JEE Main 2023 — Mathematics Calculus

Given,

f(x+y)=f(x)+f(y)&f(1)=\frac{1}{5}

Now solving the given functional equation by taking x=1&y=1 we get,

$$f(1+1)=f(1)+f(1)$$f(1+1)=f(1)+f(1)

$$\Rightarrow f(2)=\frac{2}{5}{\text{as }f(1)=\frac{1}{5}}$$⇒f(2)=52​as f(1)=51​

Now taking x=2&y=1 we get,

$$f(3)=\frac{2}{5}+\frac{1}{5}\Rightarrow f(3)=\frac{3}{5}$$f(3)=52​+51​⇒f(3)=53​

Hence, $f(x)=\frac{x}{5}$f(x)=5x​

Now solving $\displaystyle \sum _{n=1}^{m}\frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$n=1∑m​n(n+1)(n+2)f(n)​=121​

$$\Rightarrow \sum _{n=1}^{m}\frac{\frac{n}{5}}{n(n+1)(n+2)}=\frac{1}{12}$$⇒n=1∑m​n(n+1)(n+2)5n​​=121​

$$\Rightarrow \frac{1}{5}\sum _{n=1}^{m}\frac{1}{(n+1)(n+2)}=\frac{1}{12}$$⇒51​n=1∑m​(n+1)(n+2)1​=121​

$$\Rightarrow \frac{1}{5}\sum _{n=1}^{m}\frac{1}{(n+1)}-\frac{1}{(n+2)}=\frac{1}{12}$$⇒51​n=1∑m​(n+1)1​−(n+2)1​=121​

$$\Rightarrow \frac{1}{5}[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}.....-\frac{1}{m+2}]=\frac{1}{12}$$⇒51​[21​−31​+31​−41​+41​.....−m+21​]=121​

$$\Rightarrow \frac{1}{5}[\frac{1}{2}-\frac{1}{m+2}]=\frac{1}{12}$$⇒51​[21​−m+21​]=121​

$$\Rightarrow [\frac{m}{2(m+2)}]=\frac{5}{12}$$⇒[2(m+2)m​]=125​

$$\Rightarrow 12m=10(m+2)$$⇒12m=10(m+2)

$$\Rightarrow m=10$$⇒m=10