(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\).
(b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value of x and y.
(b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value of x and y.
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Correct Answer: Option n
Explanation:
(a) \(t = \sqrt{\frac{pq}{r} - r^{2}q}\)
Squaring both sides,
\(t^{2} = \frac{pq}{r} - r^{2}q\)
\(t^{2} = \frac{pq - r^{3}q}{r}\)
\(t^{2}r = pq - r^{3}q \implies t^{2}r = q(p - r^{3})\)
\(q = \frac{t^{2}r}{p - r^{3}}\).
(b) \(9^{(1 - x)} = 27^{y}\)
\(\implies 3^{2(1 - x)} = 3^{3y}\)
\(2 - 2x = 3y ..... (1)\)
\(x - y = -1\frac{1}{2} ....... (2)\)
From (2), \(x = -1\frac{1}{2} + y\)
\(\therefore 2 - 2(-1\frac{1}{2} + y) = 3y\)
\(2 + 3 - 2y = 3y \implies 5 = 3y + 2y = 5y\)
\(\implies y = 1\)
\(x = -1\frac{1}{2} + 1 = -\frac{1}{2}\)
\(\therefore x + y = -\frac{1}{2} + 1 = \frac{1}{2}\)
(a) \(t = \sqrt{\frac{pq}{r} - r^{2}q}\)
Squaring both sides,
\(t^{2} = \frac{pq}{r} - r^{2}q\)
\(t^{2} = \frac{pq - r^{3}q}{r}\)
\(t^{2}r = pq - r^{3}q \implies t^{2}r = q(p - r^{3})\)
\(q = \frac{t^{2}r}{p - r^{3}}\).
(b) \(9^{(1 - x)} = 27^{y}\)
\(\implies 3^{2(1 - x)} = 3^{3y}\)
\(2 - 2x = 3y ..... (1)\)
\(x - y = -1\frac{1}{2} ....... (2)\)
From (2), \(x = -1\frac{1}{2} + y\)
\(\therefore 2 - 2(-1\frac{1}{2} + y) = 3y\)
\(2 + 3 - 2y = 3y \implies 5 = 3y + 2y = 5y\)
\(\implies y = 1\)
\(x = -1\frac{1}{2} + 1 = -\frac{1}{2}\)
\(\therefore x + y = -\frac{1}{2} + 1 = \frac{1}{2}\)