(IC/M)
Saturday, October 31, 2009
Saturday, October 17, 2009
Thursday, October 8, 2009
Monday, October 5, 2009
Sunday, October 4, 2009
SOLUTION OF PROBLEM: 11
Plot the curve of y = |tan(x)| (DC/M)
SOLUTION : We already know how to plot the curve of tan(x). |tanx| will always yield positive values, which means wherever tan(x) is negative we have to make it positive in order get |tan(x)| curve.
SOLUTION : We already know how to plot the curve of tan(x). |tanx| will always yield positive values, which means wherever tan(x) is negative we have to make it positive in order get |tan(x)| curve.
Saturday, October 3, 2009
Friday, October 2, 2009
SOLUTION OF PROBLEM: 10
Let function f is from R to R be defined by f(x) =2x + sin(x) for x belongs to R. Then f is
a) one-to-one and onto
b) one-to-one but not onto.
c) onto but not one-to-one
d) neither one-to-one nor onto. ( DC/CH)SOLUTION: The given function has co-domain as R( set of all real numbers)
For the function f(x) = 2x + sin(x) , 2x ranges from negative infinity to positive infinity whereas sin(x) varies from [-1,1] so ultimately the f(x) will have its range as R. Therefore, the f(x) is Onto function.
Also, f'(x) = 2 + cos(x) , cos(x) will vary from [-1,1], So 2+cos(x) varies from [1,3].
Therefore, f'(x) is always positive. Hence f(x) will always be an increasing function.Therefore, f(x) will be one-to-one.
So, the correct option is (a)Thursday, October 1, 2009
SOLUTION OF PROBLEM: 9
Find the area bounded by the curve y = e(x) and the lines x = 0 and y = e. ( IC/E)
SOLUTION:
NOTE: e(x) is e to the power x
SOLUTION OF PROBLEM: 8
Prove that the f(x) = 2(x)cube + 3(x)sq+6x +10 will always be an increasing function. (DC/M)
Solution: If we differentiate the given function we will get f'(x) = 6(x)sq + 6x +6 = 6( (x)sq +x +1)
Now, we can write f'(x) = 6[(x + 1/2)sq +3/4] which is always positive.So f(x) will always be an increasing function.
Alternatively, we can also say the for the quadratic expression (x)sq +x +1 coefficient of (x)sq is positive and discriminant is negative (-3), so the expression is always positive
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