Differentiation Test 40

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Differentiation Test 40

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$^{47}C_{4}+\displaystyle \sum _{ j=1 }^{ 5 }\ ^{ \left( 52-j \right) } { C }_{ 3 }$$ is

A
$$^{47}C_{5}$$

B
$$^{52}C_{5}$$

C
$$^{52}C_{4}$$

D
$$^{52}C_{3}$$

Solution

We have,
$$\begin{array}{l} ^{ 47 }{ C_{ 4 } }{ +^{ 51 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 47 } }{ C_{ 3 } } \\ { =^{ n } }{ C_{ r } }{ +^{ n } }{ C_{ r-1 } }{ =^{ n+1 } }{ C_{ r } } \\ { =^{ 48 } }{ C_{ 4 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \\ { =^{ 49 } }{ C_{ 4 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \end{array}$$
Similarly,
$$=^{52}{C_4}$$.
Question 2
1 / -0

Find the missing number.

A
$$74$$

B
$$62$$

C
$$91$$

D
$$97$$

Solution

$$ \textbf{ Step 1: Observing the difference between the adjacent numbers}$$

$$ \text{If we notice the adjacent numbers, we see } 4\times 2 -1 =7$$

$$ \text{Similarly, } 7 \times 2-1=13$$

$$ \text{Similarly, } 13\times2 -1=25$$

$$ \text{Similarly, } 25\times2-1=49$$

$$ \textbf{Step 2: Calculating the missing number}$$

$$ \text{Thus, looking at the above observations, we can conclude the missing number will be:}$$

$$ \Rightarrow 49\times 2-1=97$$

$$\textbf{Thus, the missing number is D 97}$$
Question 3
1 / -0

A committee of $$4$$ persons is to be formed from $$2$$ ladies, $$2$$ old men and $$4$$ young men such that it includes at least $$1$$ lady. at least $$1$$ old man and at most $$2$$ young men. Then the total number of ways in which this committee can be formed is :

A
$$40$$

B
$$41$$

C
$$16$$

D
$$32$$

Solution
Question 4
1 / -0

Find $$x$$, if $$\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}$$.

A
$$5$$

B
$$4$$

C
$$30$$

D
$$None$$

Solution
Question 5
1 / -0

solve

A
$$225$$

B
$$175$$

C
$$335$$

D
$$350$$

Solution

Pattens is as follows:-

$$ (20 - 9)^{2} = 121 , $$ $$(24 - 11)^{2} = 169 $$

So $$ (28 - 13)^{2} = 15^{2} = 225$$

Hence missing place = 225
Question 6
1 / -0

There is a certain relationship between the pair of figure on either side of ::. Identify the relationship on the left side and find the missing figure.

A

B

C

D

Solution

As per the first relation, in the first figure, there is a square on which there is 3 circle and in the second figure, there is a circle on which there is $$3$$ square.

Following the same in second relation, the correct option is $$B$$.
Question 7
1 / -0

Which number replaces the question mark ?

A
10

B
12

C
14

D
16

Solution

In each triangle, the central value equals the average of the 3 values around the outside.

Hence, the required value = $$\dfrac{15+17+4}{3}=12$$

Therefore, option B is correct.
Question 8
1 / -0

Rahul told Anand, "Yesterday I defeated the only brother of the daughter of my grandmother." Whom did Rahul defeat ?

A
Son

B
Father

C
Brother

D
Cousin

Solution

Daughter of Grandmother $$\rightarrow$$ Aunt Aunt's only brother $$\rightarrow$$ Father
Question 9
1 / -0

Find $$f^{\prime} (3) $$ if $$f(x)=x^3+5x^2-3x+5$$

A
$$28$$

B
$$54$$

C
$$32$$

D
None

Solution

$$f(x)=x^3+5x^2-3x+5\\f^{\prime}(x)=3x^2+10x-3\\f^{\prime }(x)=3(3)^2+10(3)-3=54$$
Question 10
1 / -0

If $$ y = \sin ^ { - 1 } x$$ then $$ \frac { d y } { d x } $$ is equal to

A
$$\sec y$$

B
$$\cos x$$

C
$$\tan x$$

D
$$1$$

Solution

$$y=\sin ^{-1}x\\x=\sin y \\\dfrac{dx}{dx}=\dfrac{d}{dx}\sin y\\1=\cos y \dfrac{dy}{dx}\\\dfrac{dy}{dx}=\sec y$$