Differentiation Test 46

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Differentiation Test 46

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Weekly Quiz Competition

Question 1
1 / -0

If $$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$$, then $$\dfrac {dy}{dx}$$ is equal to

A
$$\cfrac{1}{(1+x)^{2}}$$

B
$$\cfrac{-1}{(1+x)^{2}}$$

C
$$\cfrac{1}{(1-x)^{2}}$$

D
None of these

Solution

Given $$x\sqrt{1+y}+y\sqrt{1+x}=0$$
$$x\sqrt{1+y}=-y\sqrt{1+x}$$
$$\cfrac{\sqrt{1+y}}{y}=-\cfrac{\sqrt{1+x}}{x}$$
$$\left (\cfrac{\sqrt{1+y}}{y} \right )^{2}=\left ( -\cfrac{\sqrt{1+x}}{x} \right )^{2}$$
$$\cfrac{1+y}{y^{2}}=\cfrac{1+x}{x^{2}}$$
$$x^{2}+x^{2}y=y^{2}+y^{2}x$$
$$x^{2}-y^{2}=y^{2}x-x^{2}y$$
$$(x+y)(x-y)=xy(y-x)$$
$$\therefore x+y+xy=0\Rightarrow y=\cfrac{-x}{1+x}$$ $$(\because x\neq y)$$
Differentiate on both sides w.r.t x
$$1+\cfrac{\mathrm{d} y}{\mathrm{d} x}+x\cfrac{\mathrm{d} y}{\mathrm{d} x}+y=0$$
$$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{-(1+y)}{1+x}$$
$$=\cfrac{-\left ( 1-\cfrac{x}{1+x} \right )}{1+x}$$
$$=\cfrac{-1}{(1+x)^{2}}$$
$$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{-1}{(1+x)^{2}}$$
Question 2
1 / -0

Seven person $$P_1,P_2......, P_7$$ initially seated at chairs $$C_1,C_2,.....C_7$$ respectively.They all left there chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and $$P_1$$, sits on $$C_2$$ and $$P_2$$ sits on $$C_3$$ ?

A
$$52$$

B
$$53$$

C
$$54$$

D
$$55$$

Solution
Question 3
1 / -0

If $$m$$ denotes the number of $$5$$ digit numbers if each successive digits are in their descending order of magnitude and $$n$$ is the corresponding figure. When the digits and in their ascending order of magnitude then $$(m-n)$$ has the value

A
$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

B
$${ _{ }^{ 9 }{ C } }_{ 5 }$$

C
$${ _{ }^{ 10 }{ C } }_{ 3 }$$

D
$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

$$ {\textbf{Step 1: Find m}} $$

$$ {\text{For m,}} $$

$$ {\text{First we select any 5 digits from 0,1,2,}}...{\text{,9}} $$

$$ {\text{Number of ways = }}{}^{10}{{\text{C}}_5} $$

$$ {\text{Now after selection there is only 1 way to arrange these selected digits, i}}{\text{.e}}{\text{., in descending order}}{\text{.}} $$

$$ {\text{Therefore m = }}{}^{10}{{\text{C}}_5}\times{\text{ 1 = }}{}^{10}{{\text{C}}_5} $$

$$ {\textbf{Step 2: Find n}} $$

$$ {\text{For n,First we select any 5 digits from 1,2,}}...{\text{,9}} $$

$$ {\text{We can't select zero as first digit because then the number won't be a 5 - digit number}}{\text{.}} $$

$$ {\text{Therefore number of ways = }}{}^9{{\text{C}}_5} $$

$$   \Rightarrow {\text{n = }}{}^9{{\text{C}}_5}\times{\text{ 1 = }}{}^9{{\text{C}}_5} $$

$$   \Rightarrow {\text{m - n = }}{}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5} $$

$$ {\text{We know that,}}{}^n{{\text{C}}_r}{\text{ + }}{}^n{{\text{C}}_{r - 1}}{\text{ = }}{}^{n + 1}{{\text{C}}_r} $$

$$   \Rightarrow {}^{n + 1}{{\text{C}}_r}{\text{ - }}{}^n{{\text{C}}_r}{\text{ = }}{}^n{{\text{C}}_{r - 1}} $$

$$   \Rightarrow {}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}{\text{ = }}{}^9{{\text{C}}_4} $$

$$ {\text{Hence, m - n = }}{}^9{{\text{C}}_4} $$

$$ {\textbf{Hence, the correct answer is option A}} $$
Question 4
1 / -0

There are $$2$$ identical white balls, $$3$$ identical red balls and $$4$$ green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is

A
$$6(7! - 4!)$$

B
$$7(6! - 4!)$$

C
$$8! - 5!$$

D
None

Solution

These are totally $$9$$ balls of which $$2$$ are identical of one kind, $$3$$ are a like of another kind $$and$$ $$4$$ district ones.

At least one ball of same color separated $$=$$ Total $$-$$ No ball of same color is separated
Total permutation $$=\dfrac{9!}{2!3!}$$

For no ball is separated : we consider all balls of same color as $$1$$ entity, so there are $$3$$ entities which can be placed in $$3!$$ ways.

