 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2002

If the sum of the coefficients in the expansion of $$\,{\left( {a + b} \right)^n}$$ is 4096, then the greatest coefficient in the expansion is
A
1594
B
792
C
924
D
2924

## Explanation

We know, $$\,{\left( {a + b} \right)^n}$$ = $${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$$

Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable.

So put $$a$$ = b = 1

$$\therefore$$ 2n = $${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$$

According to question, 2n = 4096 = 212

$$\Rightarrow n = 12$$

So $$\,{\left( {a + b} \right)^n}$$ = $$\,{\left( {a + b} \right)^{12}}$$

Here n = 12 is even so formula for greatest term is
$${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$$

For n = 12, greatest term $${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$$

$$\therefore$$ Coefficient of the greatest term = $${}^{12}{C_6}$$ = $${{12!} \over {6!6!}}$$ = 924
2

### AIEEE 2002

The positive integer just greater than $${\left( {1 + 0.0001} \right)^{10000}}$$ is
A
4
B
5
C
2
D
3

## Explanation

$${\left( {1 + 0.0001} \right)^{10000}}$$ = $${\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}$$

= 1 + 10000$${ \times {1 \over {{{10}^4}}}}$$ + $${{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}$$+......$$\infty$$

< 1 + 1 + $${1 \over {2!}}$$ + $${1 \over {3!}}$$ + ...... $$\infty$$ = e = 2.71828 < 3
3

### AIEEE 2002

$$r$$ and $$n$$ are positive integers $$\,r > 1,\,n > 2$$ and coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ term and $$3{r^{th}}$$ term in the expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal, then $$n$$ equals
A
$$3r$$
B
$$3r + 1$$
C
$$2r$$
D
$$2r + 1$$

## Explanation

$$\,{\left( {r + 2} \right)^{th}}$$ term = $${}^{2n}{C_{r+1}}{\left( x \right)^r}$$

And coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ = $${}^{2n}{C_{r+1}}$$

$$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$$

And coefficient of $$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}$$

According to the question,
$${}^{2n}{C_{r+1}}$$ = $${}^{2n}{C_{3r - 1}}$$

$$\Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n$$

[As if $${}^n{C_p} = {}^n{C_q}$$ then p + q = n]

$$\Rightarrow 4r = 2n$$

$$\Rightarrow n = 2r$$
4

### AIEEE 2002

The coefficients of $${x^p}$$ and $${x^q}$$ in the expansion of $${\left( {1 + x} \right)^{p + q}}$$ are
A
equal
B
equal with opposite signs
C
reciprocals of each other
D
none of these

## Explanation

Here in this expansion $${\left( {1 + x} \right)^{p + q}}$$

The general term = $${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$$

$$\therefore$$ $${x^p}$$ will be present in the term = $${}^{p + q}{C_p}.{\left( x \right)^p}$$

So coefficient of $${x^p}$$ = $${}^{p + q}{C_p}$$

And $${x^q}$$ will be present in the term = $${}^{p + q}{C_q}.{\left( x \right)^q}$$

$$\therefore$$ coefficient of $${x^q}$$ = $${}^{p + q}{C_q}$$

We know $${}^n{C_r}$$ = $${}^n{C_{n - r}}$$

$$\therefore$$ $${}^{p + q}{C_q}$$ = $${}^{p + q}{C_{\left( {p + q} \right) - q}}$$ = $${}^{p + q}{C_p}$$

So coefficients of $${x^p}$$ and $${x^q}$$ are equal.

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