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
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
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
MCQ (Single Correct Answer)
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
[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]
= $$ - {3 \over 8}{x^2}$$
4
AIEEE 2005
MCQ (Single Correct Answer)
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.