Let for a differentiable function f:(0, \infty) arrow R, f(x)-f(y) \geqslant \log _e(\fracxy)+x-y, \forall x, y \in(0, \infty) . Then \sum\limits_n=1^20 f^\prime(\frac1n^2) is equal to \underline\hspace2cm .

Application of Derivatives PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q55 with Solution

View written solutionFree

Correct answer: 2890

f(x)-f(y)\ge \ln\left(\frac{x}{y}\right)+x-y.

We need to find

\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right).

f(x)-\ln x -x \ge f(y)-\ln y -y.

Indeed,

f(x)-f(y)\ge \ln x-\ln y +x-y

which gives

\bigl(f(x)-\ln x-x\bigr)\ge \bigl(f(y)-\ln y-y\bigr).

But this holds for all $x,y>0$ .

If a quantity $g(x)$ satisfies $g(x)\ge g(y)$ for all $x,y$ , then also by swapping $x$ and $y$ we get $g(y)\ge g(x)$ . Hence $g(x)=g(y)$ for all $x,y$ , so $g$ is constant.

Let

g(x)=f(x)-\ln x-x.

Then $g(x)$ is constant on $(0,\infty)$ . Therefore,

f(x)=\ln x+x+C

for some constant $C$ .

f'(x)=\frac1x+1.

So,

f'\left(\frac1{n^2}\right)=\frac{1}{1/n^2}+1=n^2+1.

\sum_{n=1}^{20} f'\left(\frac1{n^2}\right)=\sum_{n=1}^{20}(n^2+1) =\sum_{n=1}^{20}n^2+\sum_{n=1}^{20}1.

Now,

\sum_{n=1}^{20} n^2 = \frac{20\cdot 21\cdot 41}{6}=2870,

and

\sum_{n=1}^{20}1=20.

Therefore,

\sum_{n=1}^{20} f'\left(\frac1{n^2}\right)=2870+20=2890.

\boxed{2890}