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Let $f(x)=a^{x}(a>0)$ be written as $f(x)=f_{1}(x)+f_{2}(x)$, where $f_{1}(x)$ is an even function and $f_{2}(x)$ is an odd function. Then $f_{1}(x+y)+f_{1}(x-y)$ equals:

• Option 1)

$2f_{1}(x)f_{1}(y)$

• Option 2)

$2f_{1}(x+y)f_{1}(x-y)$

• Option 3)

$2f_{1}(x)f_{2}(y)$

• Option 4)

$2f_{1}(x+y)f_{2}(x-y)$

Now,     Option 1) Option 2) Option 3) Option 4)

Let A , B and C be sets such that

$\phi \neq A\cap B\subseteq C.$  Then which of the following

statements is not true ?

• Option 1)

$B\cap C\neq \phi$

• Option 2)

$If\: \: (A-B)\subseteq C, then\: \: A\subseteq C$

• Option 3)

$(C\cup A)\cap(C\cup B)=C$

• Option 4)

$If\: \: (A-C)\subseteq B, then\: \: A\subseteq B$

As  =>  as  =>  So, option (1) is true let  => =>  let  => =>  Hence     =>  So, option (2) is  true. Let  ................(1) Now, ..............(2) From (1) and (2) => option (3) is  correct. For A = C ,                   but   So, option (4) is not true.   Option 1) Option 2) Option 3) Option 4)

If $[x]$ denotes the greatest integer $\leq x$ , then the system

of linear equations $[sin\theta]x+[-cos\theta]y=0$

$[cot\theta]x+y=0$

• Option 1)

Have infinitely many solutions if $\theta\epsilon (\frac{\pi}{2},\frac{2\pi}{3})$ and has a unique solution if $\theta\epsilon (\pi,\frac{7\pi}{6})$.

• Option 2)

has a unique solution if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})\cup (\pi,\frac{7\pi}{6})$

• Option 3)

has a unique solution if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})$ and have infinitely many solutions if $\theta\epsilon (\pi,\frac{7\pi}{6})$

• Option 4)

infinitely many solutions if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})\cup$$(\pi,\frac{7\pi}{6})$

linear equations  &                              For infinite many solution, i.e. ...................(1) * when                         so,     * when                         so,     *when   *when                          Option 1) Have infinitely many solutions if  and has a unique solution if . Option 2) has a unique solution if  Option 3) has a unique solution if  and have infinitely many...

For $x\equiv \left ( 0,\frac{3}{2} \right ),$ let $f(x)=\sqrt{x},g(x)=\tan x$ and $h(x)=\frac{1-x^{2}}{1+x^{2}}$. If $\phi \left ( x \right )=\left ( \left (hof \right )og\left ( x \right ) \right ),$ then $\phi \left ( \frac{\pi }{3} \right )$ is equal to :

• Option 1)

$\tan \frac{11\pi }{12}$

• Option 2)

$\tan \frac{7\pi }{12}$

• Option 3)

$\tan \frac{5\pi }{12}$

• Option 4)

$\tan \frac{\pi }{12}$

Option 1)     Option 2)              Option 3)        Option 4)

For $x\equiv R$, let $\left [ x \right ]$ denote the greatest integer $\leq x,$ then the sum of the series

$\left [ -\frac{1}{3} \right ]+\left [ -\frac{1}{3}-\frac{1}{100} \right ]+\left [ -\frac{1}{3}-\frac{2}{100} \right ]+\cdots +\left [ -\frac{1}{3}-\frac{99}{100} \right ]$ is :

• Option 1)

$-131$

• Option 2)

$-133$

• Option 3)

$-135$

• Option 4)

$-153$

Option 1) Option 2) Option 3) Option 4)

The number of real roots of the equation

$5+|2^{x}-1|=2^{x}(2^{x}-2)$ is :

• Option 1)

3

• Option 2)

2

• Option 3)

4

• Option 4)

1

Case 1:  Case 2:  LHS = +ve  &  RHS = - ve (no solution)  no. of solution = 1 option(4) will be correct answer. Option 1) 3 Option 2) 2 Option 3) 4 Option 4) 1

Let $f(x)=x^{2},x\epsilon R.$ For any $A\subseteq R$ , define

$g(A)={\left \{ x\epsilon R : f(x)\epsilon A}\right \}$. If $S=[0,4]$ , then which

one of the following statements is not true ?

