f ( x ) = s i n x x and g ( x ) = t a n x x where 0 < x ≤ 1 . Then in this interval

f ′ ( x ) = s i n 2 x s i n x − x c o s x Now h ( x ) = sin x − x cos x h ′ ( x ) = x sin x > 0 ∀0 0 for 0 0 Hence f ( x ) is increasing. Similarly it can be done for g ( x ) .

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-Increasing-DecreasingMedium2 minQB

MathematicsMediumsingle choice

$f (x) = \frac{x}{s i n x}$ and $g (x) = \frac{x}{t a n x}$ where $0 < x \leq 1$ . Then in this interval

Options:

Answer:

Solution:

$f^{'} (x) = \frac{s i n x - x c o s x}{s i n ^{2} x}$

Now $h (x) = sin x - x cos x$ $h^{'} (x) = x sin x > 0 \forall0 < x \leq 1$ $h (x)$ is increasing in (0, 1] $h (0) < h (x) \Rightarrow sin x - x cos x > 0 for 0 < x \leq 1 \Rightarrow f^{'} (x) > 0$

Hence $f (x)$ is increasing. Similarly it can be done for $g (x)$ .

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-Increasing-Decreasing

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...