Functions Test 19

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Functions Test 19

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
1 / -0

Suppose f and g both are linear function with $$\displaystyle f(x)=-2x+1$$ and $$\displaystyle f \left ( g\left ( x \right ) \right )=6x-7$$ then slope of line $$y=g(x)$$ is

A
$$3$$

B
$$-3$$

C
$$6$$

D
$$-2$$

Solution

Given, $$f(x)=-2x+1$$

$$\therefore f(g(x))=-2g(x)+1=6x-7$$

$$\Rightarrow g(x)=-3x+4$$

$$\Rightarrow g'(x)=-3$$

Hence slope of $$g(x)$$ is $$-3$$
Question 5
1 / -0

If $$f(x) = x^3 $$ and $$g(x) = sin2x$$, then

A
$$g[f(1)] = 1$$

B
$$f(g(\pi/12) = 1/8$$

C
$$g{f(2)} = \sin 2$$

D
none of these

Solution

Given,
$$ f(x)=x^3 $$ and $$g(x)=\sin 2x$$

Clearly,

$$ f(1)=1 \Rightarrow g[f(1)]=\sin 2 $$

$$ g(\cfrac{\pi}{12}) =\dfrac{1} {2} \Rightarrow f[g(\dfrac{\pi}{12})]=\dfrac{1}{8}$$

and,$$ f(2)=8 \Rightarrow g[f(2)]= \sin 16 $$

So, the correct option is (B).
Question 6
1 / -0

If $$f(x) = (a x^n)^{1/n},$$ where $$\ n \in N$$, then $$f\{f(x)\}$$ equals

A
$$0$$

B
$$x$$

C
$$x^n$$

D
none of these

Solution

We have, $$f(x) = (a x^n)^{1/n}=a^{1/n}x=kx$$ where $$k = a^{1/n}$$

$$\therefore f\{f(x)\} = kf(x) = k^2x =a^{2/n} x$$
Question 7
1 / -0

If $$f(x) =\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$, then $$f[g(x)]$$ equals.

A
$$f(x)$$

B
$$[f(x)]^3$$

C
$$3f(x)$$

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

Solution

Given $$f(x)=\ln {\displaystyle \frac { 1+x }{ 1-x } } $$ and $$g(x)=\displaystyle \frac {3x+x^3}{1+3x^2}$$

then $$f[g(x)]=\ln {\displaystyle \frac { 1+\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } }{ 1-\displaystyle\frac { 3x+x^{ 3 } }{ 1+3x^{ 2 } } } } $$

$$=\ln {\displaystyle \frac { 1+3x^{ 2 }+3x+x^{ 3 } }{ 1+3x^{ 2 }-3x-x^{ 3 } } } $$

$$=\ln\left[ {\displaystyle \frac { 1+x }{ 1-x } }\right]^3 $$

$$=3\ln {\displaystyle \frac { 1+x }{ 1-x } } $$

$$\Rightarrow f[g(x)]=3f(x)$$

Hence, option C.
Question 8
1 / -0

Let $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$, then $$fog (x)$$ is

A
$$3x$$

B
$$x^3$$

C
$$\log_{10}3x$$

D
$$\log3x$$

Solution

Given $$f(x) = e^{3x}, g(x) = \log_ex, x > 0$$

$$fog(x)=f(\log_ex)=e^{3\log_ex}=x^3$$

Hence, option B.
Question 9
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 10
1 / -0

The domain of the function, $$\displaystyle y=f(x)=\sqrt {\log_{10} \left ( \frac {5x-x^2}{4} \right )}$$ is

A
$$[1, 4]$$

B
$$(1, 4)$$

C
$$[1, 4)$$

D
$$(1, 4]$$

Solution

Given,
$$\displaystyle y=f(x)=\sqrt {\log_{10} \left ( \frac {5x-x^2}{4} \right )}$$

For given function to be defined,
$$\displaystyle \log_{10} \left ( \frac {5x-x^2}{4} \right )\geq 0 \Rightarrow \left ( \frac {5x-x^2}{4} \right ) \geq 10^0=1$$

$$\Rightarrow 5x-x^2\geq 4\Rightarrow x^2-5x+4\leq 0$$

$$\Rightarrow (x-1)(x-4) \leq 0\Rightarrow x \in [1,4]$$

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Re-Attempt Weekly Quiz Competition

Functions Test 19

Result

Functions Test 19

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SHARING IS CARING

Question 1

$$121$$

$$144$$

$$112$$

$$11$$

Solution

Question 2

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

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

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

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

Solution

Question 3

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

$$3f(ab)$$

$$g[f(ab)]$$

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

Solution

Question 4

$$3$$

$$-3$$

$$6$$

$$-2$$

Solution

Question 5

$$g[f(1)] = 1$$

$$f(g(\pi/12) = 1/8$$

$$g{f(2)} = \sin 2$$

none of these

Solution

Question 6

$$0$$

$$x$$

$$x^n$$

none of these

Solution

Question 7

$$f(x)$$

$$[f(x)]^3$$

$$3f(x)$$

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

Solution

Question 8

$$3x$$

$$x^3$$

$$\log_{10}3x$$

$$\log3x$$

Solution

Question 9

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 10

$$[1, 4]$$

$$(1, 4)$$

$$[1, 4)$$

$$(1, 4]$$

Solution

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