JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 18)
If the constant term in the binomial expansion
of
$${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$ is 405, then |k| equals :
$${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$ is 405, then |k| equals :
3
9
1
2
Explanation
$${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$$
rth term of the expansion,
Tr+1 = 10Cr$${\left( {\sqrt x } \right)^{10 - r}}{\left( {{{ - k} \over {{x^2}}}} \right)^r}$$
= 10Cr.$${x^{{{10 - r} \over 2}}}.{\left( { - k} \right)^r}.{x^{ - 2r}}$$
= 10Cr.$${x^{{{10 - 5r} \over 2}}}.{\left( { - k} \right)^r}$$
If it is constant term then
$${{{10 - 5r} \over 2}}$$ = 0
$$ \Rightarrow $$ r = 2
T3 = 10C2.(-k)2 = 405
$$ \Rightarrow $$ k2 = $${{405} \over {45}}$$ = 9
$$ \Rightarrow $$ k = $$ \pm $$ 3
$$ \Rightarrow $$ |k| = 3
rth term of the expansion,
Tr+1 = 10Cr$${\left( {\sqrt x } \right)^{10 - r}}{\left( {{{ - k} \over {{x^2}}}} \right)^r}$$
= 10Cr.$${x^{{{10 - r} \over 2}}}.{\left( { - k} \right)^r}.{x^{ - 2r}}$$
= 10Cr.$${x^{{{10 - 5r} \over 2}}}.{\left( { - k} \right)^r}$$
If it is constant term then
$${{{10 - 5r} \over 2}}$$ = 0
$$ \Rightarrow $$ r = 2
T3 = 10C2.(-k)2 = 405
$$ \Rightarrow $$ k2 = $${{405} \over {45}}$$ = 9
$$ \Rightarrow $$ k = $$ \pm $$ 3
$$ \Rightarrow $$ |k| = 3
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