• Option 1)

$g(f(S)) \neq S$

• Option 2)

$f(g(S)) = S$

• Option 3)

$g(f(S)) = g(S)$

• Option 4)

$f(g(S)) \neq f(S)$

g(S) = [-2,2] So, f(g(S)) = [0,4] = S And f(S) = [0,16] => f(g(S)) f(S) Also, g(f(S)) = [-4,4]  g(S) So, g(f(S)) S correct option is (3) Option 1) Option 2) Option 3) Option 4)

• Option 1)

$13.9$

• Option 2)

$12.8$

• Option 3)

$13$

• Option 4)

$13.5$

Let us consider total percent be 100 . The from venn diagram  Total person took into advertisement = Option 1) Option 2) Option 3) Option 4)

The domain of the definition of the function

$f(x)=\frac{1}{4-x^{2}}+ \log_{10}(x^{3}-x)$  is :

• Option 1)

$(-1,0)\cup (1,2)\cup (3,\infty )$

• Option 2)

$(-2,-1)\cup (-1,0)\cup (2,\infty )$

• Option 3)

$(-1,0)\cup (1,2)\cup (2,\infty )$

• Option 4)

$(1,2)\cup (2,\infty )$

Domain :  Option 1) Option 2) Option 3) Option 4)

If the function $f:\mathbf{R}-\left \{ 1,-1 \right \}\rightarrow A\; defined$ by $f\left ( x \right )=\frac{x^{2}}{1-x^{2}} \; ,$ is surjective , then $A$ is equal to :

• Option 1)

$\textbf{R}-\left \{ -1 \right \}$

• Option 2)

$\left [ 0,\infty )$

• Option 3)

$\textbf{R}-\left [ -1,0 )$

• Option 4)

$\textbf{R}-\left ( -1,0 \right )$

SIMPLY TO SAVE YOUR TIME DO OPTION VERIFICATION

If $f(x)=\log_{e}\left ( \frac{1-x}{1+x} \right ),\left | x \right |<1,$ then $f\left ( \frac{2x}{1+x^{2}} \right )$ is equal to :

• Option 1)

$2f(x^{2})$

• Option 2)

$2f(x)$

• Option 3)

$-2f(x)$

• Option 4)

$(f(x))^{2}$

the     Option 1)   Option 2) Option 3)   Option 4)

The number of functions f from {1,2,3,........,20} onto {1,2,3,........,20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is:

• Option 1)

$(15)! \times 6!$

• Option 2)

$5! \times 6!$

• Option 3)

$6^{5}\times(15)!$

• Option 4)

$5^{6} \times 15$

Onto function - If  f:AB is such that each & every element in B is the f image of atleast one element in A.Then it is Onto function.   - wherein The range of f is equal to Co - domain of f.   f(k) = 3,6,9,12,15,18 for k = 4,8,12,16,20 ways = 6 x 5 x 4 x 3 x 2 x 1 = 6! For remaining numbers = (20-5)! = 15! Total ways = 15! x 6!  Option 1)Option 2)Option 3)Option 4)

Let a function $f:(0,\infty)\rightarrow (0,\infty)$ be defined by $f(x)=\left | 1-\frac{1}{x} \right |.$

Then $f$ is:

• Option 1)

not injective but it is surjective

• Option 2)

both injective as well as surjective

• Option 3)

neither injective nor surjective

• Option 4)

injective only

One - One or Injective function - A function in which every element of range of function corresponds to exactly one elements. - wherein A line parallel to x - axis cut the curve at most one point. Curve of       Option 1)  not injective but it is surjectiveOption 2)  both injective as well as surjectiveOption 3)neither injective nor surjectiveOption 4)injective only

Let $f:R\rightarrow R$  be defined by $f(x)=\frac{x}{1+x^{2}},\: x\epsilon R.$   Then the range of $f$ is :

