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Let f be a polynomial function such that $\dpi{100} \small f(3x)= {f}'\left ( x \right ) \cdot{f}''\left ( x \right ),$ for all $\small x\epsilon R$ Then :
Option: 1 $f\left ( 2 \right )+{f}' \left ( 2 \right ) =28$

Option: 7 ${f}''\left ( 2 \right )-{f}'\left ( 2 \right )=0$

Option: 13 ${f}''\left ( 2 \right )-f\left ( 2 \right )=4$

Option: 19 $f\left ( 2 \right )-{f}'\left ( 2 \right )+{f\left }''\left ( 2 \right )=10$

$f(x)$ is given a polynomial function

Let $f(x)$ having $n^{th}$ degree

$f'(x)$ becomes $(n-1)^{th}$ degree

$f''(x)$ becomes $(n-2 )^{th}$ degree

Now it is given that

$f(3x)=f'(x)\cdot f''(x)$

If we consider only degree

$n=(n-1)+(n-2)\\\Rightarrow n=3$

Assume

$f(x)=ax^3+bx^2+cx+d$

we get

$f'(x)=3ax^2+2bx+c$ and $f''(x)=6ax+2b$

$f(3x)=f'(x)\cdot f''(x)$

$a(3x)^3+b(3x)^2+c(3x)+d=(3ax^2+2bx+c)\cdot(6ax+2b)$

$27ax^3+9bx^2+3cx+d=18a^2x^3+18abx^2+x(6ac+4b^2)+2bc$

Now Comparing coefficient of $x^3$

$27a=18a^2$

$a=\frac32$

Now Comparing coefficient of $x^2$

$9b=18ab$

Here a=0 or b=0

a cannot be zero so b=0

Now Comparing coefficient of $x$

c=0

Similarly d=0

Hence

$f(x)=\frac32x^3$

$f'(x)=\frac323x^2=\frac92x^2$ and

$f''(x)=\frac326x=9x$

$f''(2)-f'(2)=\frac92\times4-9\times2=0$

Option 2 is correct

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Posted by

vishal kumar

The function $f$ : N → N defined by $f\left ( x \right ) = x -5 \left [ \frac{x}{5} \right ]$ , where N is the set of natural numbers and $\left [ x \right ]$ denotes the greatest integer less than or equal to $x$ , is :
Option: 1 one-one and onto.
Option: 2 one-one but not onto.
Option: 3 onto but not one-one.
Option: 4 neither one-one nor onto.

$\\f(x)=x-5\left[\frac{x}{5}\right]\\$

Taking x in an interval of five natural numbers, we have the following:

$f(x)=\left\{\begin{array}{ll} x-5(0), & 0 \leq x < 5 \\ x-5(1), & 5 \leq x < 10 \\ x-5(2), & 10 \leq x < 15 \\ x-5(3), & 15 \leq x < 20 \end{array}\right.$

Therefore, here, x ∈N. hence, f(x) is neither a one-one function nor onto function.

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Posted by

vishal kumar

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The inverse function of $f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x\epsilon (-1,1),$ is
Option: 1 $\frac{1}{4}(\log _{8}e)\log _{e}\left ( \frac{1-x}{1+x} \right )$
Option: 2 $\frac{1}{4}(\log _{8}e)\log _{e}\left ( \frac{1+x}{1-x} \right )$
Option: 3 $\frac{1}{4}\log _{e}\left ( \frac{1+x}{1-x} \right )$
Option: 4 $\frac{1}{4}\log _{e}\left ( \frac{1-x}{1+x} \right )$

$\\f(x)=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}} \\ f(x)=\frac{8^{4 x}-1}{8^{4 x}+1} \text{put }8^{4 x}=t \\ y=\frac{t-1}{t+1}\\ {y t+y=t-1} \\ {\frac{y+1}{1-y}=t} \\ {\frac{y+1}{1-y}=8^{4 x}}\\ ln\left(\frac{y+1}{1-y}\right)=4 x \ln 8 \quad$

$\\x=\frac{1}{4 \ln 8} \ln \left(\frac{y+1}{1-y}\right)\\\\f^{-1}(x)=\frac{1}{4 \ln 8} \ln \left(\frac{x+1}{1-x}\right)\\$

Correct Option (2)

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Posted by

Kuldeep Maurya

If $g(x)=x^{2}+x-1$ and $(gof)(x)=4x^{2}-10x+5,$ then $f\left ( \frac{5}{4} \right )$ is equal to:
Option: 1 $-\frac{3}{2}$
Option: 2 $-\frac{1}{2}$
Option: 3 $\frac{1}{2}$
Option: 4 $\frac{3}{2}$

