Relations and Functions Test

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Relations and Functions Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x) = 3x - 5$$, then $$f^{-1}\, (x) $$

A
is given by $$\displaystyle \frac{1}{3x\, -\, 5}$$

B
is given by $$\displaystyle \frac{x\, +\, 5}{3}$$

C
does not exist because $$f$$ is not one-one

D
does not exist because $$f$$ is not onto

Solution

Let $$y = f(x) = 3x - 5$$

Clearly $$f(x)$$ is one-one onto function so it is invertible.

$$\Rightarrow y = 3x - 5\Rightarrow x = \cfrac{y+5}{3}$$

$$\therefore \displaystyle f^{-1}(x)= \frac{y+5}{3}$$
Question 2
1 / -0

If $$f(x)=\sqrt [ n ]{ a-{ x }^{ n } } ,x>0,n\geq 2,n\in N$$, then find the inverse of $$f(x)$$.

A
$$f^{-1} (x) = (a - x^{n})^{1/n}$$

B
$$f^{-1} (x) = a-x$$

C
$$f^{-1} (x) = (a + x^{n})^{1/n}$$

D
None of these

Solution

First method:
$$f(x) = (a - x^{n})^{1/n}$$

$$(fof)(x)=f[f(x)] $$

$$=f[(a - x^{n})^{1/n}]$$

$$ =(a - (f(x))^{n})^{1/n} $$

$$= (a - a + x^{n})^{1/n} = x$$

$$\Rightarrow (fof)(x) = x$$

$$\Rightarrow f^{-1}(x)=f(x)=\sqrt [ n ]{ a-{ x }^{ n } }$$

Alternative Method :
$$f(x)=\sqrt [ n ]{ a-{ x }^{ n } }$$

Put $$f(x)=y$$

$$y={ (a-{ x }^{ n }) }^{ 1/n }$$

Taking nth power ,
$${ y }^{ n }={ a-{ x }^{ n } }$$

$${ x }^{ n }={ a-{ y }^{ n } }$$

$$\Rightarrow x={ { (a-{ y }^{ n }) } }^{ 1/n }$$

$$\Rightarrow f^{-1}(y)={ { (a-{ y }^{ n }) } }^{ 1/n }$$

$$\Rightarrow f^{-1}(x)={ { (a-{ x }^{ n }) } }^{ 1/n }$$
Question 3
1 / -0

If $$g(x) = 2x + 1$$ and $$h(x) = 4x^{2} + 4x + 7$$, find a function $$f$$ such that $$f o g = h$$

A
$$f(x) = x^{3} - 6$$

B
$$f(x) = x^{2} + 6$$

C
$$f(x) = x^{2} - 6$$

D
$$f(x) = (2x+1)^2 + 6$$

Solution

$$f(g(x)) = h = 4x^2+4x+7=(2x+1)^2+6=(g(x))^2+6$$

$$\Rightarrow f(x) =x^2+6$$
Question 4
1 / -0

Which of the following are two distinct linear functions which map the interval $$[-1, 1]$$ onto $$[0, 2]$$

A
$$f(x) = 1 + x$$ or $$1 - x$$

B
$$f(x) = 1 + 2x$$ or $$1 - x$$

C
$$f(x) = 1 + x$$ or $$1 - 2x$$

D
$$f(x) = 1 + x$$ or $$2 - x$$

Solution

Out of two linear functions one will be increasing and other will be decreasing.

Let $$f(x) = ax+b$$

For increasing: $$f(-1)=0=> b-a=0\Rightarrow b=a$$ and $$f(1) = 2\Rightarrow a+b=2\Rightarrow a=b=1\therefore f(x) = 1+x$$

For decreasing: $$f(-1)=2=> b-a=2$$ and $$f(1) = 0\Rightarrow a+b=2\Rightarrow b=1, a=-1\therefore f(x) = 1-x$$
Question 5
1 / -0

A function $$f : \left[ \displaystyle \frac{1}{2}, \infty \right) \rightarrow \left[ \displaystyle \frac{3}{4}, \infty \right)$$ defined as, $$f(x) = x^{2} - x + 1$$.

Then for what value of $$x$$ does equation $$f(x) = f^{-1} (x)$$ hold right.

A
$$x=3$$

B
$$x=2$$

C
$$x=1$$

D
None of these

Solution

$$f(x) = x^{2} - x + 1$$

Substitute $$f(x)=y$$

$$y= x^{2} - x + 1$$

$$\Rightarrow x^2-x+1-y=0$$

$$\Rightarrow x=\dfrac{1\pm\sqrt{4y-3}}{2}$$

$$\Rightarrow f^{-1}(y)=\dfrac{1\pm\sqrt{4y-3}}{2}$$

$$\Rightarrow f^{-1}(x)=\dfrac{1\pm\sqrt{4x-3}}{2}$$

Now, we will check which value of $$x$$ from the options gives $$f(x)=f^{-1}(x)$$

Here, $$f(3)=7$$

$$f^{ -1 }(3)=\dfrac { 1\pm 3 }{ 2 } =2,-1$$

$$f(x)\neq f^{-1}(x)$$ for $$x=3$$

Similarly for $$x=2$$, $$f(x)\ne f^{-1}(x)$$

Now, for $$x=1$$

$$f(1)=1$$

$$f^{-1}=\dfrac{1\pm{1}}{2}=1,0$$

Hence, for $$x=1$$ , $$f(x)= f^{-1}(x)$$
Question 6
1 / -0

$$f(x)\, >\, x;\, \forall\, x\, \epsilon\, R.$$ The equation $$f (f(x)) -x = 0$$ has

A
Atleast one real root

B
More than one real root

C
No real root if f(x) is a polynomial & one real root if f(x) is not a polynomial

