Functions Test 55

Result

Functions Test 55

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Weekly Quiz Competition

Question 1
1 / -0

If $$f:R\rightarrow R\quad defined\quad by\quad f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ { -x }^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,\quad then\quad f\quad is$$

A
one-one but not onto

B
not one-one but onto

C
one-one and onto

D
neither one-one noronto

Solution
Question 2
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 3
1 / -0

The domain of definition of the function $$f(x)=\sqrt{log_{x^2-1}x}$$ is-

A
$$(\sqrt2, \infty)$$

B
$$(0, \infty)$$

C
$$(1, \infty)$$

D
None of these

Solution
Question 4
1 / -0

If $$\left|x\right|$$ denotes the integral part of $$x$$ and $$f(x)$$ has domain $$\left[ -\dfrac { 5 }{ 2 } ,2 \right] $$, the domain of $$f\left( \left[ \left| x \right| \right] \right) $$ is

A
$$\left[-3,2\right]$$

B
$$\left(-3,2\right]$$

C
$$\left[-2,2\right]$$

D
$$\left(-3,3\right)$$

Solution
Question 5
1 / -0

The domain of the real valued function $$f(x)$$ for which $$4^{f(x)+4^{1-f(x)}}= 4^{x}$$ is

A
$$(-1,1)$$

B
$$(-\infty ,1)$$

C
$$[1,\infty)$$

D
R

Solution

Given,
$$4^{f(x)}+4^{(1-f(x)}=4^{x}$$
$$\Rightarrow \quad 4^{f(x)}+\frac{4}{4^{f(x)}}=4^{x}$$
$$\Rightarrow \frac{4^{f(x)} \cdot 4^{f(x)}+4}{4^{f(x)}}=4^{x}$$
$$\Rightarrow y^{2 f(x)}+4=4^{x+f(x)}$$

Dividing above equation by 4 on both sides
$$\begin{aligned} & \frac{4^{2 f(x)}}{4}+1=\frac{y^{x+f(x)}}{4} \\ \Rightarrow & 4^{2 f(x)-1}+1=4^{x+f(x)-1}-(i) \\ &\left.1+4^{-(1-2 f(x)}=4^{x-f(x)}-(ii\right)\end{aligned}$$

$$\text {both equation are same, } \\$$
$$\text { Then, }$$
$$4^{2 f(x)-1} =4^{1-2 f(x)} \\$$
$$2 f(x)-1 =1-2 f(x) \\$$
$$f(x)+3 f(x) =2 \\$$
$$\Rightarrow 4 f(x)=2 \\$$
$$7 f(x)=\frac{1}{2}$$

$$\quad 4^{x+f(x)-1}=4^{x-f(x)}$$
$$\Rightarrow x+f(x)-1=x-f(x)$$
$$\Rightarrow \quad 2 f(x)=1$$
$$\Rightarrow \quad f(x)=\frac{1}{2}$$
$$Therefore = f(x)$$ is a constant function.

but from above equation
$$4^{x}=4^{f(x)}+4^{1-f(x)}$$

$$\text { = Putting the value of } f(x)=\frac{1}{2} \\$$
$$\text { in above equation. } \\$$
$$\Rightarrow \quad 4^{x}=4^{\frac{1}{2}}+4^{1-\frac{1}{2}} \\$$
$$\Rightarrow \quad 4^{x}=4$$

$$x=1$$
$$x$$ has one value in i.e. $$x=1$$
So, $$x \in[1, \infty)$$
Option C
Question 6
1 / -0

The domain of the function $$f(x) =\log_{\left[x+\dfrac{1}{2}\right]} |x^{2}-5x+6|$$ is

A
$$\left[\dfrac{3}{2}, 2\right) \cup (2,3) \cup (3, \infty)$$

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

C
$$\left[\dfrac{1}{2}, \infty\right)$$

D
$$None\ of\ these$$

Solution
Question 7
1 / -0

The function $$f:R\rightarrow R$$ defined by $$f\left(x\right)=6^ {x}+6$$ is

A
one-one and onto

B
many-one and onto

C
one-one and into

D
mauny-one and into

Solution

$$f: R \rightarrow R \\$$
$$f(x)=6^{x}+6$$
$$\text { For one - one, } \\$$
$$\qquad f\left(x_{1}\right)=f\left(x_{2}\right) \\$$
$$\Rightarrow 6^{x_{1}}+6$$ &=$$6^{x_{2}}+$$
$$\Rightarrow \quad 6^{x_{1}}=6^{x_{2}} \\$$
$$\Rightarrow \quad x_{1}=x_{2}$$
Therefore f is one - one.

Now,
$$\begin{aligned} f(x) &=y=6^{x}+6 \\ & \Rightarrow(y-6)=6^{x} \end{aligned}$$

Taking log both side.
$$\Rightarrow \log _{6}(y-6)=x$$

Now,
$$f\left(\log _{6}(y-6)\right)=6^{\log _{6}(y-6)}+6$$
$$=y-6+6$$
$$=y$$
$$\therefore f \text { is onto }$$

Hence, f is one - one and onto.
Question 8
1 / -0

Directions For Questions

$$f:R \rightarrow R$$ defined by, $$f(x) = x^3 + x^2 f'(1) + x.f^n(2) + f^m(3)$$ for all $$x \in R$$

...view full instructions

$$f(x)$$ is

A
One-one and onto

B
One-one and into

C
Many-one and onto

D
Many-one and into
Question 9
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 10
1 / -0

The domain of $$\frac{x+1}{\sqrt{x^{2}-5x+6}}$$ is

A
R-{2,3}

B
$$(3,\infty )$$

C
$$(-\infty ,\infty )$$

D
$$(-\infty, 2)\cup (3,\infty )$$

Solution

$$\dfrac{x+1}{\sqrt{x^{2}-5 x+6}}$$

$$x^{2}-5 x+6>0$$

$$x^{2}-3 x-2 x+6>0$$

$$x(x-3)-2(x-3)>0$$

$$(x-3)(x-2)>0$$

$$x \in[-\infty, 2) \cup(3, \infty)$$

Domain of

$$x+1 \in R$$

$$\therefore$$ Domain of $$\dfrac{x+1}{\sqrt{x^{2}-5 x+6}}$$ is $$R-(2,3)$$

Option $$A=R-(2,3)$$