1

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)
A value of $$b$$ for which the equations $$$\matrix{ {{x^2} + bx - 1 = 0} \cr {{x^2} + x + b = 0} \cr } $$$

have one root in common is

A
$$ - \sqrt 2 $$
B
$$ - i\sqrt 2 $$
C
$$i\sqrt 5 $$
D
$$\sqrt 2 $$
2

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)
Let $$\left( {{x_0},{y_0}} \right)$$ be the solution of the following equations
$$\matrix{ {{{\left( {2x} \right)}^{\ell n2}}\, = {{\left( {3y} \right)}^{\ell n3}}} \cr {{3^{\ell nx}}\, = {2^{\ell ny}}} \cr } $$
Then $${x_0}$$ is
A
$${1 \over 6}$$
B
$${1 \over 3}$$
C
$${1 \over 2}$$
D
$$6$$
3

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)
Let $$\alpha $$ and $$\beta $$ be the roots of $${x^2} - 6x - 2 = 0,$$ with $$\alpha > \beta .$$ If $${a_n} = {\alpha ^n} - {\beta ^n}$$ for $$\,n \ge 1$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is
A
1
B
2
C
3
D
4
4

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)
Let $$p$$ and $$q$$ be real numbers such that $$p \ne 0,\,{p^3} \ne q$$ and $${p^3} \ne - q.$$ If $${p^3} \ne - q.$$ and $$\,\beta $$ are nonzero complex numbers satisfying $$\alpha \, + \beta = - p\,$$ and $${\alpha ^3} + {\beta ^3} = q,$$ then a quadratic equation having $${\alpha \over \beta }$$ and $${\beta \over \alpha }$$ as its roots is
A
$$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} + 2q} \right)x + \left( {{p^3} + q} \right) = 0$$
B
$$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} - 2q} \right)x + \left( {{p^3} + q} \right) = 0$$
C
$$\left( {{p^3} - q} \right){x^2} - \left( {5{p^3} - 2q} \right)x + \left( {{p^3} - q} \right) = 0$$
D
$$\left( {{p^3} - q} \right){x^2} - \left( {5{p^3} + 2q} \right)x + \left( {{p^3} - q} \right) = 0$$

Explanation

Sum of roots = $${{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}$$ and product = 1

Given, $$\alpha$$ + $$\beta$$ = $$-$$ p and $$\alpha$$3 + $$\beta$$3 = q

$$ \Rightarrow (\alpha + \beta )({\alpha ^2} - \alpha \beta + {\beta ^2}) = q$$

$$\therefore$$ $${\alpha ^2} + {\beta ^2} - \alpha \beta = {{ - q} \over p}$$ ..... (i)

and $${(\alpha + \beta )^2} = {p^2}$$

$$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \beta = {p^2}$$ ..... (ii)

From Eq. (i) and (ii), we get

$${\alpha ^2} + {\beta ^2} = {{{p^3} - 2q} \over {3p}}$$

and $$\alpha \beta = {{{p^3} + q} \over {3p}}$$

$$\therefore$$ Required equation is

$${x^2} - {{({p^2} - 2q)x} \over {({p^3} + q)}} + 1 = 0$$

$$ \Rightarrow ({p^3} + q){x^2} - ({p^3} - 2q)x + ({p^3} + q) = 0$$

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