Relations and Functions Test

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Relations and Functions Test - 77

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f: R\rightarrow R$$ be a function such that $$f(x) = ax + 3\sin x + 4\cos x$$. Then $$f(x)$$ is invertible if

A
$$a\in (-5,5)$$

B
$$a\in (-\infty,5)$$

C
$$a\in (-5,\infty)$$

D
None of the above

Solution

Given : $$f\left( x \right) =ax+3\sin { x+4\cos { x } } $$
$$\rightarrow $$As $$\sin { x }$$ and $$\cos { x } $$ are periodic function of period : $$2\pi $$

$$\rightarrow $$ The function $$' f '$$ is invertible provided that the domain is restricted to an interval of length $$2\pi $$ or less than that.

Hence the option given $$(-5,5),(-\infty ,5) and (-5,\infty )$$ does not fall under the interval $$2\pi $$

Hence none of the above is the correct answer.

Therefore option D is correct Answer
Question 2
1 / -0

$$f:\left( 0,\infty \right) \rightarrow \left( 0,\infty \right) $$ is defined by $$f(x)=\begin{cases} { 2 }^{ x },\quad x\in \left( 0,1 \right) \\ { 5 }^{ x },\quad x\in [1,\infty ) \end{cases}$$ is

A
one-one but not onto

B
onto but not one-one

C
neither one-one nor onto

D
bijective

Solution
Question 3
1 / -0

In the set $$Q^{+}$$ of all positive rational numbers, the operation $$\ast$$ is defined by the formula $$a\ast b = \dfrac {ab}{6}$$. Then, the inverse of $$9$$ with respect to $$\ast$$ is

A
$$4$$

B
$$3$$

C
$$\dfrac {1}{9}$$

D
$$\dfrac {1}{3}$$

Solution

Let $$e$$ be the identity,
Therefore, $$ a\ast e = a$$
$$\Rightarrow \dfrac {ae}{6} = a$$
$$\Rightarrow e = 6$$
Since $$ a\ast a^{-1} = e$$,
thus $$a\ast a^{-1} = 6$$
$$\Rightarrow \dfrac {aa^{-1}}{6} = 6$$
$$\Rightarrow a^{-1} = \dfrac {6\times 6}{a}$$
Therefore, $$ 9^{-1} = \dfrac {6\times 6}{9} = 4$$
Question 4
1 / -0

Let $$A=\left\{ { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }{ a }_{ 5 },{ a }_{ 6 } \right\} $$ and $$B=\left\{ { b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 } \right\} $$. The number of functions of $$f:A\rightarrow B$$ such that it is onto and there are exactly three elements in $$A$$ such that $$f(A)=b$$, is

A
$$75$$

B
$$90$$

C
$$100$$

D
$$120$$

Solution

Since there should be exactly 3 elements in $$ A $$ such that
$$ f(A) = b $$
the function $$ f$$ is a bijection from a subset $$ A_1 $$ of $$ A $$ to $$ B $$., each having 3 elements.

Number of ways to select 3 elements from $$ A $$ $$ =\; ^6C_3 $$

Number of bijection from $$ A_1 $$ to $$ B $$ $$ = 3 ! $$

Hence, the number of functions $$ =\; ^6C_3 \; 3!$$ $$ = 120 $$
Question 5
1 / -0

Let $$f(x) = \dfrac {x}{1 - x}$$ and let $$\alpha$$ be a real number. If $$x_{0} = \alpha, x_{1} = f(x_{0}), x_{2} = f(x_{1}), ....$$ and $$x_{2011} = - \dfrac {1}{2012}$$ then the value of $$\alpha$$ is

A
$$\dfrac {2011}{2012}$$

B
$$1$$

C
$$2011$$

D
$$-1$$

Solution

$$f(x)=\dfrac {x}{1-x}$$ also $$x_0=\alpha$$
$$x_1 =f(x_0)=f(\alpha)=\dfrac {\alpha}{1-\alpha}$$
$$x_2 =f(x_1)=f\left (\dfrac {\alpha}{1-\alpha}\right)=\dfrac {\dfrac {\alpha}{1-\alpha}}{1-\dfrac {\alpha}{1-\alpha}}=\dfrac {\alpha}{1-2\alpha}$$
$$x_3 =f(x_2)=f\left (\dfrac {\alpha}{1-2\alpha}\right)=\dfrac {\dfrac {\alpha}{1-2\alpha}}{1-\dfrac {\alpha}{1-2\alpha}}=\dfrac {\alpha}{1-3\alpha}$$
Analysis the values $$x_1, x_2, x_3$$, we can say that $$x_{2011}=\dfrac {\alpha}{1-20011\alpha}=-\dfrac {1}{2012}$$
$$\Rightarrow \ 2012\alpha =-1+2011\alpha$$
$$\Rightarrow \ 2012\alpha -2011\alpha =-1$$
$$\Rightarrow \ \boxed {\alpha =-1}$$
Question 6
1 / -0