The white and red balls are identical so they will be placed in $$1$$ way whereas green balls are different so they can be placed in $$4!$$ ways
$$\Rightarrow Req=3!\times 4!$$

At least one ball is separated $$=\dfrac{9!}{2!3!}-3!4!$$

                                                  $$=\dfrac{9\times 8\times 7!}{2\times 6}-6\times 4!$$

$$=6\left(7!\right)-6\left(\times 4!\right) $$

                                                  $$=6\left( 7!-4!\right ).$$

Hence, the answer is $$6\left( 7!-4!\right ).$$
Question 5
1 / -0

If $$\int_{0}^{x}(t^{2}+2t+2)$$ dt, $$2\leq x\leq 4$$

A
The maximum value of f(x) is $$\dfrac{136}{3}$$

B
The minimum value of f(x) is 10

C
The maximum value of f(x) is 26

D
None of these
Question 6
1 / -0

If $$y^{y^{y^{....{^\infty}}}} = \log_e(x+\log_e(x+....))$$, then $$\dfrac{dy}{dx}$$ at $$(x= e^2-2, y= \sqrt2)$$ is

A
$$\dfrac{\log\left(\dfrac{e}{2}\right)}{2\sqrt2(e^2-1)}$$

B
$$\dfrac{\log2}{2\sqrt2(e^2-1)}$$

C
$$\dfrac{\sqrt2 \log\dfrac{e}{2}}{(e^2-1)}$$

D
None of these

Solution
Question 7
1 / -0

Let $$y=x^{x^x}$$, then differentiate $$y$$ w.r.t $$x$$.

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

B
$$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x-(\log x)^2\right)$$

C
$$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$$

D
$$x^{x^x}(x^x)\left(\dfrac{1}{x}-\log x-(\log x)^2\right)$$

Solution

$$y=x^{x^x}$$
diffrentiate it w.r.t.$$x$$
$$\dfrac{dy}{dx}=(e^{lnx^{x^x}})^{'}$$
$$\dfrac{dy}{dx}=(e^{x^x lnx})^{'}$$
$$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})^{'}$$
$$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}lnx)^{'}$$
$$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(xlnx)^{'}ln x$$+$$\dfrac{e^{xlnx}}{x})$$
$$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(xlnx)^{'}ln x$$+$$\dfrac{e^{xlnx}}{x})$$
$$\dfrac{dy}{dx}=(e^{e^{x lnx}lnx)})(e^{xlnx}(lnx+\dfrac{x}{x})ln x$$+$$\frac{e^{xlnx}}{x}$$
$$\dfrac{dy}{dx}=x^{x^{x}}$$$$(x^x(lnx+1)lnx+\dfrac{x^x}{x}$$
$$\dfrac{dy}{dx}=x^{x^{x}}$$$$(x^x)(\dfrac{1}{x}+lnx+(lnx)^2)$$
so option C is correct.
Question 8
1 / -0

if $$\int \dfrac {\cos 4x+1}{\cos x-\tan x}dx=A \cos 4x+B;$$ where $$A$$ & $$B$$ are constants, then

A
$$A=-1/4$$ & $$B$$ may have any value

B
$$A=-1/8$$ & $$B$$ may have any value

C
$$A=-1/2$$ & $$B=-1/4$$

D
$$A=B=1/2$$

Solution
Question 9
1 / -0

How many $$10-digit$$ numbers can be formed by using the digits $$1$$ and $$2$$?

A
$$^{10}{P}_{2}$$

B
$$^{10}{C}_{2}$$

C
$${2}^{10}$$

D
$$10\ !$$

Solution

Each place of a ten digit number can be fixed by any of the two digits. So, the number of ways to form a ten digit number is $${2^{10}}$$.
Question 10
1 / -0

What is the value of $$^nC_n$$?

A
zero

B
$$1$$

C
$$n$$

D
$$n!$$

Solution

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Re-Attempt Weekly Quiz Competition

Differentiation Test 46

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Differentiation Test 46

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SHARING IS CARING

Question 1

$$\cfrac{1}{(1+x)^{2}}$$

$$\cfrac{-1}{(1+x)^{2}}$$

$$\cfrac{1}{(1-x)^{2}}$$

None of these

Solution

Question 2

$$52$$

$$53$$

$$54$$

$$55$$

Solution

Question 3

$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

$${ _{ }^{ 9 }{ C } }_{ 5 }$$

$${ _{ }^{ 10 }{ C } }_{ 3 }$$

$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

Question 4

$$6(7! - 4!)$$

$$7(6! - 4!)$$

$$8! - 5!$$

None

Solution

Question 5

The maximum value of f(x) is $$\dfrac{136}{3}$$

The minimum value of f(x) is 10

The maximum value of f(x) is 26

None of these

Question 6

$$\dfrac{\log\left(\dfrac{e}{2}\right)}{2\sqrt2(e^2-1)}$$

$$\dfrac{\log2}{2\sqrt2(e^2-1)}$$

$$\dfrac{\sqrt2 \log\dfrac{e}{2}}{(e^2-1)}$$

None of these

Solution

Question 7

$$x^{x^x}\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$$

$$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x-(\log x)^2\right)$$

$$x^{x^x}(x^x)\left(\dfrac{1}{x}+\log x+(\log x)^2\right)$$

$$x^{x^x}(x^x)\left(\dfrac{1}{x}-\log x-(\log x)^2\right)$$

Solution

Question 8

$$A=-1/4$$ & $$B$$ may have any value

$$A=-1/8$$ & $$B$$ may have any value

$$A=-1/2$$ & $$B=-1/4$$

$$A=B=1/2$$

Solution

Question 9

$$^{10}{P}_{2}$$

$$^{10}{C}_{2}$$

$${2}^{10}$$

$$10\ !$$

Solution

Question 10

zero

$$1$$

$$n$$

$$n!$$

Solution

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