• Option 1)

$R-\begin{bmatrix} -\frac{1}{2}\: ,\frac{1}{2} \end{bmatrix}$

• Option 2)

$R-\begin{bmatrix} -{1}\: ,\1\end{bmatrix}$

• Option 3)

$(-1,1)- {\left { 0 \right }}$

• Option 4)

$\begin{bmatrix} -\frac{1}{2}\: ,\frac{1}{2} \end{bmatrix}$

Range of function - All possible values of  domain   is known as Range  -                       Range of Option 1)  Option 2)  Option 3)  Option 4)

In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by3 opted Physics course  and those whose number is divisible by 5 opted Chemistry course. Then the number of the students who did not opt for any of the three courses is;

• Option 1)

102

• Option 2)

42

• Option 3)

1

• Option 4)

38

Number of Elements in Union A , B & C - n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) + n (A ∩ B ∩ C) - wherein Given A, B and C be any finite sets. then Number of Elements in union A , B & C is given by this formula.   From the concept, Let n(A) = no. of students opted maths = 70 n(B) = student opted physics = 40 n(C) = student opted chemistry = 28 Total no. of...

Let N be the set if natural numbers and two functions $f$ and $g$ be difined as $f,g :N\rightarrow N$such that

$f(n) = \left\{\begin{matrix} \frac{n+1}{2}\;if\;n\;is\;odd\\ \frac{n}{2}\;if\;n\;is\;even \end{matrix}\right.$ and $g(n) = n - (-1)^n$ . Then $fog$ is:

• Option 1)

Onto but not one-one

• Option 2)

Neither one-one nor onto

• Option 3)

one-one but not onto

• Option 4)

both one-one and onto

Onto function - If  f:AB is such that each & every element in B is the f image of atleast one element in A.Then it is Onto function.   - wherein The range of f is equal to Co - domain of f.     One - One or Injective function - A function in which every element of range of function corresponds to exactly one elements. - wherein A line parallel to x - axis cut the curve at most one...

Let $S= \left \{ 1,2,3,\cdots ,100 \right \}.$ The number of non-empty subsets A of S such that the product of elements in A is even is :

• Option 1)

$2^{50}\left ( 2^{50}-1 \right )$

• Option 2)

$2^{100}-1$

• Option 3)

$2^{50}-1$

• Option 4)

$2^{50}+1$

Number of sub set of a set - If a set has n elements, then it has 2n sub set. - Subsets = Total Subsets - Number of subsets which are odd                =                =          Option 1)      Option 2)Option 3)Option 4)

Let $A = {x\epsilon }$ R : x is not a positive integer. Define a function $f:A\rightarrow R$ as$f(x) = 2x/x-1,$then f is:

• Option 1)

injective but not surjective

• Option 2)

neither injective nore surjective

• Option 3)

not injective

• Option 4)

surjective but not injective

One - One or Injective function - A function in which every element of range of function corresponds to exactly one elements. - wherein A line parallel to x - axis cut the curve at most one point.   This can be written as f is one-one i.e injective but not surjective.Option 1)  injective but not surjectiveOption 2)  neither injective nore surjectiveOption 3)  not injectiveOption...

For $x\in \mathbb R - \{0,1\}$ , let  $f_1(x ) = \frac{1}{x}, f_2(x)= 1 -x$ and $f_3(x) = \frac{1}{1-x}$ be three given functions. If a function, $J(x)$ satisfies $(f_2\circ J\circ f_1) (x) = f_3(x)$ then $J(x)$ is equal to:

• Option 1)

$f_3 (x)$

• Option 2)

$\frac{1}{x}f_3 (x)$

• Option 3)

$f_2 (x)$

• Option 4)

$f_1 (x)$

COMPOSITION OF FUNCTIONS - Let  f? A → B and g? B → C be two functions. composition of f and g, denoted by g o f, then g o f (x) = g (f (x)), ∀ x ∈ A.   - Given that  from the concept of composition.  Now,   Option 1)Option 2)Option 3)Option 4)
Engineering
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Let where       Then   equals :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Option 2

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