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Let f be a polynomial function such that for all  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 $f\left ( 2 \right )-{f}'\left ( 2 \right )+{f\left }''\left ( 2 \right )=10$

is given a polynomial function

Let having degree

becomes degree

becomes  degree

Now it is given that

If we consider only degree

Assume

we get

and

Now Comparing coefficient of

Now Comparing coefficient of

Here a=0 or b=0

a cannot be zero so b=0

Now Comparing coefficient of

c=0

Similarly d=0

Hence

and

Option 2 is correct

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

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

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

Correct Option (2)

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 Option: 2 Option: 3 Option: 4

\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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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 Option: 2   Option: 3  Option: 4

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

Correct Option 2

Let  be a function defined by  where  denotes the greatest integer  Then the range of f is : Option: 1           Option: 2  Option: 3  Option: 4

Piecewise function -

Greatest integer function

The function f: R R defined by f(x) = [x], x 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 x < 0

[x] = 0 for 0 x < 1

[x] = 1 for 1 x < 2

[x] = 2 for 2 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)

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.

-

Correct Option (4)

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Let S be the set of all real roots of the equation,  $3^{x}(3^{x}-1)+2=\left | 3^{x}-1 \right |+\left | 3^{x}-2 \right |.$ Then S _____. Option: 1 is a singleton set Option: 2 is an empty set Option: 3 contains at least four elements Option: 4 contains exactly two elements

$\text{Put }3^x=t$

$\\t\left(t-1\right)+2=\left|t-1\right|+\left|t-2\right|\\t^2-t+2=\left|t-1\right|+\left|t-2\right|$

Only one point of intersection to the right of y-axis (we need postive t as 3x > 0)

Hence, singleton set

Correct Option (1)

Let $X=\left \{ n\; \epsilon\; N:1\leq n\leq 50 \right \}.$ If $A=\left \{ n\; \epsilon\; X:n\; is\; multiple\; of\; 2 \right \}$ and $B=\left \{ n\; \epsilon\; X:n\; is\; multiple\; of\; 7 \right \}$, then the number of elements in the smallest subset of X containing both A and B is Option: 1 29 Option: 2 45 Option: 3 32 Option: 4 16

A = {2, 4, 6, 8, 10, ....., 50}

B = {7, 14, 21, 28, 35, 42, 49}

n(A $\cap$ B) = {14, 28, 42}

Smallest subset of X which has all elements of A and B is the union of A and B

And we know

$\\\mathrm{n(A\cup B)=n(A)+n(B)-n(A\cap B)}= 25+7-3=29$

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The sum of the roots of the equation, , is :Option: 1Option: 2Option: 3Option: 4

$x+1-2\log _{2}\left ( 3+2^{x} \right )+2\log_{4}\left ( 10-2^{-x} \right )= 0\\$

$\Rightarrow log_{2}\left ( 3+2^{x} \right )^{2}-\log_{2}\left ( 10 -2^{-x}\right )=x+1\\$

$\Rightarrow \log_{2}\left ( \frac{\left ( 2^{x} \right )^{2}+6.2^{x}+9}{10-2^{-x}} \right )= x+1\\$

Raise to the power of 2, take  $2^{x}=t$

$\Rightarrow \frac{t^{2}+6t+9}{10-\frac{1}{t}}=2t\\$

$\Rightarrow t^{2}+6t+9=20t-2\\$

$\Rightarrow t^{2}-14t+11=0< ^{t_{1}=2^{x_{1}}}_{t_{2}=2^{x_{2}}}\\$

$t_{1}\cdot t_{2}=2^{x_{1}+x_{2}}=11\\$

$\Rightarrow x_{1}x_{2}=\log_{2}11$

Let be a function such that for every , then is equal to :Option: 1Option: 2Option: 3Option: 4

$f(m+n)=f(m)+f(n),f(6)=18\\$

$Put\; m=n=3;\\$

$f(6)=f(3)+f(3)=18\\$

$\Rightarrow f(3)=9\\$          ....(1)

$Put\; m=4,n=2\\$

$f(6)=f(4)+f(2)=18\\$

Also put $m=n=2\\$

$\Rightarrow f(4)=f(2)+f(2)=2f(2)\\$

$\Rightarrow f(6)=f(2)+f(2)+f(2)=18\\$

$\Rightarrow 3f(2)=18\Rightarrow f(2)=6\\$     ..........(2)

$f(2)\cdot f(3)=9\times6=54$

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