The number of integral values of k for which the equation 7cosx+5sinx=2k+1 has a solution is
A
8
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B
9
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C
10
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D
12
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Solution
The correct option is A8 7cosx+5sinx=2k+1
This is acosx+bsinx=c form.
So, solution is possible iff −1≤c√a2+b2≤1⇒−1≤2k+1√72+52≤1⇒−√74≤2k+1≤√74⇒−1−√742≤k≤−1+√742∴k=−4,−3,−2,−1,0,1,2,3