2
To solve this problem, we need to manipulate the given equation to find the value of the expression:
Given:
\(\frac{1}{x} = \frac{1}{p} + \frac{1}{q}\)
First, find a common denominator on the right-hand side:
\(\frac{1}{x} = \frac{q + p}{pq}\)
This implies:
\(x = \frac{pq}{p + q}\)
Substitute \(x\) in the given expression:
\(\frac{pq}{p^2-q^2}\left(\frac{x+p}{x-p}-\frac{x+q}{x-q}\right)\)
This equals:
\(\frac{pq}{p^2-q^2}\left(\frac{\left(\frac{pq}{p+q}\right)+p}{\left(\frac{pq}{p+q}\right)-p}-\frac{\left(\frac{pq}{p+q}\right)+q}{\left(\frac{pq}{p+q}\right)-q}\right)\)
First handle the two fractions separately:
- Let's simplify the first part:
\(\frac{\left(\frac{pq}{p+q}\right)+p}{\left(\frac{pq}{p+q}\right)-p} = \frac{\frac{pq + p(p + q)}{p+q}}{\frac{pq - p(p+q)}{p+q}}\) - Now simplify the second part:
\(\frac{\left(\frac{pq}{p+q}\right)+q}{\left(\frac{pq}{p+q}\right)-q} = \frac{\frac{pq + q(p + q)}{p+q}}{\frac{pq - q(p+q)}{p+q}}\)
Subtract these simplified terms:
\(-\frac{2pq + p^2}{p^2} + \frac{2pq + q^2}{q^2} = -1 +1\)
Finally, plug these back into the main expression:
\(\frac{pq}{p^2-q^2}\times 2 = 2\)
The correct answer is:
Option 2
Important Questions from Algebra
If 2x – y = 2 and xy = \(\frac{3}{2}\) , then what is the value of x 3– \(\frac{{{y^3}}}{8}\) ?
If (10a 3+ 4b 3) : (11a 3- 15b 3) = 7 : 5, then (3a + 5b) : (9a - 2b) =?
- If 4sin 2 θ = 3(1+ cos θ), 0° < θ < 90°, then what is the value of (2tan θ + 4sin θ - sec θ)?
The value of:
\(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)
If (x + y) 3+ 27(x - y) 3= (Ax - 2y)(Bx 2+ Cxy + 13y 2), then the value of A - B - C is:
