Relations and Functions Test

Result

Relations and Functions Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The real value of $$\theta $$for which the expression $$\dfrac { 1+icos\theta }{ 1-2icos\theta } $$ is a real number __________.

A
$$n\pi $$

B
$$2n\pi +\dfrac { \pi }{ 2 } $$

C
$$2n\pi -\dfrac { \pi }{ 2 } $$

D
none of these

Solution
Question 2
1 / -0

$$f(x)=4-(6-x)^{2/3}$$ in $$[5,7]$$

A
Lagrange's theorem is applicable

B
Rolle's theorem is applicable

C
Lagrange's theorem is applicable but not the Roll's theorem

D
Both theorem are not applicable

Solution

Let $$a=5,b=7$$
$$f(a)=f(5)=4-(6-5)^{2/3}=3$$
$$f(b)=f(7)=4-(6-7)^{2/3}=3$$
Since $$f(a)=f(b)$$
we can say that Rolle's theorem is applicable.
Question 3
1 / -0

If $$S$$ is the set of all real $$x$$ such that $$\dfrac{2x}{2x^{2}+5x+2}>\dfrac{1}{x+1}$$, then $$S$$ is equal to

A
$$\left(-\infty,\dfrac{-1}{2}\right)$$

B
$$\left(-\infty,\dfrac{-2}{3}\right)$$

C
$$\left(-\infty,\dfrac{2}{3}\right)$$

D
$$\left(\dfrac{2}{3},\infty\right)$$

Solution
Question 4
1 / -0

Let $$f(x)=x^{3}-2x+2$$. If real numbers $$a,b$$ and $$c$$ such that $$|f(a)|+|f(b)|+|f(c)|=0$$, then the value of $$f^{2}\left(a^{2}+\dfrac {2}{a}\right)+f^{2}\left(b^{2}+\dfrac {2}{b}\right)-f^{2}\left(c^{2}+\dfrac {2}{c}\right)$$ equal to

A
$$6$$

B
$$24$$

C
$$36$$

D
$$48$$
Question 5
1 / -0

If $$\sqrt { \log _ { 4 } \left( \log _ { 2 } \left( \log _ { 2 } \left( x ^ { 2 } - 2 x + a \right) \right) \right. } )$$ is defined $$\forall x \in R$$, then

A
$$[ 10 , \infty )$$

B
$$[ 9 , \infty )$$

C
$$[ 2 , \infty )$$

D
$$[ 5 , \infty )$$

Solution
Question 6
1 / -0

If $$f(x) = (x - 1)^{100} (x - 2)^{2(99)} (x - 3)^{3(98)} .... (x - 100)^{100}$$, then the value of $$ \dfrac{f' (101)}{f(101)}$$ is

A
5050

B
2575

C
3030

D
1250

Solution
Question 7
1 / -0

Let A be a finite set and n(A) = 5, then the number of relations which are not symmetric is _____________________.

A
$$2^{25}$$

B
$$2^{15}$$

C
$$2^{25} - 2^{10}$$

D
$$2^{25} - 2^{15}$$

Solution
Question 8
1 / -0

Let $$y=x^{x+1}+\left(1+\dfrac{1}{x}\right)^{x}$$ then $$y'(1)$$ equal

A
$$(ln\ 2)+1$$

B
$$(2\ ln\ 2)+1$$

C
$$(ln\ 2)-1$$

D
$$(2\ ln\ 2)-1$$

Solution
Question 9
1 / -0

If $$f(x)=4x^{3}+3x^{2}+3x+4$$ , then $$x^{3}f(\frac{1}{x})$$ is equal to

A
f(-x)

B
f(x+2)

C
$$[f(\frac{1}{x})]^{2}$$

D
f(x)

Solution

Given equation is $$f(x)=4x^3+3x^2+3x+4$$
We have to find $$x^3f\left(\dfrac {1}{3}\right)$$
Substituting $$x=\dfrac {1}{x}$$ in the given equation:
Thus $$f\left (\dfrac {1}{x}\right )=4\left(\dfrac {1}{x}\right)^3+3\left(\dfrac {1}{x}\right)^2+3\left(\dfrac {1}{x}\right)+4$$
$$=\dfrac {4}{x^3}+\dfrac {3}{x^2}+\dfrac {3}{x}+4$$
Therefore, $$x^3f\left(\dfrac {1}{x}\right)$$ $$=x^3\left(\dfrac {4}{x^3}+\dfrac {3}{x^2}+\dfrac {3}{x}+4\right)$$
$$=4+3x+3x^2+4x^3$$
$$=4x^3+3x^2+3x+4=f(x)$$
Option D is correct.
Question 10
1 / -0

If $$x\ \epsilon \ R$$, the number of solutions of $$\sqrt{2x+1}-\sqrt{2x-1}-1$$ is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$infinite$$

Solution

We have,

$$x\in R$$

Then,

$$\sqrt{2x+1}-\sqrt{2x-1}-1$$

Now,

Put $$x=1,2,3,4.........$$

So,

$$ \sqrt{2\left( 1 \right)+1}-\sqrt{2\left( 1 \right)-1}-1=\sqrt{3}-2 $$

$$ \sqrt{2\left( 2 \right)+1}-\sqrt{2\left( 2 \right)-1}-1=\sqrt{5}-\sqrt{3}-1 $$

And so on….

Then,

Infinite number of solution.

Hence, this is the answer.