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $;
For a polynomial f, let f' and f'' denote its first and second order derivatives, respectively. Then the minimum possible value of (mf' + mf''), where f $ \in $ S, is ..............
Explanation:
$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $
and for a polynomial $f \in S$, Let
$f(x) = {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3})$
it have $-$1 and 1 as repeated roots twice, so graph of f(x) touches the X-axis at x = $-$1 and x = 1, so f'(x) having at least three roots x = $-$1, 1 and $\alpha $. Where $\alpha $$ \in $($-$1, 1) and f''(x) having at least two roots in interval ($-$1, 1)
So, mf' = 3 and mf'' = 2
$ \therefore $ Minimum possible value of (mf' + mf'') = 5
The number of bijective functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$ such that $f(x+y)=f(x)+f(y) \forall x, y \in \mathbf{Z}$, is
two
four
zero
infinitely many
For each $n \in \mathbf{N}$, let $A_n=\{(n+1) k / k \in \mathbf{N}\}$ and $X=\bigcup_{n \in \mathbf{N}} A_n \cdot A$ mapping $f: X \rightarrow N$ defined by $f(x)=x$, $\forall x \in \mathbf{X}$, is
one-one and onto
one-one but not onto
onto but not one-one
neither one-one nor onto
If $f: Z \rightarrow N$ is defined by
$ f(n)=\left\{\begin{array}{cll} 2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0, \text { then } f \text { is } \\ -2 n-1, & \text { if } & n<0 \end{array}\right. $
one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto
If $\frac{x^5-5}{x^3+x^2}=f(x)+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$, then the larger value of $K$ for which $f(K)+A+B+C=1$, is
3
2
-2
4
If $f(x)=x-\frac{1}{x}, x \neq 0$, then $3 f(x)=$
$3[f(x)]^2-f\left(x^2\right)$
$[f(x)]^2-f\left(x^3\right)$
$f\left(x^3\right)-[f(x)]^3$
$f\left(x^3\right)-f\left(x^2\right)$
Let $[\cdot]$ denote greatest integer function. If $f(x)=[x]$ and $g(x)=3\left[\frac{x}{3}\right]$, then the set of all real $x$ such that $f(x)=g(x)$ is
$\mathbf{R}$
$\{x \in \mathbf{R} / x=3 k, k \in \mathbf{Z}\}$
$\{x \in \mathbf{R} / 3 k-1
$\{x \in \mathbf{R} / 3 k \leq x<3 k+1, k \in \mathbf{Z}\}$
A function $f: \mathbf{R} \rightarrow \mathbf{R}$ is such that $f(\mathrm{l})=2$ and $f(x+y)=f(x) \cdot f(y) \forall x, y$. The area (in square units) enclosed by the lines $2|x|+5|y| \leq 4$ expressed interms of $f(1), f(2)$ and $f(4)$ is
$\frac{f(4)}{f(1)+2 f(2)}$
$\frac{f(4)}{1+f(2)}$
$\frac{2 f(4)}{2 f(1)+f(2)}$
$\frac{f(4)}{2 f(1)+f(2)}$
$ \begin{aligned} & =\frac{16}{1+(2)^2} \\ =\frac{(2)^4}{1+(2)^2}= & \frac{f(4)}{1+f(2)} \end{aligned} $Let $f:[0,10] \rightarrow[1,20]$ be a function defined as
$ f(x)=\left\{\begin{array}{ll} \frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10 \end{array} \text { then } f\right. \text { is } $
bijective function
one-one but not onto function
onto but not one-one function
neither one-one nor onto function
The domain of the function, $f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}$ is
$[0,1]$
$[1,4]$
$[4,5]$
$(-\infty, \infty)$
$ \begin{aligned} &\therefore \quad x \in[1,4]\\ &\text { So, domain of } f(x) \text { is }[1,4] \text {. } \end{aligned} $If $2f(xy) = {(f(x))^x} + {(f(y))^x}$ for all $x,y \in R$ and $f(1) = a( \ne 1)$. Then $\sum\limits_{k = 1}^n {f(k) = } $
Let f(x) = x $-$ 3, g(x) = 4 $-$ x. Then the set of values of x for which $|f(x) + g(x)|\, < \,|f(x)| + |g(x)|$ is true, is given by :
$\left\{ {x \in R:{{2x - 1} \over {{x^3} + 4{x^2} + 3x}} \in R} \right\}$ is equal to
The solution set of ${{|x - 2|\, - 1} \over {|x - 2|\, - 2}} \le 0$ is
Let $f(x) = {x \over {\sqrt {1 + {x^2}} }}$, $\underbrace {fofofo.....of(x)}_{x\,times}$ is
$f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)$ is
Æ’(x + y) = Æ’(x)Æ’(y) for all natural numbers x, y and Æ’(1) = 2. then the natural number 'a' is
Æ’(x) = ${{{x^2}} \over {1 - {x^2}}}$ , is surjective, then A is equal to
Æ’(x) = Æ’1 (x) + Æ’2 (x), where Æ’1 (x) is an even function of Æ’2 (x) is an odd function.
Then ƒ1 (x + y) + ƒ1 (x – y) equals
