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Let $$x, y$$ be real numbers such that $$xy=1$$. Let $$T$$ and $$t$$ be the largest and the smallest values of the expression $$\frac{(x+y)^{2}-(x-y)-2}{(x+y)^{2}+(x-y)-2}$$. If $$T+t$$ can be expressed in the form $$\frac{m}{n}$$ where $$m, n$$ are nonzero integers with $$\gcd(m,n)=1$$, find the value of $$m+n$$.
Correct Answer: 25
Let
$$E=\frac{(x+y)^{2}-(x-y)-2}{(x+y)^{2}+(x-y)-2}, \qquad xy=1.$$
Introduce the symmetric and skew-symmetric combinations
$$S=x+y, \qquad D=x-y.$$
Solving $$x=\dfrac{S+D}{2}, \; y=\dfrac{S-D}{2}$$ and using $$xy=1$$ gives
$$\frac{(S+D)(S-D)}{4}=1 \;\Longrightarrow\; S^{2}-D^{2}=4 \;.\;\; -(1)$$
Rewrite the numerator and denominator of $$E$$ in terms of $$S$$ and $$D$$:
Numerator $$=S^{2}-D-2,$$
Denominator $$=S^{2}+D-2.$$
From $$(1)$$, $$S^{2}=D^{2}+4$$. Substitute this into $$E$$:
$$E=\frac{D^{2}+4-D-2}{D^{2}+4+D-2}=\frac{D^{2}-D+2}{D^{2}+D+2}.$$
Thus $$E$$ depends only on $$D$$ (which can take any real value). Define
$$f(D)=\frac{D^{2}-D+2}{D^{2}+D+2}, \qquad D\in\mathbb{R}.$$
To find the extreme values, differentiate:
$$f'(D)=\frac{(2D-1)(D^{2}+D+2)-(2D+1)(D^{2}-D+2)}{(D^{2}+D+2)^{2}} =\frac{2(D^{2}-2)}{(D^{2}+D+2)^{2}}\;.$$
Critical points occur where $$D^{2}-2=0 \;\Longrightarrow\; D=\pm\sqrt{2}.$$
Evaluate $$f(D)$$ at these points:
For $$D=\sqrt{2}:$$ $$f_{\min}=f(\sqrt{2})=\frac{4-\sqrt{2}}{4+\sqrt{2}}.$$
For $$D=-\sqrt{2}:$$ $$f_{\max}=f(-\sqrt{2})=\frac{4+\sqrt{2}}{4-\sqrt{2}}.$$
(As $$|D|\to\infty$$, $$f(D)\to1,$$ confirming that these are indeed the minimum and maximum.)
Denote $$T=f_{\max},\; t=f_{\min}.$$ Observe that
$$T=\frac{A}{B},\; t=\frac{B}{A}, \quad\text{where } A=4+\sqrt{2},\; B=4-\sqrt{2}.$$
Hence
$$T+t=\frac{A}{B}+\frac{B}{A}=\frac{A^{2}+B^{2}}{AB}.$$
Compute the required quantities:
$$A^{2}=(4+\sqrt{2})^{2}=18+8\sqrt{2},$$ $$B^{2}=(4-\sqrt{2})^{2}=18-8\sqrt{2},$$ $$A^{2}+B^{2}=36,$$ $$AB=(4+\sqrt{2})(4-\sqrt{2})=16-2=14.$$
Therefore
$$T+t=\frac{36}{14}=\frac{18}{7}.$$
Here $$m=18,\; n=7,\; \gcd(18,7)=1.$$ The required sum is
$$m+n=18+7=25.$$
Answer: 25
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