 ### JEE Mains Previous Years Questions with Solutions

4.5     (100k+ )
1

### AIEEE 2006

If the expansion in powers of $$x$$ of the function $${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$ is $${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....$$ then $${a_n}$$ is
A
$${{{b^n} - {a^n}} \over {b - a}}$$
B
$${{{a^n} - {b^n}} \over {b - a}}$$
C
$${{{a^{n + 1}} - {b^{n + 1}}} \over {b - a}}$$
D
$${{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}$$

## Explanation

$${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$

= $${\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}$$

= $$\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$$ -
$$\left[ {1 + \left( { - 1} \right)\left( { - bx} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - bx} \right)}^2} + ...} \right]$$

= $$\left[ {1 + ax + {a^2}{x^2} + ... + {a^{n - 1}}{x^{n - 1}} + {a^n}{x^n}}+.... \right]$$ -
$$\left[ {1 + bx + {b^2}{x^2} + ... + {b^{n - 1}}{x^{n - 1}} + {b^n}{x^n}}+.... \right]$$

Coefficient of xn =

$${a^n} + {a^{n - 1}}b + {a^{n - 2}}{b^2} + .... + {b^n}$$

= $${a^n}\left[ {1 + {b \over a} + {{{b^2}} \over {{a^2}}} + ..... + {{{b^n}} \over {{a^n}}}} \right]$$

= $${a^n}\left[ {{{{{\left( {{b \over a}} \right)}^{n + 1}} - 1} \over {{b \over a} - 1}}} \right]$$

= $${a^n}\left[ {{{{b^{n + 1}} - {a^{n + 1}}} \over {{a^{n + 1}}\left( {{{b - a} \over a}} \right)}}} \right]$$

= $${{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}}$$
2

### AIEEE 2005

If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation
A
$${m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0$$
B
$${m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0$$
C
$${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$
D
$${m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0$$

## Explanation

Let r = 2

$$\therefore$$ 2nd, 3rd and 4th terms are in AP.

2nd term = T2 = $${}^m{C_1}.y$$

Coefficient of T2 = $${}^m{C_1}$$

3rd term = T3 = $${}^m{C_2}.{y^2}$$

Coefficient of T3 = $${}^m{C_2}$$

4th term = T4 = $${}^m{C_3}.{y^3}$$

Coefficient of T2 = $${}^m{C_3}$$

$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$

$$\Rightarrow$$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$

$$\Rightarrow$$ 6m2 - 6m = 6m +m(m2 - 3m + 2)

$$\Rightarrow$$ 6m2 - 6m = 6m + m3 - 3m2 + 2m

$$\Rightarrow$$ 6m - 6 = 6 + m2 - 3m + 2

$$\Rightarrow$$ m2 - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get

m2 - 9m + 14 = 0. So Option C is correct.
3

### AIEEE 2005

If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as
A
$$1 - {3 \over 8}{x^2}$$
B
$$3x + {3 \over 8}{x^2}$$
C
$$- {3 \over 8}{x^2}$$
D
$${x \over 2} - {3 \over 8}{x^2}$$

## Explanation

$${\left( {1 + x} \right)^{{3 \over 2}}}$$ = 1 + $${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$$

= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)

$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$

= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$

= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$

= $$- {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$

[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]

= $$- {3 \over 8}{x^2}$$
4

### AIEEE 2005

If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation
A
$$a - b = 1$$
B
$$a + b = 1$$
C
$${a \over b} = 1$$
D
$$ab = 1$$

## Explanation

General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is Tr+1.

Tr+1 = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$

For the coefficient of x7,

$$\Rightarrow$$ 22 - 3r = 7

$$\Rightarrow$$ r = 5

So coefficient of x7 = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)

Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is Tr+1.

Tr+1 = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$

For the coefficient of x-7,

11 - 3r = -7

$$\Rightarrow$$ r = 6

$$\therefore$$ Coefficient of x-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

According to question,

Coefficient of x7 = Coefficient of x-7

$$\Rightarrow$$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

$$\Rightarrow$$ $$ab = 1$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12