Differentiation Test 35

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Differentiation Test 35

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\dfrac{a^x}{x^a}$$ then $$f'(a)=$$?

A
$$log a-1$$

B
$$log a-a$$

C
$$a log a-a$$

D
$$a log a+a$$

Solution

According to question,

$$ f' (x)= \dfrac{ a^{x} loga x^{a} - x^{a+1} a^{x} }{ x^{2a} } $$

$$ \Rightarrow f' (a)= \dfrac{ a^{a} loga. a^{a} - a^{a+1} a^{a} }{ a^{2a} }$$

Hence,

$$ \Rightarrow f' (a)=loga-a$$
Question 2
1 / -0

If the letters of the word "VARUN" are written in all possible ways and then are arranged as in a dictionary, then the rank of the word VARUN is?

A
$$98$$

B
$$99$$

C
$$100$$

D
$$101$$

Solution

$${\textbf{Step 1}}\;:\;{\mathbf{Finding}}\;{\mathbf{number}}\;{\mathbf{of}}\;{\mathbf{words}}\;{\mathbf{starting}}\;{\mathbf{with}}\;{\mathbf{the}}\;{\mathbf{alphabets}}\;{\mathbf{in}}\;{\mathbf{the}}\;{\mathbf{given}}\;{\mathbf{word}}$$

$${\text{No}}.{\text{ of words beginning with A }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{No}}.{\text{ of words beginning with N }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{No}}.{\text{ of words beginning with R }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{No}}.{\text{ of words beginning with U }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{So}},{\text{ in total }}96{\text{ words will be formed while beginning with letter A}},{\text{ N}},{\text{ R and U}}.$$

$${\textbf{Step 2}}\;\;:\;{\mathbf{Finding}}\;{\mathbf{the}}\;{\mathbf{order}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{word}}$$

$${\text{Order of }} 97{\text { th word }} - {\text{ VANRU}}$$

$${\text{Order of }}98{\text{th word }} - {\text{ VANUR}}$$

$${\text{Order of }}99{\text{th word }} - {\text{ VARNU}}$$

$${\text{Order of }}100{\text{th word}} - {\text{ VARUN}}.$$

$${\mathbf{Therefore}},\;{\mathbf{rank}}\;{\mathbf{of}}\;{\mathbf{word}}\;'{\mathbf{VARUN}}'\;{\textbf{is 100. Option C is correct.}}$$
Question 3
1 / -0

If the vertices of a triangle are $$(1,2),(4,-6)$$ and $$(3,5)$$, then its area is

A
$$\cfrac{23}{2}$$ sq. units

B
$$\cfrac{25}{2}$$ sq. units

C
$$12$$ sq.unit

D
none of these

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore, area of required triangle is given by
Area $$ = \dfrac{1}{2} \times [ 1 (-6 - 5) + 4(5 - 2) + 3(2 + 6) ] $$
$$=\dfrac{1}{2}[-11+12+24]$$
$$=\dfrac{1}{2}\times25$$
$$=\dfrac{25}{2}\ sq.unit$$
Question 4
1 / -0

The unit's place digit in $$(1446)^{4n + 3}$$ is

A
$$n$$

B
$$0$$

C
$$6$$

D
None of these

Solution
Question 5
1 / -0

Number of rectangles in the grid shown which are not squares is?

A
$$160$$

B
$$162$$

C
$$170$$

D
$$185$$

Solution
Question 6
1 / -0

Number of ways in which $$7$$ green bottles and $$8$$ blue bottles can be arranged in a row if exactly $$1$$ pair of green bottles is side by side is (Assume all bottles to be a like except for the colour).

A
$$84$$

B
$$360$$

C
$$504$$

D
None of the above

Solution

Consider the two green bottles as one entity. Let's call it $$GG$$.

We will also refer to the green bottle as: $$G$$, and to the blue bottle as: $$B$$.

So now we have $$14$$ units to arrange ( 1 green pair bottle + the remaining 5 green bottles + the 8 blue bottles):

$$GG, G, G, G, G, G, B, B, B, B, B, B, B, B$$

Now place the 8 blue bottles in a row, such that between each bottle has a gap before and after it:

_B_B_B_B_B_B_B_B_

Now we need to take the six green units (the 1 pair and 5 bottles) and place them in six of the nine gaps.

Therefore, the solution is: $$^9C_6 = 84 ways$$.
Question 7
1 / -0

The last three digits in $$10!$$ are ?

A
$$800$$

B
$$700$$

C
$$500$$

D
$$600$$

Solution

The last three digits in$$ 10!$$ are,
The $$10!$$ is,
$$=10\times 9\times 8\times 7 \times 6\times 5\times 4\times 3\times 2\times 1\\$$
$$=3628800.$$

So, the last three digits of$$ 10!$$ is
$$=3628(800),$$
$$=800$$.
Question 8
1 / -0

The number of zeroes, at the end of $$50!$$, is

A
$$5$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Solution-
50!$$\rightarrow $$ Number of zeroes of the end
Zeroes can be obtained by multiples of 10
as multiples of 5 and 2.
5 zeroes there are many powers of 2 in 50!
we need to find power of 5 in 50!
Power $$ = 1(5)+1(15)+2(25)+1(35)+1(45)+1(50)$$
$$ = 7 $$
7 zeroes caused by 5, 15, 25, 35, 45, 50
Total zeroes = 5+7 = 12
Question 9
1 / -0

Garlands are formed using 6 red roses and 6 yellow roses of different sizes. The number of arrangements in garland which have red roses and yellow roses come alternately is

A
$$5!\times 6!$$

B
$$6! \times 6!$$

C
$$\dfrac{5!}{2!}\times 6!$$

D
$$2(6!\times 6!)$$

Solution
Question 10
1 / -0

The three digits in 10! are:

A
$$800$$

B
$$700$$

C
$$500$$

D
$$600$$

Solution

As we know that $$10!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$$
$$10!=10^2(3\times 4\times 6\times 7\times 8\times 9)$$
$$10!=10^2(3^4\times 2^6\times 7)$$
$$10!=10^2(81\times 64\times 7)$$
The last three digit in $$10!$$ is $$800$$
The correct option is A.

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Re-Attempt Weekly Quiz Competition

Differentiation Test 35

Result

Differentiation Test 35

Score

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Rank

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SHARING IS CARING

Question 1

$$log a-1$$

$$log a-a$$

$$a log a-a$$

$$a log a+a$$

Solution

Question 2

$$98$$

$$99$$

$$100$$

$$101$$

Solution

Question 3

$$\cfrac{23}{2}$$ sq. units

$$\cfrac{25}{2}$$ sq. units

$$12$$ sq.unit

none of these

Solution

Question 4

$$n$$

$$0$$

$$6$$

None of these

Solution

Question 5

$$160$$

$$162$$

$$170$$

$$185$$

Solution

Question 6

$$84$$

$$360$$

$$504$$

None of the above

Solution

Question 7

$$800$$

$$700$$

$$500$$

$$600$$

Solution

Question 8

$$5$$

$$10$$

$$11$$

$$12$$

Solution

Question 9

$$5!\times 6!$$

$$6! \times 6!$$

$$\dfrac{5!}{2!}\times 6!$$

$$2(6!\times 6!)$$

Solution

Question 10

$$800$$

$$700$$

$$500$$

$$600$$

Solution

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