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Yesterday, I received a letter from an IIT-JEE 2004 aspirant. It is shown here along with my comments as it is of general interest.

Respected Sir,

I will be highly greatful if you please aware me about the suggestions of iitjee 2004 or sample sugesstive questions for iit jee 2004, if there any.

Help me soon. Thanking you!

Sugata


Comments: I can suggest the following:

Complete thorough revision by 31'st of March. This should be done according to the mains standards. Objective questions may be done occasionally.

From 1'st April till the date of screening exam, do only objective questions. Take time bound tests often during this period. For test taking techniqe refer to FAQ.

There are original questions framed by IIT for JEE, hence there is no point in looking for guess questions.

Refrain from any entertainment like T.V., sports etc.

Think positive.

Manish Verma

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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, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$
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