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number of integral values of k for which the equation has a solution 7cosx+5sinx=2k+1. number of integral values of k for which the equation has a solution 7cosx+5sinx=2k+1.
number of integral values of k for which the equation has a solution
7cosx+5sinx=2k+1.
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..
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..
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)/741/2 sin(x+@) lies bw -1 to 1 so +1>= (2k+1)/741/2 >=-1 +3.8 >= k >=-4.8 integral values of k are [-4,-3,-2,-1,0,1,2,3] , 8 values of k
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)/741/2
sin(x+@) lies bw -1 to 1 so
+1>= (2k+1)/741/2 >=-1
+3.8 >= k >=-4.8
integral values of k are [-4,-3,-2,-1,0,1,2,3] , 8 values of k
7/2=3.56/2=32/2=15/2=2.58/2=46ujahakk8nbzbaka8nsbjak8anh jdjjsb jsjbs hjsi 8929w???????¿¿¿¿¿¿¿¿□♤♡●◇♧●¿●¿◇▪◇•}♡•¡(#¥£'nwwk♤♡○♤~}◇°¡○♤\992♡•♤*#€&$*@(+(-9/2=4.5
Dear student,Please find the answer to your question. Maximum value of 7cosx+5sinx=√(72+52)=√74=8.6and minimum value is – 8.6so it will vary between – 8.6 to 8.6– 8.6 – 4.8 Hence, values of k will be -4,-3,-2,-1,0,1,2,3hence there are 8 values of k Thanks and regards,Kushagra
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