Last Activity: 13 Years ago
Q2; we can find the points of discontinuity by plotting graph of [4sinx]... this can be done by drawing horizontal line parallel to x axis and terminating it whenever [4sinx ] becomes integer ...
2nd method : just find the no. of points where 4 sinx becomes integer because [] graph is discontinuous at integers so ans is like sin^(-1) 4,.... and so on...
Q1: plot both graphs and find there intersection points there is no solution because of foll:
we devide it into intervals
1st: frm (- infinity) to 1: e^x graph is above x axis while lnx graph is below x axis so no solution
2nd:frm 1 to infinity : the slope of e^x is e^x while that of lnx is 1/x( we can findit by differentiating the curves) and at 1 the slope of e^x is e which is nearly 2.7 while that of lnx is 1 and the slope of e^x is continuously increasing while that of lnx is decreasing so...lnx will never be able to catch up e^x and hence no intersection and hence no solution
pls approve the answer if u like the answer ...
Last Activity: 13 Years ago
Dear student,
Draw both the graphs and where there is sharp turn that is the point of discontinuity...
BEST OF LUCK..!!!!
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