Relations and Functions Test

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Relations and Functions Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If $$f\left( 0 \right) = 0,f\left( 1 \right) =1 ,f\left( 2 \right) = 2$$ and $$f\left( x \right) = f\left( {x - 2} \right) + f\left( {x - 3} \right)$$ for $$x = 3,4,5.....$$ then $$f\left( 9 \right) = ?$$

A
$$12$$

B
$$13$$

C
$$14$$

D
$$10$$
Question 2
1 / -0

If $$a\ne R$$ and the equation $$-3(x-[x])^2+2(x-[x])+a^2=0$$ ( where $$[x]$$ denotes the greatest integer $$\le x$$) has no integral solution, then all possible values of a lie in the interval :

A
$$(-2,-1)$$

B
$$(-\infty ,-2)\cup(2,\infty)$$

C
$$(-1 ,0)\cup(0,1)$$

D
$$(1,2)$$

Solution

REF.Image
As $$x-[x]=\left \{ x \right \}$$, let $$\left \{ x \right \}=y$$

As x is non-integral, $$0<y<1$$ $$(y\neq 0)$$
equation becomes, $$-3y^{2}+2y+a^{2}=0$$

$$\Rightarrow a^{2}=3y^{2}-2y=y(3y-2)$$

Drawing graph of $$3y^{2}-2y=0$$ (parabola)

Minima $$\Rightarrow \frac{d}{dy}(3y^{2}-2y)=0\Rightarrow 6y-2=0\Rightarrow y=1/3$$

Taking positive solutions $$(a^{2}\geq 0)$$

we get $$0 < a^{2} < 1$$

$$\Rightarrow a\epsilon (-1,0)\cup (0,1)$$

Option (C)
Question 3
1 / -0

$${x\epsilon R:\frac{14x}{x+1}-\frac{9x-30}{x-4}<0}$$ is equal to

A
(-1,4)

B
(1,4) $$\cup $$(5,7)

C
(1,7)

D
(-1,1) $$\cup $$ (4,6)

Solution

$$ x\epsilon R :\frac{14x}{x+1}-\frac{9x-30}{x-4}< 0 $$
$$ x+1\neq 0 $$ $$ x-4 \neq 0 $$
$$ \Rightarrow x \neq -1 $$ $$ x \neq 4 $$
$$ 14(x-4)-(x+1)(7x-30)< 0 $$
$$ 14x^{2}-56x-(9x^{2}-30x+9x-30)< 0 $$
$$ 14x^{2}-56x-9x^{2}+30x-9x+30< 0 $$

$$ 5x^{2}-35x+30< 0 $$
$$ x^{2}-7x+6< 0 $$
$$ x^{2}-6x+6< 0 $$
$$ x(x-6)-1(x-6)< 0 $$
$$ (x-1)(x-6)< 0 $$
$$ \Rightarrow x> 1 $$ & $$ x< 6 $$
$$ \Rightarrow x \epsilon ; (1,4)\cup (4,6) $$
Question 4
1 / -0

$$f:\left( 0,\infty \right) \rightarrow R$$ is continuous. If $$F\left(x\right)$$ is a differentiable function such that $$F\left(x\right)= f\left(x\right), \forall x>0$$ and $$ f\left( { x }^{ 2 } \right) ={ x }^{ 2 }+{ x }^{ 3 }$$, then $$f\left(4\right)$$ equals

A
$$12$$

B
$$7$$

C
$$4$$

D
$$2$$

Solution

Given $$t (x^{2})= x^{2}+ x^{3}$$
Put $$x=2$$
$$t(4)= 4+8$$
$$t (4)= 12$$
Question 5
1 / -0

If $$\left| { z }_{ 1 }-a \right| <a,\left| { z }_{ 2 }-a \right| <b,\left| { z }_{ 3 }-a \right| <c$$, $$(a,b,c\in R)$$ then $$\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| $$ is

A
less than $$(a+b+c)$$

B
more than $$(a+b+c)$$

C
less than $$2(a+b+c)$$

D
more than 2$$(a+b+c)$$

Solution

Given$$,$$
$$\left| {{z_1} - a} \right| < a$$
$$\left| {{z_2} - b} \right| < b$$
$$\left| {{z_3} - c} \right| < c$$
Then$$,$$
$$\left| {{z_1} - a} \right| + \left| {{z_2} - b} \right| + \left| {{z_3} - c} \right| < a + b + c$$
$$\left| {{z_1} + {z_2} + {z_3}} \right| < \left| {{z_1}} \right| + \left| {{z_2}} \right| + \left| {{z_3}} \right|$$
and$$,$$
$$\left| {\left( {{z_1} - a} \right) + \left( {{z_2} - b} \right) + \left( {{z_3} - c} \right) + a + b + c} \right|$$ $$ < \left| {{z_1} - a} \right| + \left| {{z_2} - b} \right| + \left| {{z_3} - c} \right| + \left| {a + b + c} \right|$$
$$\left| {{z_1} + {z_2} + {z_3}} \right|$$ $$ < a + b + c + a + b + c$$
$$\left| {{z_1} + {z_2} + {z_3}} \right|$$ $$ <2 \left( {a + b + c} \right)$$
$$\left| {{z_1} + {z_2} + {z_3}} \right|$$ less then $$2\left( {a + b + c} \right)$$
Hence$$,$$ Option $$(C)$$ is correct$$.$$
Question 6
1 / -0

The domain of $$\dfrac { 10 ^ { x } + 10 ^ { - x } } { 10 ^ { x } - 10 ^ { - x } }$$ is

