Relations and Functions Test

Result

Relations and Functions Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

If $$k\ \epsilon\ R^+$$ and the middle term of $$(\dfrac{k}{2} + 2)^8$$ is $$1120$$, then value of k is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$4$$

Solution

$$K\epsilon { R }^{ + }$$

The general term, $${ \left( r+1 \right) }^{ th }$$ term of the expansion
$${ \left( \dfrac { K }{ 2 } +2 \right) }^{ 8 }$$ is $${ T }_{ r+1 }={ 8 }_{ Cr }{ \left( \dfrac { K }{ 2 } \right) }^{ 8-r }{ 2 }^{ r }$$

In the above expansion there are $$9$$ terms, hence the middle term will be the $$5th$$ term $$(r=4)$$

$$\therefore$$ $${ T }_{ 5 }={ 8 }_{ { C }_{ 4 } }{ \left( \dfrac { K }{ 2 } \right) }^{ 8-4 }{ 2 }^{ 4 }$$

$$=\dfrac { 8! }{ 4!4! } \dfrac { { K }^{ 4 } }{ { 2 }^{ 4 } } .{ 2 }^{ 4 }$$

$$=\dfrac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2 } { K }^{ 4 }$$

$$=70{ K }^{ 4 }$$

Given,

$${ 70K }^{ 4 }=1120$$

$$\Rightarrow { K }^{ 4 }=112/7$$

$$\Rightarrow { K }^{ 4 }=16$$

$$\Rightarrow { K }^{ 4 }={ 2 }^{ 4 }$$

$$\therefore$$ $$K=2$$
Question 2
1 / -0

If $$f(x)=\sin^{2}x+x\sin 2x.\log x$$ then $$f(x)=0$$ has:

A
exactly one root in $$(0, 2\pi]$$

B
at least two roots in $$(0, 2\pi]$$

C
at most one root in $$(0, 2\pi]$$

D
no root in $$(0, 2\pi]$$

Solution
Question 3
1 / -0

$$f(n) = \begin{cases} e^{-y_2}, n \neq 0 & f'(0) = ?\\ 0, n = 0\end{cases}$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
Does not exist

Solution
Question 4
1 / -0

If $$f\left(x\right)=\dfrac{x}{x-1}$$, then $$\dfrac{ f\left(a\right)}{ f\left(a+1\right)}$$ is equal to

A
$$f\left(-a\right)$$

B
$$f\left(1/a\right)$$

C
$$f\left(a^{2}\right)$$

D
$$f\left(\dfrac{-a}{a-1}\right)$$

Solution

Given,

$$f(x)=\frac{x}{x-1}$$

$$f(a)=\frac{a}{a-1}$$
$$f(a+1)=\frac{a+1}{a+1-1}$$
$$=\frac{a+1}{a}$$

Then,
$$\frac{f(a)}{f(a+1)}=\frac{\frac{a}{a-1}}{\frac{a+1}{a}}$$
$$=\frac{a}{a-1} \times \frac{a}{a+1}=\frac{a^{2}}{a^{2}}$$

$$f\left(a^{2}\right)=\frac{a^{2}}{a^{2}-1}$$

$$\text {Therefore option } C=f\left(a^{2}\right)$$
Question 5
1 / -0

If $$f(x) = -x^2 - 3x^2 - 2x + a:a \in R$$, then the real values of $$x$$ satisfying $$f(x^2 + 1) > f(2x^2 + 2x + 3)$$ will be:-

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

B
$$(0, \infty)$$

C
$$(\infty, 0)$$

D
$$\phi$$

Solution
Question 6
1 / -0

If $$f\left(y\right)=\log y$$, then $$f\left(y\right)+f\left(1/y\right)$$ is equal to

A
$$2$$

B
$$1$$

C
$$0$$

D
$$-1$$

Solution

$$\text { Griven, } f(y)=\log y \\$$
$$\qquad f(y)+f\left(\frac{1}{y}\right)$$

$$\Rightarrow \quad \log y+\log \left(\frac{1}{y}\right) \\$$
$$\Rightarrow \quad \log y+\log \left((y)^{-1}\right) \\$$
$$\Rightarrow \quad \log y-\log y \\$$
$$\Rightarrow 0$$

$$\therefore \text { option } c=0$$
Question 7
1 / -0

$$f(x)=\begin{cases} 5,\ \ \ \ \ \ \ x \le 1 \\ a+bx,\ \ 1 < x < 3 \\ b+ax , \ \ \ 3 \le x < 5 \end{cases}$$ then what is the value of $$a,b$$.

A
$$a=0,\ b=5$$

B
$$none$$

C
$$a=10,\ b=10$$

D
$$a=-5,\ b=5$$
Question 8
1 / -0

If $$fxln\left(1+\dfrac{1}{x}\right)dx=p(x)ln\left(1+\dfrac{1}{x}\right)+\dfrac{1}{2}x-\dfrac{1}{2}ln(1+x)+c$$, being arbitary costant, then

A
$$p(X)=\dfrac{1}{2}x^{2}$$

B
$$p(x)=0$$

C
$$p(x)=1$$

D
$$none\ of\ these$$

Solution
Question 9
1 / -0

If $$\emptyset (x)=\emptyset '(x)$$ and $$\emptyset (1)=2,$$ then $$\emptyset (3)=?$$

A
$${e}^{2}$$

B
$$2{e}^{2}$$

C
$$3{e}^{2}$$

D
$$2{e}^{3}$$

Solution

In the above question it is given that
$$\phi(x)=\phi^{\prime}(x)$$
This condition is only followed by
$$\phi(x)=a e^{x}$$
Given that $$\phi(1)=2$$
$$\Rightarrow \phi(1)=a e^{\prime}$$
$$\Rightarrow \quad 2=a e$$
$$\Rightarrow \quad a=\dfrac{2}{e}$$
Therefore to find $$\phi(3)$$ put $$x=3$$
$$\Rightarrow \phi(3)=a e^{3}$$
$$\Rightarrow \quad \phi(3)=\dfrac{2}{e} e^{3}$$
$$\Rightarrow \phi(3)=2 e^{2}$$
hence, (B) is the correct option.
Question 10
1 / -0

If the function $$f:[2,\infty ]\rightarrow [1,\infty ]$$ is defined by $$f(x)=3^{x}(x-2)$$ then $$f^{-1}(x)$$ is

A
$$1+\sqrt{1+ log_{3^{x}}}$$

B
$$1-\sqrt{1+ log_{3^{x}}}$$

C
$$1+\sqrt{1- log_{3^{x}}}$$

D
Does not exist

Solution