D
No real root at all

Solution

Given $$f(x) > x $$ $$\forall x \epsilon R$$
Hence, $$ f(x)-x$$ has no real roots, because $$f(x)-x >0$$
Since, $$x \epsilon R \implies f(x) \epsilon R$$
$$f(f(x))>x$$
$$\implies f(f(x))-x>0$$
Hence it will have no real root.
Question 7
1 / -0

If $$ f : R \rightarrow R, f(x) = (x + 1)^2$$ and $$g : R \rightarrow R, g(x) = x^2 + 1 $$ then $$(fog)(3)$$ is equal to

A
$$121$$

B
$$144$$

C
$$112$$

D
$$11$$

Solution

Given,
$$ f : R \rightarrow R, f(x) = (x + 1)^2$$ and $$g : R \rightarrow R,g(x) = x^2 + 1 $$

$$g(3)=3^2+1=10$$

$$f(10)=(10+1)^2=121$$

$$\therefore fog(3)=f\left(g(3)\right)=f(10)=121$$

Hence, option A.
Question 8
1 / -0

If $$f(x) = \sqrt{| x-1|}$$ and $$g(x) = \sin x$$, then $$(fog) (x)$$ equals

A
$$\sin \sqrt{| x-1|}$$

B
$$\left|\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right|$$

C
$$\left|\sin x + \cos x\right|$$

D
$$\left|\sin\dfrac{x}{2} + \cos\dfrac{x}{2}\right|$$

Solution

$$\displaystyle fog\left( x \right) =f\left( g\left( x \right) \right) =\sqrt { \left| \sin { x } -1 \right| } $$

$$=\sqrt { \left| 1-\cos { \left( \dfrac { \pi }{ 2 } -x \right) } \right| } =\sqrt { 2 } \left| \sin { \left( \dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right) } \right| =\left| \sin { \dfrac { x }{ 2 } } -\cos { \dfrac { x }{ 2 } } \right| $$
Question 9
1 / -0

If $$f(x) = \log x$$, $$g(x) = x^3$$, then $$f[g(a)] + f[g(b)]$$ equals

A
$$f[g(a) + g(b)]$$

B
$$3f(ab)$$

C
$$g[f(ab)]$$

D
$$g[f(a) + f(b)]$$

Solution

We have $$f(x) = \log x$$

$$g(x) = x^3$$

$$g(a) = a^3 $$

$$f(g(a)) = \log (a^3) $$

$$f(g(a)) =3 \log a $$

Similarly,

$$f(g(b)) = 3 \log b $$

$$\Rightarrow f(g(a)) + f(g(b)) = 3 \log a + 3 \log b $$

$$\Rightarrow f(g(a)) + f(g(b)) = 3 \log ab $$

$$\Rightarrow f(g(a)) + f(g(b)) = 3 f(ab)$$
Question 10
1 / -0

from the given statement $$N$$ denotes the natural number and $$W$$ denotes the whole number, so which statement in the following is correct

A
N=W

B
N $$\subset$$ W

C
W $$\subset$$ N

D
N $$\cong$$ W

Solution

Natural numbers are naturally occurring counts of an object $$1, 2, 3, 4,… $$forming an infinite list in which the first number is $$1$$ and every next number is equal to the preceding number $$+1.$$
Whole number include the set of all Natural numbers and also $$0.$$
Let $$N$$ and $$W$$ elements of natural and whole numbers.
$$N=\{1,2,3,4,....\infty\}$$
$$W=\{0,1,2,3,4,...\infty\}$$
So, here we can say $$N$$ is subset of $$W.$$
$$\therefore$$ $$N\subset W$$ is correct statement.

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Relations and Functions Test - 49

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Relations and Functions Test - 49

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SHARING IS CARING

Question 1

is given by $$\displaystyle \frac{1}{3x\, -\, 5}$$

is given by $$\displaystyle \frac{x\, +\, 5}{3}$$

does not exist because $$f$$ is not one-one

does not exist because $$f$$ is not onto

Solution

Question 2

$$f^{-1} (x) = (a - x^{n})^{1/n}$$

$$f^{-1} (x) = a-x$$

$$f^{-1} (x) = (a + x^{n})^{1/n}$$

None of these

Solution

Question 3

$$f(x) = x^{3} - 6$$

$$f(x) = x^{2} + 6$$

$$f(x) = x^{2} - 6$$

$$f(x) = (2x+1)^2 + 6$$

Solution

Question 4

$$f(x) = 1 + x$$ or $$1 - x$$

$$f(x) = 1 + 2x$$ or $$1 - x$$

$$f(x) = 1 + x$$ or $$1 - 2x$$

$$f(x) = 1 + x$$ or $$2 - x$$

Solution

Question 5

$$x=3$$

$$x=2$$

$$x=1$$

None of these

Solution

Question 6

Atleast one real root

More than one real root

No real root if f(x) is a polynomial & one real root if f(x) is not a polynomial

No real root at all

Solution

Question 7

$$121$$

$$144$$

$$112$$

$$11$$

Solution

Question 8

$$\sin \sqrt{| x-1|}$$

$$\left|\sin\dfrac{x}{2} - \cos\dfrac{x}{2}\right|$$

$$\left|\sin x + \cos x\right|$$

$$\left|\sin\dfrac{x}{2} + \cos\dfrac{x}{2}\right|$$

Solution

Question 9

$$f[g(a) + g(b)]$$

$$3f(ab)$$

$$g[f(ab)]$$

$$g[f(a) + f(b)]$$

Solution

Question 10

N=W

N $$\subset$$ W

W $$\subset$$ N

N $$\cong$$ W

Solution

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