Differentiation Test 32

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Differentiation Test 32

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Question 1
1 / -0

The number of ways in which we can select 5 letters of the word INTERNATIONAL is equal to

A
$$200$$

B
$$220$$

C
$$242$$

D
$$256$$

Solution

We have thirteen letters.
$$2l, 3N, 2T, 2A$$ and $$(E, O, L,R)8$$ (types)
We can select 5(five) letters in the following manner:

1. All different $${^8C}_5 = \dfrac{8.7.5}{1.2.3} = 56$$ ways

2. 2 alike, 3 different $${^4C}_1 . {^7C}_3 = 435 = 140$$ ways (we have 4 sets of alike letters)

3. 3 alike, 2 different $${^1C}_1 . {^7C}_2 = 1.21 = 21$$ ways (we have only one set of 3 alike)

4. 3 alike and 2 alike $${^1C}_1 . {^3C}_1 = 1.3 = 3$$ ways

5.Two sets of alike and one different
$${^4C}_2 . {^6C}_1 = 6.6 = 36$$ ways

$$\therefore$$ Total number of selection is
$$56 + 140 + 21 + 3 + 36 = 256$$ ways $$\Longrightarrow$$ (d)
Question 2
1 / -0

$$\frac{d^n}{dx^n}[log(ax+b)]$$ is equal to:

A
$$\frac{(-1)^n n! a^n}{(ax+b)^{n+1}}$$

B
$$\frac{(-1)^{n-1} {(n-1)}! a^n}{(ax+b)^{n+1}}$$

C
$$\frac{(-1)^{n+1} ({n+1)}! a^{n-1}}{(ax+b)^{n+1}}$$

D
$$\frac{(-1)^{n-1} {(n-1)}! a^n}{(ax+b)^n}$$

Solution

Let $$y=\log { \left( ax+b \right) } $$, let $$D=\cfrac { d }{ dx } $$
$$\cfrac { dy }{ dx } =Dy=\cfrac { a }{ ax+b } $$
$${ D }^{ 2 }y=\cfrac { -{ a }^{ 2 } }{ { \left( ax+b \right) }^{ 2 } } ,{ D }^{ 3 }y=\cfrac { 2{ a }^{ 3 } }{ { \left( ax+b \right) }^{ 3 } } $$
$${ D }^{ 4 }y=\cfrac { -3.2.1.{ a }^{ 4 } }{ { \left( ax+b \right) }^{ 4 } } $$
$$\therefore { D }^{ n }y=\cfrac { { \left( -1 \right) }^{ n-1 }\left( n-1 \right) !{ a }^{ n } }{ { \left( ax+b \right) }^{ n } } $$
$$\therefore \cfrac { { d }^{ n } }{ d{ x }^{ n } } \left( \log { \left( ax+b \right) } \right) =\cfrac { { \left( -1 \right) }^{ n-1 }\left( n-1 \right) !{ a }^{ n } }{ { \left( ax+b \right) }^{ n } } $$
Question 3
1 / -0

The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is

A
$$75 \ sq. units$$

B
$$37.5 \ sq. units$$

C
$$45 \ sq. units$$

D
$$22.5 \ sq. units$$

Solution

Let $$A(3,8), B(-4,2),C(5,-1)$$ be the vertices of the given $$\triangle ABC$$.

Then,

$$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$$

Area of $$\triangle ABC$$ = $$\dfrac{1}{2}|[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$$

$$=\dfrac{1}{2}|3[2-(-1)]-4(-1-8)+5(8-2)|$$

$$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$$
Question 4
1 / -0

Statement I: The function f(x) in the figure is differentiable at x = a
Statement II: The function f(x) continuous at x = a

A
Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I.

B
Both Statement I and Statement II are true and the Statement II is not the correct explanation of the Statement I.

C
Statement I is true but Statement II is false

D
Statement I is false but Statement II is true.

Solution

$$f(x)$$ is continuous at as it can be observed from the figure there is no discontinuous at $$x=a$$ but coming to differentiability $$f(x)$$ is not differentiable at $$x=a$$ since the curve is sharp[i.e the left slope and right slope are not equal]
therefore the answer is statement I is false but statement II is true
Question 5
1 / -0

The total number of non-negative integer n satisfying the
equation $$ {n^2} = p + q\,and\,{n^3} = {p^2} + {q^2}$$, where p and q are
integers, is

A
$$1$$

B
$$2$$

C
$$3$$

D
infinite

Solution
Question 6
1 / -0

Given $$y=\dfrac {3}{x}, \dfrac {dy}{dx}=$$

A
$$3$$

B
$$\dfrac {3}{x^{2}}$$

C
$$\dfrac {-3}{x^{2}}$$

D
$$3x$$

Solution

$$y=\dfrac{3}{x}$$
differentiating with respect to x
$$\dfrac{dy}{dx}=\dfrac{-3}{x^2}$$
Question 7
1 / -0

If $$ f(x) = \bigg[ \frac {a+x}{1+x} \bigg]^{a+1+2x} $$ then $$ {a^{a+1}} \bigg [ 2 \ log \ a + {\frac {1-a ^2}{a}} \bigg]$$ is

A
$$ f^\prime (1) $$

B
$$ f^\prime (0) $$

C
$$ f^\prime (2) $$

D
$$ f^{\prime \prime} $$

Solution
Question 8
1 / -0

Pam likes the numbers 1689 and 6891. Knowing this, which pair of numbers will she like among the ones below?

A
1981 and 1891

B
19 and 91

C
190 and 160

D
1198911 and 1168611

Solution
Question 9
1 / -0

If $$(1+3+5+....+p)+(1+3+5+....+q)=(1+3+5+.....+r),$$ where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of $$p+q+r$$, (where $$P>6$$ ) is :

A
$$12$$

B
$$21$$

C
$$27$$

D
$$24$$

Solution

$$ 1+3+5+...+2n-1 = n^{2} $$
no. of terms $$ = \dfrac{n+1}{2} $$
According to question we can write
$$\left (\dfrac{p+1}{2}\right)^{2}+\left(\dfrac{q+1}{2}\right)^{2}\pm \left(\dfrac{r+1}{2}\right)^{2} $$
It forms Pythagoras triplet Thus for min value
take 3, 4 and 5 as our Pythagoras triplet
$$ \therefore \dfrac{p+1}{2} = 4 $$ $$ \dfrac{q+1}{2} = 3 $$ $$ \dfrac{r+1}{2} = 5 $$
$$ p = 7 $$ $$ q = 5 $$ $$ r = 9 $$
$$ p + q + r = 21 $$
Question 10
1 / -0

The total number of different combinations of one
or more letters which can be made from the letter of the word MISSISSIPPI is,

A
$$150$$

B
$$148$$

C
$$149$$

D
$$146$$

Solution

We have 1M, 4I ,4S , 2P
Therefore total number of selection of one or more letters=(1+1)(4+1)(4+1)(2+1)-1=149