1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

The coefficient of t4 in the expansion of $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$ is :
A
14
B
15
C
10
D
12

Explanation

$${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$

= (1 $$-$$ t6)3 (1 $$-$$ t)$$-$$3

= (1 $$-$$ 3C1t6 + 3C2t12 $$-$$ 3C3t18) $$ \times $$ (1 $$-$$ t)$$-$$3

coefficient of t4 is 1 $$ \times $$ coefficient of t4 in (1 $$-$$ t)$$-$$3

= 1 $$ \times $$ 3+4$$-$$1C4 (By multinomial theorem)

= 6C4 = 15
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

If  $${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$  then k is equal to
A
100
B
200
C
50
D
400

Explanation

$${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$$

$$ \Rightarrow 100 = k$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

If the third term in the binomial expansion
of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -
A
$$2\sqrt 2 $$
B
$$4\sqrt 2 $$
C
$${1 \over 8}$$
D
$${1 \over 4}$$

Explanation

$${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$

$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$

$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$

$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$

$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$

$$ \Rightarrow 2{({\log _2}x)^2} = 8$$

$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$

$$ \Rightarrow \,\,{\log _2}x = 2$$  or  $$-$$ 2

$$x = 4$$   or  $${1 \over 4}$$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

The positive value of $$\lambda $$ for which the co-efficient of x2 in the expression x2 $${\left( {\sqrt x + {\lambda \over {{x^2}}}} \right)^{10}}$$ is 720, is -
A
4
B
$$2\sqrt 2 $$
C
3
D
$$\sqrt 5 $$

Explanation

$${x^2}\left( {{}^{10}{C_r}{{\left( {\sqrt x } \right)}^{10 - r}}{{\left( {{\lambda \over {{x^2}}}} \right)}^r}} \right)$$

$${x^2}\left[ {{}^{10}{C_r}{{\left( x \right)}^{{{10 - r} \over 2}}}{{\left( \lambda \right)}^r}{{\left( x \right)}^{ - 2r}}} \right]$$

$${x^2}\left[ {{}^{10}{C_r}{\lambda ^r}{x^{{{10 - r} \over 2}}}} \right]$$

$$ \therefore $$  r = 2

Hence, $${}^{10}{C_2}{\lambda ^2} = 720$$

$${\lambda ^2} = 16$$

$$\lambda = \pm 4$$

Questions Asked from Mathematical Induction and Binomial Theorem

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