Differentiation Test 36

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Differentiation Test 36

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Question 1
1 / -0

A question paper on mathematics consists of twelve questions divided into three parts A, B and C, each containing four questions. In how many ways can an examinee answer five questions, selecting atleast one from each part.

A
$$624$$

B
$$208$$

C
$$2304$$

D
None

Solution

Total no. of ways
$$A$$ $$B$$ $$ C$$ $$A$$ $$B$$ $$ C$$
$$=(^4C_1\times \ ^4C_1 \times \ ^4C_3)\times 3+ (^4C_1\, ^4C_2\, ^4C_2)\times 3$$ ($$\therefore$$ multiple by 3 as for each A,B,C different combination)
$$=192+432$$
$$=624$$
Question 2
1 / -0

Area of the triangle formed by $$(x_{1,}y_{1}),(x_{2},y_{2}), (3y_{2},(-2y_{1}))$$

A
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

B
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

C
$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

D
$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

We have,

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right) $$

We know that the area of triangle is

$$ Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}+2{{y}_{1}} \right)+{{x}_{2}}\left( -2{{y}_{1}}-{{y}_{1}} \right)+3{{y}_{2}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$
Hence, the is the answer
Question 3
1 / -0

If $$f(x)=1-x+x^2-x^3+....-x^{15}+x^{16}+x^{17}$$, then the coefficient of $$x^2$$ in $$f(x-1)$$ is?

A
$$826$$

B
$$816$$

C
$$822$$

D
None of these

Solution

$$f\left( x \right) = 1 - \left( {x - 1} \right) + {\left( {x - 1} \right)^2} - {\left( {x - 1} \right)^3} + {\left( {x - 1} \right)^4}.... - {\left( {x - 1} \right)^{15}} + {\left( {x - 1} \right)^{16}} + {\left( {x - 1} \right)^{17}}$$
The coefficient of $${x^2}$$ by using binomial expansion
$$^2{c_0}{x^2}{ + ^3}{c_1}{x^2}{ + ^4}{c_2}{x^2}{ + ^5}{c_3}{x^2}{ + ^6}{c_4}{x^2}{ + ^7}{c_5}{x^2}{ + ^8}{c_6}{x^2}{ + ^9}{c_7}{x^2}$$
$${ + ^{10}}{c_8}{x^2}{ - ^{11}}{c_9}{x^2}{ + ^{12}}{c_{10}}{x^2}{ - ^{13}}{c_{11}}{x^2}{ + ^{14}}{c_{12}}{x^2}{ - ^{15}}{c_{13}}{x^2}{ + ^{16}}{c_{14}}{x^2}{ - ^{17}}{c_{15}}{x^2}$$
$$ \Rightarrow {x^2}\left( {1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 + 91 + 105 + 120 + 136} \right)$$
= 372 + 444 = 816
Question 4
1 / -0

Replace the question mark (?) with the correct option.

A
$$2$$

B
$$7$$

C
$$4$$

D
$$9$$

Solution
Question 5
1 / -0

A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the element of P.A subset Q of A is again chosen. The number of way of choosing P and Q so that P Q =$$\phi $$ is :-

A
$$2^{2n}-^{2n}C_{n}$$

B
$$2^{n}$$

C
$$2^{n}-1$$

D
$$3^{n}$$

Solution

Let A = $${a_{1},a_{2},a_{3},....a_{n}}$$. For $$a_{1}$$ $$\epsilon $$ For $$a_{1} \epsilon $$A we have following choices:
(i) $$a_{1} \epsilon $$ P and $$a_{1} \epsilon $$ Q
(ii) $$a_{1} \epsilon $$P $$\notin $$
(iii) $$_{1}\notin P and a_{1}\epsilon Q$$
(Iv)$$_{1}\notin P and a_{1}\notin Q$$
Out of these only (Ii), and (iii) and (iv) imply $$a_{1}\notin P\cap Q $$ therefore, the number of ways in which none of $$a_{1},a_{2}.....a_{n}$$ belong $$P\cap Q$$ is $$3^{n}$$.
Question 6
1 / -0

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are $$(2, 1)$$, $$(-2, 3)$$, $$(4, -3)$$.

A
$$1.5$$ sq. units

B
$$3$$ sq. units

C
$$6$$ sq. units

D
$$20$$ sq. units
Question 7
1 / -0

Find: $$6,25,62,123,(?),341$$

A
$$216$$

B
$$214$$

C
$$215$$

D
$$218$$

Solution

$$6=2^{3}-2=6$$
$$25=3^{3}-2=25$$
$$62=4^{3}-2=62$$
$$123=5^{3}-2=123$$
$$214=6^{3}-2=214$$
$$341=7^3-2=341$$

Missing term is $$214$$
Question 8
1 / -0

If $$y=|\cos x|+|\sin x|$$, then $$\dfrac {dy}{dx}$$ at $$x=\dfrac {2\pi}{3}$$ is

A
$$\dfrac {1-\sqrt {3}}{2}$$

B
$$0$$

C
$$\dfrac {\sqrt {3}-1}{2}$$

D
$$\dfrac {\sqrt {3}+1}{2}$$

Solution

$$ y = |cos\,x|+|sin\,x| $$ $$ , x = \dfrac{2n}{3} $$
$$ y = -cos\,x+sin\,x $$ $$ \left \{ x\sum (\dfrac{\pi }{2},\pi ) \right \} $$
$$ y^{1} = sinx+cosx $$
$$ y^{1} = sin(\dfrac{2n}{3})+cos(\dfrac{2n}{3}) $$
$$ = \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} = \dfrac{\sqrt{3}-1}{2} $$ Option C is correct
Question 9
1 / -0

Select the missing number from the given responses ?

A
$$15$$

B
$$10$$

C
$$12$$

D
$$17$$

Solution
Question 10
1 / -0

Number of five-digit numbers divisible by 5 that can be formed from the digits $$0,1, 2, 3, 4, 5$$ without repetition of digits are

A
$$240$$

B
$$360$$

C
$$148$$

D
$$216$$

Solution