JEE Advance - Mathematics (1998 - No. 45)
If$$\,\,\,$$ $$y = {{a{x^2}} \over {\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}} + {{bx} \over {\left( {x - b} \right)\left( {x - c} \right)}} + {c \over {x - c}} + 1$$,
prove that $${{y'} \over y} = {1 \over x}\left( {{a \over {a - x}} + {b \over {b - x}} + {c \over {c - x}}} \right)$$.
prove that $${{y'} \over y} = {1 \over x}\left( {{a \over {a - x}} + {b \over {b - x}} + {c \over {c - x}}} \right)$$.
Differentiate both sides with respect to x, and simplify.
Multiply both sides by (x-a)(x-b)(x-c) and then differentiate.
Use partial fraction decomposition and then differentiate.
Take the natural logarithm of both sides and then differentiate.
Substitute x = a, b, c into the equation and solve for y.
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