Consider set $$A={1,2,3,4}$$ and set $$B={0,2,4,6,8}$$, then the number of one-one function set $$A$$ to set $$B$$ in which $$f(i)\neq i$$ is,

A
$$84$$

B
$$78$$

C
$$42$$

D
$$24$$
Question 7
1 / -0

If $$f:R\rightarrow R$$ is defined by $$f(x)={x}^{2}-6x-14$$, then $$f^{-1}(2)$$ equals to

A
$${2, 8}$$

B
$${\phi }$$

C
$${-2, 8}$$

D
$${2, -8}$$

Solution
Question 8
1 / -0

If $$f:R\rightarrow S$$ defined by
$$f(x)=4\sin { x } -3\cos { x } +1$$ is onto, then $$S$$ is equal to

A
$$[-5,5]$$

B
$$(-5,5)$$

C
$$(-4,6)$$

D
$$[-4,6]$$

Solution

We have $$f(x)=4sin\:x-3cos\:x+1$$

Therefore
maximum of $$f=\sqrt{4^2+(-3)^2}+1=\sqrt{16+9}+1=\sqrt{25}+1=5+1=6$$ and
minimum of $$f=-\sqrt{4^2+(-3)^2}+1=-\sqrt{16+9}+1=-\sqrt{25}+1=-5+1=-4$$

$$S=$$ Range of $$f$$=$$[minimum\:f,maximum\:f]=[-4,6]$$
Question 9
1 / -0

Let for $$a \neq a_{1} \neq 0,\ f(x)=ax^{2}+bx+c,\ g(x)=a_{1}x^{2}+b_{1}x+c_{1}$$ and $$p(x)=f(x)-g(x)$$. If $$p(x)=0$$ only for $$x=-1$$ and $$p(-2)=2$$, then the value of $$p(2)$$ is

A
$$6$$

B
$$18$$

C
$$3$$

D
$$9$$

Solution

Given, $$f\left(x\right)=ax^{2}+bx+c,g\left(x\right)=ax^{2}+bx+c,$$

and $$p\left(x\right)=f\left(x\right)-g\left(x\right)$$

$$p\left(x\right)=0$$ for only $$x=-1,p\left(-2\right)=2$$ and $$a\neq a_{1}\neq 0$$

$$\therefore p\left(x\right)=(a-a_{1})x^{2}+(b-b_{1})x+(c-c_{1})$$

The standard quadratic equation is $$ax^2+bx+c=0$$

For roots to be equal $$ \displaystyle \Rightarrow b^{2}-4ac= 0 $$

Then Sum of roots = $$-\dfrac {b}{a}$$

Here, Only root is $$-1 \Rightarrow D=0 \Rightarrow (b-b_{1})^{2}=4(a-a_{1})(c-c_{1})\rightarrow(1)$$

Sum of roots $$\Rightarrow (-1)+(-1)=\dfrac{-(b-b_{1})}{(a-a_{1})} \Rightarrow (b-b_{1})=2(a-a_{1}) \rightarrow(2)$$

From (1) and (2), $$c-c_{1}=(a-a_{1})\rightarrow(3)$$

$$\Rightarrow p\left(x\right)=(a-a_{1})x^{2}+2(a-a_{1})x+(a-a_{1})$$

Now, $$p\left(-2\right)=2$$

$$\Rightarrow 2=(a-a_{1})(-2)^{2}-2(a-a_{1})(-2)+(a-a_{1})$$

$$\Rightarrow (a-a_{1})=2$$

Now $$p(2)=2(2)^{2}+2(2)(2)+2=18$$
Question 10
1 / -0

Given that $$f'(x) > g'(x)$$ for all real x, and $$f(0) = g(0)$$. Then $$f(x) < g(x)$$ for all x belongs to

A
$$(0, \infty)$$

B
$$(- \infty, 0)$$

C
$$(- \infty, \infty)$$

D
none of these

Solution

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 77

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Relations and Functions Test - 77

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SHARING IS CARING

Question 1

$$a\in (-5,5)$$

$$a\in (-\infty,5)$$

$$a\in (-5,\infty)$$

None of the above

Solution

Question 2

one-one but not onto

onto but not one-one

neither one-one nor onto

bijective

Solution

Question 3

$$4$$

$$3$$

$$\dfrac {1}{9}$$

$$\dfrac {1}{3}$$

Solution

Question 4

$$75$$

$$90$$

$$100$$

$$120$$

Solution

Question 5

$$\dfrac {2011}{2012}$$

$$1$$

$$2011$$

$$-1$$

Solution

Question 6

$$84$$

$$78$$

$$42$$

$$24$$

Question 7

$${2, 8}$$

$${\phi }$$

$${-2, 8}$$

$${2, -8}$$

Solution

Question 8

$$[-5,5]$$

$$(-5,5)$$

$$(-4,6)$$

$$[-4,6]$$

Solution

Question 9

$$6$$

$$18$$

$$3$$

$$9$$

Solution

Question 10

$$(0, \infty)$$

$$(- \infty, 0)$$

$$(- \infty, \infty)$$

none of these

Solution

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