Let f be a real valued function satisfying f ( y x ) = f ( x ) − f ( y ) and x → 0 L t x f ( 1 + x ) = 3 . The area bounded by the curve y = f ( x ) , the y -axis and the line y = 3 is

f 1 ( x ) = h → 0 Lt h f ( x + h ) − f ( x ) = h → 0 Lt x h . x f ( 1 + x h ) = x 3 \

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by Miscellaneous CurvesHard2 min

MathematicsHardsingle choice

Let f be a real valued function satisfying $f (\frac{x}{y}) = f (x) - f (y)$ and $x \to 0 L t \frac{f ( 1 + x )}{x} = 3$ . The area bounded by the curve $y = f (x)$ , the $y$ -axis and the line $y = 3$ is

Options:

Answer:

Solution:

$f^{1} (x) = h \to 0 Lt \frac{f ( x + h ) - f ( x )}{h} = h \to 0 Lt \frac{f ( 1 + \frac{h}{x} )}{\frac{h}{x} . x} = \frac{3}{x}$ \

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by Miscellaneous Curves

⏱ 2m

Doubts & Discussion

Loading discussions...