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यदि $p$ तथा $q$ दो वास्तविक संख्याऐं इस प्रकार है, कि $p + q =3$ तथा $p ^4+ q ^4=369$ है, तो $\left(\frac{1}{ p }+\frac{1}{ q }\right)^{-2}$ का मान होगा-
$2$
$1$
$4$
$5$
Solution
$p + q =3 \quad p ^{4}+ q ^{4}=369$
$\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$
$(p+q)^{2}=9$
$p ^{2}+ q ^{2}=9-2 pq$
$\frac{1}{\left(\frac{1}{p}+\frac{1}{q}\right)^{2}}=\frac{(q p)^{2}}{(q+p)^{2}}=\frac{(q p)^{2}}{9}$
$p ^{4}+ q ^{4}=\left( p ^{2}+ q ^{2}\right)^{2}-2 p ^{2} q ^{2}$
$369=(9-2 p q)^{2}-2(p q)^{2}$
$369=81+4 p ^{2} q ^{2}-36 pq -2 p ^{2} q ^{2}$
$288=2 p ^{2} q ^{2}-36 pq$
$144= p ^{2} q ^{2}-18 pq$
$( pq )^{2}-2 \times 9 \times pq +9^{2}=144+9^{2}$
$(p q-9)^{2}=225$
$p q-9=\pm 15$
$p q=\pm 15+9$
$p q=24,-6$
($24$ is rejected because $p ^{2}+ q ^{2}=9-2 pq$ is negative)
$\frac{(q p)^{2}}{9}=\frac{16 \times 16}{9}=4$
