JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Application of Derivatives

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

Answer: 1

Solution

Let

f (x) = \int_{0}^{x} \frac{t ^{2}}{1 + t ^{4}} d t - 2 x + 1

.

Then

f^{'} (x) = \frac{x ^{2}}{1 + x ^{4}} - 2 < 0

for all

x \in [0, 1]

(since

\frac{x ^{2}}{1 + x ^{4}} < \frac{x ^{2}}{x ^{4}} = \frac{1}{x ^{2}} \leq 1 < 2

for

x \in (0, 1]

).

So

f

is strictly decreasing on

[0, 1]

f (0) = 0 - 0 + 1 = 1 > 0

.

For

x \in [0, 1]

\frac{t ^{2}}{1 + t ^{4}} < \frac{1}{2}

, so

\int_{0}^{1} \frac{t ^{2}}{1 + t ^{4}} d t < \frac{1}{2}

.

Thus

f (1) = \int_{0}^{1} \frac{t ^{2}}{1 + t ^{4}} d t - 2 + 1 < \frac{1}{2} - 1 = - \frac{1}{2} < 0

.

By the Intermediate Value Theorem,

f (x) = 0

has exactly 1 solution in

[0, 1]

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer