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### JEE Changes - One Side of the Coin

The type of reaction that can be seen from the student community within 24 hours of press release points towards serious restrictions in eligibility criteria for JEE 2006. The group of people involved in bringing out changes in JEE system are no doubt eminent people having sound minds. However, the reforms suggested by them and accepted by HRD say the strory from their perspective. Things could have been much better if ground realities would have been accounted for - not just conferences in AC rooms.

The reforms in JEE need to be made by allowing participation from IIT aspirants and their parents, IIT students and IIT alumni. This way the ground realities will emerge more clearly and a better policy can be made.

This type of thoughts can be taken from different people through internet by creating feedback areas / forums in IIT websites.

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$