JEE Main 2025 Jan 29 Shift 2, Mathematics Q8: Maxima & Minima

$A=(1,0)$, $B=(0,1)$, $O=(0,0)$. Area of $△OAB=\frac{1}{2}$, so area of $△AMN=\frac{4}{9}\cdot \frac{1}{2}=\frac{2}{9}$.

Let $\angle MAN=\theta$ (measured from $AM$, with $M$ on the $y$-axis, toward $AN$ along $AB$). Since $\angle OAB=45^{\circ }$, the angle $\angle OAM=45^{\circ }-\theta$.

From the right triangle $△AOM$, $AM=\frac{OA}{\cos (45^{\circ }-\theta )}=\sec (45^{\circ }-\theta )$, and in the right triangle $△AMN$ (right-angled at $N$):

$AN=AM\cos \theta =\sec (45^{\circ }-\theta )\cos \theta$, $MN=\sec (45^{\circ }-\theta )\sin \theta$.

Area $△AMN=\frac{1}{2}AN\cdot MN=\frac{1}{2}{\sec }^2(45^{\circ }-\theta )\sin \theta \cos \theta =\frac{2}{9}$.

Use $\cos (45^{\circ }-\theta )=\frac{\cos \theta +\sin \theta }{\sqrt{2}}$:

$\frac{\sin \theta \cos \theta }{(\cos \theta +\sin \theta {)}^2}=\frac{2}{9}$

$9\sin \theta \cos \theta =2(1+2\sin \theta \cos \theta )\Rightarrow 5\sin \theta \cos \theta =2\Rightarrow \sin 2\theta =\frac{4}{5}$.

Equivalently, $2{\tan }^2\theta -5\tan \theta +2=0\Rightarrow \tan \theta =2$ or $\tan \theta =\frac{1}{2}$.

The ratio $\frac{AN}{NB}=\cot \theta$, and $\tan \theta =2$ would make $\theta >45^{\circ }$, putting $N$ outside segment $AB$; that case is rejected.

With $\tan \theta =\frac{1}{2}$, $\lambda =\cot \theta =2$. Sum of all admissible $\lambda =2$.