JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f(x)

be a real valued function defined for all

x \geq 1

, satisfying

f(1)=1

and

f^{\prime}(x)=\frac{1}{x^2+(f(x))}

; then

\lim \limits_{{x \rightarrow \infty}} f(x)

View answer and explanation

Correct option: C

exists and less than

1+\frac{\pi}{4}

Option C, exists and less than

1 + \frac{\pi}{4}

, is correct because

f'(x) > 0

implies

f(x)

is increasing, and by comparing

f'(t)

with

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

and integrating, we find that

\lim_{x \to \infty} f(x) < 1 + \frac{\pi}{4}

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