Differentiation Test 12

Result

Differentiation Test 12

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Question 1
1 / -0

$$A$$ is twice as fast as $$B$$ is thrice as fast as $$C$$. The journey covered by $$C$$ in $$42$$ minutes, what will be covered by $$A$$ is

A
$$21$$ Min

B
$$64$$ Min

C
$$17$$ Min

D
$$40$$ Min

Solution

time taken ny C to complete the journey $$=$$ 42 minutes
time taken by A to complete the journey $$=\dfrac{42}{2}$$ minutes ($$\because $$ A is twice as fast as C)
$$=21 $$ minutes.
Question 2
1 / -0

Find $$\dfrac{dy}{dx}$$ of function $$y= e^{x^3} +\dfrac{1}{2} \log x $$

A
$$2.e^{x^3}x^2+\dfrac {1}{2x}$$

B
$$e^{x^3}x^2+\dfrac {1}{2x}$$

C
$$3.e^{x^3}x^2+\dfrac {1}{2x}$$

D
$$3.e^{x^3}x^2+\dfrac {1}{x}$$

Solution

Given y= $$e^{x^3} +\frac{logx}{x}$$
$$(\frac{dy}{dx})=(\frac{de^{x^3}}{dx})+(\frac{d(\frac{logx}{2})}{dx})\\=(\frac{de^{x^3}}{dx^3})\times(\frac{dx^3}{dx})+(\frac{1}{2})(\frac{dlogx}{dx})\\=3x^2e^{x^3}+(\frac{1}{2x})$$
Question 3
1 / -0

If $$^{n + 1}{C_3} = 4\,{\,^n}{C_2}$$ then $$n=$$

A
$$12$$

B
$$10$$

C
$$16$$

D
$$11$$

Solution

Given that,$$^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}}$$

Then $$n=?$$

$$ ^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)n!}{3\times 2!\left( n-2 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)}{3}=4 $$

$$ \Rightarrow n+1=12 $$

$$ \Rightarrow n=12-1=11\, $$

Hence, this is the answer.
Option (D) is correct.
Question 4
1 / -0

$$x^{\frac{1}{2}} + 1= t$$
differentiate w.r.t. x

A
$$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$$

B
$$\dfrac{dt}{dx} = \dfrac{1}{4\sqrt{x}}$$

C
$$\dfrac{dt}{dx} = \dfrac{1}{8\sqrt{x}}$$

D
$$\dfrac{dt}{dx} = \dfrac{1}{16\sqrt{x}}$$

Solution

$$x^{\frac{1}{2}} + 1= t$$

$$\dfrac{dt}{dx} = \dfrac{1}{2}(x) ^{\frac{1}{2}-1} + 0$$

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

$$\dfrac{1}{2x^{\frac{1}{2}}}$$

$$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$$
Question 5
1 / -0

If the letter of the word $$LATE$$ be permuted and the words so formed be arranged as in a dictionary . Then the rank of $$LATE$$ is :

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

LATE
No. of words that start with 'A'
$$\Rightarrow 3!=6$$
No. of words that start with E
$$\Rightarrow 3!=6$$
$$6+6=12$$
$$13^{th}$$ word starts with L
$$13^{th}$$ word = LAET
$$14^{th}$$ word = LATE
$$\therefore $$ Rank of LATE is $$14$$
Question 6
1 / -0

Differentiate: $$x^{100} + \sin x - 1$$

A
$$100x^{99}-\cos x$$

B
$$100x^{99}+\cos x$$

C
$$x^{99}+\cos x$$

D
$$100x^{99}+\sin x$$

Solution

$$\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} +\sin x -1 \right )$$

$$=\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} \right )+\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( \sin x \right )-\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( 1 \right )$$

$$=100x^{100-1}+\cos x -0$$

$$=100x^{99}+\cos x$$
Question 7
1 / -0

Choose the most appropriate option which fits this pattern

A

B

C

D
Question 8
1 / -0

Solve:$$\dfrac{2}{2}+\dfrac{3}{3}+\dfrac{4}{4}+$$...... + upto $$1000$$ terms= ?

A
$$1000$$

B
twice of $$500$$

C
four times of $$250$$

D
All of these

Solution

All the terms simplify to $$1$$ which is being added $$1000$$ times.
Therefore, the final answer is $$1000$$ and it can be seen that all the options are correct.
Question 9
1 / -0

$$3$$ letters are posted in $$5$$ letters boxes. If all the letters are not posted in the same box, then number of ways of posting is

A
$$120$$

B
$$125$$

C
$$130$$

D
$$124$$

Solution

According to problem,
$$3$$ letters are posted $$5$$ boxes
implies that,
we have a letter and can be posted in any of the $$5$$ boxes.
Similarly next letter can be posted in 5 ways, and the all other follow same method
implies that
$${\left(5\right)}^3$$
$$ = 5 \times 5 \times 5$$
Hence, $$ 125 $$ possible ways for posting in $$5$$ boxes.
Question 10
1 / -0

There are $$10$$ trees between two stations $$A$$ and $$B$$. Three of them are to be cut down then the total number of ways so that no two trees are to be cut consecutively, is

A
$$^{8}C_{3}$$

B
$$^{7}C_{3}$$

C
$$^{10}C_{3}$$

D
$$^{9}C_{3}$$

Solution

$$3$$ trees to be cut $$7$$ left

Total no. of ways to cut $$3$$ trees $$=$$

Total no. of was to place $$3$$ trees

in the given gaps,
$$=$$ select $$3$$ gaps from $$8$$ $$=\, ^8C_3$$