The total number of distinct x ∈ 0 , 1 for which, ∫ 0 x t 2 1 + t 4 d t = 2 x - 1 is

Let f x = ∫ 0 x t 2 1 + t 4 d t - 2 x + 1 ⇒ f ′ x = x 2 1 + x 4 - 2 As 1 + x 4 x 2 ≥ 2 ⇒ x 2 1 + x 4 ≤ 1 2 ⇒ f ′ x ≤ - 3 2 ⇒ f x is continuous and decreasing f 0 = 1 and f 1 = ∫ 0 1 t 2 1 + t 4 d t - 1 ≤ -1 2 By intermediate value theorem (IVT), f x = 0 possesses exactly one solution in [0, 1]. As f ( 0 ) f ( 1 ) < 0

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MathematicsApplication of DerivativeMonotonicity-Increasing-DecreasingHard2 minPYQ_2016

MathematicsHardnumerical

The total number of distinct $x \in [0, 1]$ for which, $\int_{0}^{x} \frac{t^{2}}{1 + t^{4}} d t = 2 x - 1$ is

Answer:

1.00

Solution:

Let $f (x) = (\int_{0}^{x} \frac{t^{2}}{1 + t^{4}} d t) - 2 x + 1$
$\Rightarrow$ $f^{'} (x) = \frac{x^{2}}{1 + x^{4}} - 2$
As $\frac{1 + x^{4}}{x^{2}} \geq 2$ $\Rightarrow \frac{x^{2}}{1 + x^{4}} \leq \frac{1}{2}$
$\Rightarrow$ $f^{'} (x) \leq - \frac{3}{2}$
$\Rightarrow$ $f (x)$ is continuous and decreasing
$f (0) = 1$ and $f (1) = (\int_{0}^{1} \frac{t^{2}}{1 + t^{4}} d t) - 1 \leq \frac{-1}{2}$
By intermediate value theorem (IVT), $f (x) = 0$ possesses exactly one solution in [0, 1]. As $f (0) f (1) < 0$

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

⏱ 2mℹ️ Source: PYQ_2016

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