$\begin{align*} g(x)&=x^2+x-1 \\ gof(x)&= 4x^2-10x+5\\ f(x)&= ax+b\\ gof(x)&= (ax+b)^2+ax+b-1\\ gof(x)&= a^2x^2+b^2+2axb+ax+b-1 = 4x^2-10x+5\\ \implies a^2x^2&+x(2ab+a)+b^2+b-1=4x^2-10x+5\\ \end{align*}$

Therefore , $a^2=4, \ 2ab+a = -10, \ b^2+b-1=5$

Solving the above equations, $a=2, \ b = -3,$

Therefore,

$f(x)=2x-3 \\ f\left(\frac{5}{4}\right)= 2 \times \frac{5}{4}-3$

$=\frac{-1}{2}$

Correct option (2)

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Posted by

Kuldeep Maurya

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If $A= \left ( x\epsilon \mathbf{\textbf{R}} :\left | x \right |<2\right )$ and $B= \left ( x\epsilon \mathbf{\textbf{R}} :\left | x-2 \right |\geqslant 3\right )$ ; then:
Option: 1 $A-B=[-1,2)$
Option: 2 $B-A=\textbf{R}-(-2,5)$
Option: 3 $A\cup B=\textbf{R}-(2,5)$
Option: 4 $A\cap B=(-2,-1)$

$\begin{array}{l}{A=\{x: x \in(-2,2)\}} \\ {B=\{x: x \in(-\infty,-1] \cup[5, \infty)\}} \\ {A \cap B=\{x: x \in(-2,-1]\}} \\ {A \cup B=\{x: x \in(-\infty, 2) \cup[5, \infty)\}} \\ {A-B=\{x: x \in(-1,2)\}} \\ {B-A=\{x: x \in(-\infty,-2] \cup[5, \infty)\}}\end{array}$

Correct Option 2

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Posted by

avinash.dongre

Let $f:(1,3)\rightarrow R$ be a function defined by $f(x)=\frac{x\left [ x \right ]}{1+x^{2}},$ where $\left [ x \right ]$ denotes the greatest integer $\leq x.$ Then the range of f is :
Option: 1 $\left ( \frac{2}{3},\frac{3}{5} \right ]\cup \left ( \frac{3}{4},\frac{4}{5} \right )$

Option: 2 $\left ( \frac{2}{5},\frac{4}{5} \right ]$

Option: 3 $\left ( \frac{3}{5},\frac{4}{5} \right )$

Option: 4 $\left (\frac{2}{5},\frac{1}{2} \right )\cup \left ( \frac{3}{5},\frac{4}{5} \right ]$

Piecewise function -

Greatest integer function

The function f: R $\small \rightarrow$ R defined by f(x) = [x], x $\small \in$ R assumes the value of the greatest integer less than or equal to x. Such a functions called the greatest integer function.

eg;

[1.75] = 1

[2.34] = 2

[-0.9] = -1

[-4.8] = -5

From the definition of [x], we

can see that

[x] = –1 for –1 $\small \leq$ x < 0

[x] = 0 for 0 $\small \leq$ x < 1

[x] = 1 for 1 $\small \leq$ x < 2

[x] = 2 for 2 $\small \leq$ x < 3 and

so on.

Properties of greatest integer function:

i) [x] ≤ x < [x] + 1

ii) x - 1 < [x] < x

iii) I ≤ x < I+1 ⇒ [x] = I where I belongs to integer.

iv) [[x]]=[x]v)

v) $[x] + [-x] = \left\{\begin{matrix} 0, &if &x & belongs \; to & integer \\ -1, & if &x & doesn't \;belongs \;to & integer \end{matrix}\right.$

vi) [x] + [-x] = 2x if x belongs to integer

2[x] + 1 if x doesn’t belongs to integer

Domain of function, Co-domain, Range of function -

All possible values of x for f(x) to be defined is known as a domain. If a function is defined from A to B i.e. f: A?B, then all the elements of set A is called Domain of the function.

If a function is defined from A to B i.e. f: A?B, then all the elements of set B are called Co-domain of the function.

The set of all possible values of f(x) for every x belongs to the domain is known as Range.

For example, let A = {1, 2, 3, 4, 5} and B = {1, 4, 8, 16, 25, 64, 125}. The function f : A -> B is defined by f(x) = x³. So here,

Domain : Set A

Co-Domain : Set B

Range : {1, 8, 27, 64, 125}

The range can be equal to or less than codomain but cannot be greater than that.

$\\f(x)=\left\{\begin{array}{ll} {\frac{x}{x^{2}+1} ;} & {x \in(1,2)} \\ {\frac{2 x}{x^{2}+1} ;} & {x \in[2,3)} \end{array}\right.\\\therefore \text{f(x) is a decreasing function }\\\begin{aligned} &\therefore \quad \mathrm{y} \in\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{6}{10}, \frac{4}{5}\right]\\ &\Rightarrow \quad y \in\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right] \end{aligned}$