A
$$R$$

B
$$R - \{ 0 \}$$

C
$$R - \{ 1 \}$$

D
$$R ^ { + }$$

Solution

For domain only the denominator must not be equal to ZERO
It means $$10^x-10^{-x} \neq 0\\or\> x\neq 0$$
So domain is R-{0}
Numerator is fine with any value of x.
Question 7
1 / -0

If $$R$$ is the relation from set $$A$$ to a set $$B$$ and $$S$$ is the relation from $$B$$ to a set $$C$$, then the relation $$SoR$$

A
If from $$A$$ to $$C$$

B
If from $$C$$ to $$A$$

C
Does not exist

D
None of these

Solution
Question 8
1 / -0

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by :$$(x,y)\in\;R\; \rightarrow x$$ is relatively prime to y. Then, domain of R is

A
{2, 3, 5}

B
{3, 5}

C
{2, 3, 4}

D
{2, 3, 4,5}

Solution

Given set $$A=\left\{ 2,3,4,5 \right\} $$ to $$B=\left\{ 3,6,7,10 \right\} $$
is defined as $$\left( x,y \right) \in R\Rightarrow x$$ is relatively prime to $$y$$
So $$(2,3)\in R$$
Similarly, $$2$$ is relatively prime to $$7$$ so that $$(2,7)\in R$$
So, we get $$R=\left\{ \left( 2,3 \right) ,\left( 2,7 \right) ,\left( 3,7 \right) ,\left( 3,10 \right) ,\left( 4,3 \right) ,\left( 4,7 \right) ,\left( 5,3 \right) ,\left( 5,6 \right) ,\left( 5,7 \right) , \right\} $$
Thus domain $$\left\{ 2,3,4,5 \right\} $$
Question 9
1 / -0

Let $$f(x)=$$max $$\{(1-x), (1+x), 2\}, \forall x\in R$$. Then?

A
$$f(x)=\left\{\begin{matrix} 1+x, & x\leq -1\\ 2, & -1 < x < 1\\ 1-x, & x\geq 1\end{matrix}\right.$$

B
$$f(x)=\left\{\begin{matrix} 1-x, & x\leq -1\\ 1, & -1 < x < 1\\ 1+x, & x\geq 1\end{matrix}\right.$$

C
$$f(x)=\left\{\begin{matrix} 1-x, & x\leq -1\\ 2, & -1 < x < 1\\ 1+x, & x\geq 1\end{matrix}\right.$$

D
None of these

Solution
Question 10
1 / -0

If $$f ( x ) + 2 f ( 1 - x ) = x ^ { 2 } + 2 , \forall x \in R$$, then find $$f ( x )$$

A
$$\dfrac { ( x - 1 ) ^ { 2 } } { 3 }$$

B
$$\dfrac { ( x - 2 ) ^ { 2 } } { 3 }$$

C
$$x ^ { 2 } - 1$$

D
$$x ^ { 2 } - 2$$

Solution

Given

$$f(x)+2f(1-x)=x^2+2 \cdots(1)$$

Replace $$x$$ with $$1-x$$

$$\implies f(1-x)+2f(1-(1-x))=(1-x)^2+2$$

$$\implies f(1-x)+2f(x)=x^2-2x+3 \cdots(2)$$

$$(1)+(2)$$

$$\implies 3f(x)+3f(1-x)=2x^2-2x+5$$

$$\implies f(x)+f(1-x)=\dfrac 23x^2-\dfrac 23 x+\dfrac 53 \cdots(3)$$

$$(2)-(1)$$

$$\implies f(x)-f(1-x)=1-2x \cdots(4)$$

$$(3)+(4)$$

$$\implies 2f(x)= \dfrac 23x^2-\dfrac 23 x+\dfrac 53 +1-2x$$

$$\implies f(x)=\dfrac 13\left(x-2\right)^2$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 41

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Relations and Functions Test - 41

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SHARING IS CARING

Question 1

$$12$$

$$13$$

$$14$$

$$10$$

Question 2

$$(-2,-1)$$

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

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

$$(1,2)$$

Solution

Question 3

(-1,4)

(1,4) $$\cup $$(5,7)

(1,7)

(-1,1) $$\cup $$ (4,6)

Solution

Question 4

$$12$$

$$7$$

$$4$$

$$2$$

Solution

Question 5

less than $$(a+b+c)$$

more than $$(a+b+c)$$

less than $$2(a+b+c)$$

more than 2$$(a+b+c)$$

Solution

Question 6

$$R$$

$$R - \{ 0 \}$$

$$R - \{ 1 \}$$

$$R ^ { + }$$

Solution

Question 7

If from $$A$$ to $$C$$

If from $$C$$ to $$A$$

Does not exist

None of these

Solution

Question 8

{2, 3, 5}

{3, 5}

{2, 3, 4}

{2, 3, 4,5}

Solution

Question 9

$$f(x)=\left\{\begin{matrix} 1+x, & x\leq -1\\ 2, & -1 < x < 1\\ 1-x, & x\geq 1\end{matrix}\right.$$

$$f(x)=\left\{\begin{matrix} 1-x, & x\leq -1\\ 1, & -1 < x < 1\\ 1+x, & x\geq 1\end{matrix}\right.$$

$$f(x)=\left\{\begin{matrix} 1-x, & x\leq -1\\ 2, & -1 < x < 1\\ 1+x, & x\geq 1\end{matrix}\right.$$

None of these

Solution

Question 10

$$\dfrac { ( x - 1 ) ^ { 2 } } { 3 }$$

$$\dfrac { ( x - 2 ) ^ { 2 } } { 3 }$$

$$x ^ { 2 } - 1$$

$$x ^ { 2 } - 2$$

Solution

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