number of integral values of k for which the equation has a solution 7cosx+5sinx=2k+1.

shiwani sharma Grade:

        number of integral values of k for which the equation has a solution

7cosx+5sinx=2k+1.

7 years ago

Answers : (3)

AKASH GOYAL AskiitiansExpert-IITD

419 Points

Dear Shiwani

Maxinmum value of 7cosx+5sinx=√(7²+5²)=√74=8.6

and minimum value is -8.6

so it will vary between -8.6 to 8.6

2k+1 is an odd integer. between -8.6 to 8.6 odd integers are -7,-5,-3,-1,1,3,5,7

and corresponding values of k will be -4,-3,-2,-1,0,1,2,3

hence there are 8 values of k

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..

7 years ago

vikas askiitian expert

509 Points

							7cosx + 5sinx =2k+1
dividing the eq  by (7²+5²)^1/2 or (74)^1/2
7/(74)^1/2 COSX + 5/(74)^1/2 SINX = 2k+1/(74)^1/2          ..............1
now put  7/(74)^1/2 = sin@
             5/(74)^1/2 = cos@
putting in eq 1
 sin@cosx + cos@sinx = (2k+1)/(74)^1/2
  sin(x+@) = (2k+1)/74^1/2
sin(x+@) lies bw -1 to 1 so
    +1>=  (2k+1)/74^1/2  >=-1
  +3.8 >=       k         >=-4.8
 integral values of k are [-4,-3,-2,-1,0,1,2,3] , 8 values of k