@thedumbheadwithnobrain          

I don't think your solution is right. Your doubt is same as mine (answered above)

Look at line number 9

It is not always that the energy stored in the capacitor is half that provided by the battery. It depends how the capacitor is connected to the battery.

@animal you are absolutely right.

I thought the same way. As x ----->infinity the function just comes into existense or as you put it x ----> x+1

You mean to say if the domain was (-1,3) or [-1,3] both would have yielded the same answer?

and also at -1 in Question 8 we consider only RHL and at 3 we consider only LHL

Thanks everybody for replying but I'm still not convinced.

@thedumbheadwithnobrain

2)But -1& 3 lie in the domain of the function as [ ] signifies that the end points are included

3) How can you say it is continuous at -1 and 3 since it is an algebraic function. (I know algebraic functions are continuous everywhere but that is precisly my doubt). From the definition of continuity it has to be discontinuous at x = -1 and x = 3. (explained in detail below)

I'll clearly frame my doubt.

I know we must have LHL = RHL = value of function at that point for a function to be continuous at that point.

But lets say a function y = x is defined for an interval [1,10] .

How can we say it is continuous at x = 1 and x =10?

Left hand Limit for x = 1 is not defined and for x = 10 RHL is not defined. so from the definition of continuity we can say the function is not continuous at x = 1 and x = 2.

Similarly in Question 8 given above we can say function is not continuous at x = -1 and x = 3.

Somebody please help.

The easiest and most practical answer:

Sit in an airplane or hot air balloon, helicopter etc.

How is it continuous at -1 and 3?

Tangent at a point on the curve is defined as the limiting value (as the the two points where the secant cuts the curve approach each other) of the secant to the curve, but since we do not have the function defined beyond 3, we cant draw a secant hence we cant draw a tangent. Is this right?

"Yeah..it is as we dont have a right & left side respectively at x=1, x = -3, but we definetly have just one tanget at these 2 points as compared to x =2."

What do you mean by "we dont have a right & left side respectively at x=1, x = -3, but we definetly have just one tanget at these 2 points as compared to x =2".

We can have only one tangent at a point. At x = 2 we can have only 1 tangent.

Thanks for clearing y doubt

I think question is incorrect. For positive values of x, numerator is less than denominator ( i.e it is a fraction) for which function is not defined