Relations and Functions Test

Result

Relations and Functions Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

If f(x)=ax+b, where a and b are integers, f(-1)=-5 and f(3)=3, then a and b are equal to

A
a=-3,b=-1

B
a=2,b=-3

C
a=0,b=2

D
a=2,b=3
Question 2
1 / -0

Let 'f' be a function defined from $$R^{+}\rightarrow R^{+}$$ .If $$(f(xy))^{2}=x(f(y))^{2}$$ for all positive numbers x and y . If f(2)=6, find f(50)=

A
20

B
30

C
5

D
40
Question 3
1 / -0

Let $$R = \left \{(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)\right \}$$ be a relation on the set $$\left \{2, 3, 4, 5\right \}$$, then

A
$$R$$ is reflexive and symmetric but not transitive

B
$$R$$ is reflexive and transitive but not symmetric

C
$$R$$ is symmetric and transitive but not reflexive

D
$$R$$ is an equivalence relation

Solution
Question 4
1 / -0

If $$g(f(x))= |sinx|$$ and $$g(f(x))= sin^2 \sqrt{x},$$ then

A
$$f(x)= \sqrt{x}, \ g(x)= sin^2x$$

B
$$f(x) = sin \sqrt{x}, \ g(x)= x^2$$

C
$$f(x)= \sqrt{sinx}, \ g(x)= x^2$$

D
$$f(x)= |x|, \ g(x)= sin^2 x$$

Solution
Question 5
1 / -0

Let $$f(x)=\left | x-x_{1} \right |+\left | x-x_{2} \right |$$ where $$x_{1} and x_{2}$$ are distinct real numbers. Then the number of points at which f(x) is minimum is:

A
more than 3

B
1

C
2

D
3
Question 6
1 / -0

If $$f(x)=\frac{1}{\sqrt{x+2\sqrt{2x-4}}}+\frac{1}{\sqrt{x-2\sqrt{2z-4}}}$$ for x > 2 then f(11) =

A
7/6

B
5/6

C
6/7

D
5/7

Solution
Question 7
1 / -0

If $$\displaystyle f(x)= \dfrac{7^{1+\ln x}}{x^{\ln 7}}$$ then $$f (2015) $$ is equal to

A
$$20$$

B
$$7$$

C
$$2015$$

D
$$1$$

Solution
Question 8
1 / -0

If $$f(x)=x+x^{2}$$ is expanded as a Fourier series in $$(-\pi ,\pi )$$, then $$a_{0}$$=

A
$$\frac{\pi ^{2}}{3}$$

B
$$\frac{2\pi ^{2}}{3}$$

C
$$\frac{-\pi ^{2}}{3}$$

D
$$\frac{-4\pi ^{2}}{3}$$

Solution
Question 9
1 / -0

If $$(x^{2}-2)+(y+3)i=7+4i$$ then x and y are

A
$$\pm 3 and 4 $$

B
$$\pm \sqrt{5} and 4 $$

C
$$\pm 3 and 1 $$

D
None of these

Solution
Question 10
1 / -0

If $$f\left(x+\dfrac{1}{x}\right)=x^2+\dfrac{1}{x^2}$$ then $$f(x)=?$$

A
$$x^2$$

B
$$(x^2-1)$$

C
$$(x^2-2)$$

D
None of these

Solution

Let $$x+\dfrac{1}{x}=z$$.

Then,
$$f(z)=f\left(x+\dfrac{1}{x}\right)=\left(x^2+\dfrac{1}{x^2}\right)=\left(x+\dfrac{1}{x}\right)^2-2=(z^2-2)$$.

$$\Rightarrow f(x)=(x^2-2)$$.