Relations and Functions Test

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

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

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 66

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Relations and Functions Test - 66

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SHARING IS CARING

Question 1

a=-3,b=-1

a=2,b=-3

a=0,b=2

a=2,b=3

Question 2

20

30

5

40

Question 3

RRR is reflexive and symmetric but not transitive

RRR is reflexive and transitive but not symmetric

RRR is symmetric and transitive but not reflexive

RRR is an equivalence relation

Solution

Question 4

f(x)=x, g(x)=sin2xf(x)= \sqrt{x}, \ g(x)= sin^2xf(x)=x​, g(x)=sin2x

f(x)=sinx, g(x)=x2f(x) = sin \sqrt{x}, \ g(x)= x^2f(x)=sinx​, g(x)=x2

f(x)=sinx, g(x)=x2f(x)= \sqrt{sinx}, \ g(x)= x^2f(x)=sinx​, g(x)=x2

f(x)=∣x∣, g(x)=sin2xf(x)= |x|, \ g(x)= sin^2 xf(x)=∣x∣, g(x)=sin2x

Solution

Question 5

more than 3

1

2

3

Question 6

7/6

5/6

6/7

5/7

Solution

Question 7

202020

777

201520152015

111

Solution

Question 8

π23\frac{\pi ^{2}}{3}3π2​

2π23\frac{2\pi ^{2}}{3}32π2​

−π23\frac{-\pi ^{2}}{3}3−π2​

−4π23\frac{-4\pi ^{2}}{3}3−4π2​

Solution

Question 9

±3and4\pm 3 and 4 ±3and4

±5and4\pm \sqrt{5} and 4 ±5​and4

±3and1\pm 3 and 1 ±3and1

None of these

Solution

Question 10

x2x^2x2

(x2−1)(x^2-1)(x2−1)

(x2−2)(x^2-2)(x2−2)

None of these

Solution

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$R$ is reflexive and symmetric but not transitive

$R$ is reflexive and transitive but not symmetric

$R$ is symmetric and transitive but not reflexive

$R$ is an equivalence relation

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

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

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

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

$20$

$7$

$2015$

$1$

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

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

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

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

$\pm 3 and 4$

$\pm \sqrt{5} and 4$

$\pm 3 and 1$

$x^2$

$(x^2-1)$

$(x^2-2)$