Observation is needed

Calculus Level 4

Let $f$ be a continuous and differentiable function in $(x_1,x_2)$ . And the following conditions hold for it: $\begin{cases} f(x) \ f'(x) \geq x \ \sqrt{1-((f(x))^4} \\ \displaystyle \lim_{x \to x_1^+} ((f(x))^2=1 \\ \displaystyle \lim_{x \to x_2^-} ((f(x))^2=\frac{1}{2} \end{cases}$

Find the minimum value of $(x_1^2-x_2^2)$

Note : $f'(x)=\dfrac{df(x)}{dx}$

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\frac{2\pi}{3}

None of the given.

\frac{\pi}{6}

\frac{\pi}{3}

2 solutions

Shivang Jindal
Mar 17, 2015

'Observe' that, derivative of $f^2(x)$ is $2f(x)f'(x)$ .

Problem becomes trivial now. . (Take $\sqrt{1-f^4(x)}$ ) To RHS ) and integrate from $x_1$ to $x_2$ .

f(x)^2=sin(t).now acc to cond.

Soumil Jain - 6 years, 2 months ago

I did the same. But i have a doubt- can we integrate in an inequality?

neelesh vij - 5 years, 1 month ago

Yes Since an inequality is true(In an interval) It's integral must also be true (As 1 function would always lie above or on the other,implying its area bounded would also be greater than or equal to the other's area) Holds irrespective of continuity of the function

Suhas Sheikh - 3 years ago

Vijay Bhaskar
Jun 24, 2015

Let y=f(x), to simplify the notation.

Hence, y.dy/dx >= x sqrt(1-y^4)

=> y dy / sqrt(1-y^4) >= x dx

Integrating along the path (x1,y1), (x2,y2),

We have:

Integral (y1 to y2) 1/2 . 2y /sqrt(1-y^4) dy >= Integral (x1 to x2) 1/2. 2x dx

=> 1/2 (lim y-> y2 arcsin(y2^2) - lim y -> y1 arcsin(y1^2)) >= 1/2(x2^2 - x1^2)

=> 1/2(pi/6-pi/2) >= 1/2(x2^2-x1^2)

Multiplying both sides by -1 (which reverses the inequality),

1/2(pi/2-pi/6) <= 1/2 (x1^2 - x2^2)

Hence, x1^2 - x2^2 >= pi/2 - pi/6 = pi/3 //

Moderator note:

Good solution. The tricky part is figuring out that we should attempt to "solve" the differential equation.

did the same way!! it was a really good question.

rahul saxena - 5 years, 8 months ago

Observation is needed

Practice the set Target JEE_Advanced - 2015 and boost up your preparation.

2 solutions

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