Relations and Functions Test

Result

Relations and Functions Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

Let f be injective map with domain {x , y, z} and range {1, 2, 4} such the exactly one of the following statement is correct and the remaining are false $$f(x)=1,f(y)\neq 1,f(z)\neq 2$$ then \\ $${ f }^{ -1 }(1)$$

A
x

B
y

C
z

D
$$\phi $$

Solution
Question 2
1 / -0

Let a, b, c be real numbers with 0 < a < 1, 0 < b < 1, 0 < c < 1 and a + b + = 2 then$$\dfrac{a}{1 - a}$$,$$\dfrac{b}{1 - b}$$, $$\dfrac{c}{1 - c}$$

A
2

B
4

C
6

D
8

Solution
Question 3
1 / -0

If $$y^2 = ax^2 +bx+c$$, then $$y^2 \dfrac{d^2y}{dx^2}$$ is

A
a constant function

B
a function of x only

C
a function of y only

D
a function of both x and y

Solution

$$y^{2}=a x^{2}+b x+c \quad$$ is
Differentiating this equ. (i),
$$2 y \dfrac{d y}{d x}=2 a x+6$$
Differentiating above equation again,
$$2 y \dfrac{d^{2} y}{d x^{2}}+2\left(\dfrac{d y}{d x}\right)\left(\dfrac{d y}{d x}\right)=2 a$$
$$\dfrac{d^{2} y}{d x^{2}}=\dfrac{2 a-2\left(\dfrac{d y}{d x}\right)^{2}}{2 y}=\dfrac{a-\left(\frac{d y}{d x}\right)^{2}}{y}$$
$$y^{2} \dfrac{\alpha^{2} y}{d x^{2}}=y^{2}\left(\dfrac{a-\left(\dfrac{d y}{d x}\right)^{2}}{y}\right)$$
$$y^{2} \dfrac{\alpha^{2} y}{d x^{2}}=y^{2}\left(\dfrac{a-\left(\dfrac{d y}{d x}\right)^{2}}{y}\right)$$
$$=y\left(a-\left(\dfrac{d y}{d x}\right)^{2}\right) \quad\left\{\begin{array}{l}\because 2 y \dfrac{d y}{d x}=2 a x+b \\ \frac{d y}{d x}=\dfrac{2 a x+b}{2 y}\end{array}\right\}$$
$$=y\left(a-\left(\dfrac{2 a x+b}{2 y}\right)^{2}\right)$$
$$=\dfrac{y\left(4 a y^{2}-\left(4 a^{2} x^{2}+b^{2}+4 a b x\right)\right.}{4 y^{2}}$$
$$=\dfrac{4 a y^{2}-4 a^{2} x^{2}-b^{2}-4 a b x}{4 y}$$
$$=\dfrac{4 a\left(a x^{2}+6 x+c\right)-4 a^{2} x^{2}-6^{2}-4 a b x}{4 y}$$
$$=\dfrac{4 a^{2} x^{2}+4 a b x+4 a c-4 a^{2} x^{2}-b^{2}-4 a b x}{4 y}$$
$$=\dfrac{4 a c-b^{2}}{4 y}$$
$$\therefore$$ It is function of $$y$$ only.
option (C) is correct.
Question 4
1 / -0

Let f be a differential function satisfying the relation f(xy) = xf(y) +yf(x) -2xy (wherer x,y > 0) and f(1) = 3 then

A
$$f(x) = x\ell nx + 3x -\dfrac{x^2}{2}$$

B
$$f(x) = x \ell nx$$

C
$$x = e^-3$$ is the abscissa of the point of inflection of f (x)

D
The equation f(x) = K has two solution if $$\in (-e^-3, 0)$$

Solution
Question 5
1 / -0

if $$\left( x \right) =\sqrt { { x+2 }\sqrt { 2x-4 } } +\sqrt { { x-2 }\sqrt { 2x-4 } } $$ then the value of $$10 { f }^{ 1 }\left( { 102 }^{ + } \right)$$

A
$$-1$$

B
$$0$$

C
$$2$$

D
$$1$$

Solution
Question 6
1 / -0

Let $$f(1)=1$$ and $$f(n)=2\sum_{r=1}^{n-1}{f(r)}$$. Then $$\sum_{n=1}^{m}{f(n)}$$. is equal to

A
$$3^{m}-1$$

B
$$3^{m}$$

C
$$3^{m-1}$$

D
$$None$$

Solution

$$ \begin{array}{l} f(1)=1 \\ f(2)=2 \sum_{x=1}^{1} f(x)=2 f(1)=2 \\ f(3)=2 \sum_{x=1}^{2} f(x)=2(f(1)+f(2))=6 \\ \text { The sequence is } 1+2+6+18+\ldots \\ \begin{aligned} \sum_{n-1} f(n) &=1+2+6+18 t \\ &=1+2(1+3+9+\ldots) \\ &=1+2\left(\frac{3^{m-1}-1}{3-1}\right) \\ &=1+3^{m-1}-1 \\ &=3^{m-1} \end{aligned} \end{array} $$
option C is correct.
Question 7
1 / -0

If $$f(x) = \dfrac{cos \, x}{(1 - sin x)}^{1/3}$$ then

A
$$\underset{x \rightarrow \dfrac{\pi^-}{2}}{lim} f(x) = - \infty$$

B
$$\underset{x \rightarrow \dfrac{\pi^+}{2}}{lim} f(x) = \infty$$

C
$$\underset{x \rightarrow \dfrac{\pi}{2}}{lim} f(x) = \infty$$

D
None of these
Question 8
1 / -0

$$f(x) = e^x \sin x$$, then $$f^n(x)$$ is equal to

A
$$8e^x \cos x$$

B
$$e^{6x} \cos 6x$$

C
$$8e^x \sin x$$

D
$$2e^x \cos x$$
Question 9
1 / -0

If $$2f(x-1)-f\left(\dfrac{1-x}{x}\right)=x$$, then $$f(x)$$ is

A
$$\dfrac{1}{3}\left\{2(1+x)+\dfrac{1}{(1+x)}\right\}$$

B
$$2(x-1)-\dfrac{(1-x)}{x}$$

C
$$x^{2}+\dfrac{1}{x^{2}}+4$$

D
$$\dfrac{1}{4}\left\{(x+2)+\dfrac{1}{(x+2)}\right\}$$

Solution
Question 10
1 / -0

Let $$f(x)=(x+1)^{2}-1, x\ge -1$$. Then the set $$\left\{x: f(x)=f^{-1}(x)\right\}$$ is

A
$$\left\{0, -1, \dfrac{-3+i\sqrt{3}}{2}, \dfrac{-3-i\sqrt{3}}{2}\right\}$$

B
$$\left\{0, 1, -1\right\}$$

C
$$\left\{0, -1\right\}$$

D
$$empty$$