Question

The number of integral values of \[K\], for which the equation \[7\cos x + 5\sin x = 2K + 1\] has a solution, is
A.4
B.8
C.10
D.12

Answer 1

Hint: Here, we will use the general form of the given equation and will find the range in which \[K\] lies. Using this range, we will be able to find the integral values which \[K\] can take. Hence, counting them, will help us to know the required number of integral values.
Formula Used: \[ - r \le a\cos \theta + b\sin \theta \le r\] where, \[r = \sqrt {{a^2} + {b^2}} \]

Complete step-by-step answer:
The given equation is: \[7\cos x + 5\sin x = 2K + 1\]
Now, we know that, if an equation is in the form of \[a\cos \theta + b\sin \theta \], then,\[ - r \le a\cos \theta + b\sin \theta \le r\] where, \[r = \sqrt {{a^2} + {b^2}} \].
Comparing this with LHS of the given equation, i.e. \[7\cos x + 5\sin x\]
Here, \[a = 7\] and \[b = 5\]
Thus, \[r = \sqrt {{7^2} + {5^2}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow r = \sqrt {49 + 25} \]
Adding the terms, we get
\[ \Rightarrow r = \sqrt {74} \]
Therefore the range becomes,
\[ - \sqrt {74} \le 7\cos x + 5\sin x \le \sqrt {74} \]
Now, we know that \[7\cos x + 5\sin x = 2K + 1\], hence, writing the RHS instead of the LHS, we get,
\[ \Rightarrow - \sqrt {74} \le 2K + 1 \le \sqrt {74} \]
We know that \[\sqrt {74} = 8.6\], therefore,
\[ \Rightarrow - 8.6 \le 2K + 1 \le 8.6\]
Now, subtracting 1 from each side of the inequality,
\[ \Rightarrow - 9.6 \le 2K \le 7.6\]
Dividing each side of the inequality by 2,
\[ \Rightarrow - 4.8 \le 2K \le 3.8\]
Hence, the integral values of \[K\] are:
\[ - 4, - 3, - 2, - 1,0,1,2,3\]
On counting, we will find that there are 8 possible integral values of \[K\].
Hence, option B is the correct answer.

 Note: In mathematics, integral is either a numerical value equal to the area under the graph of a function for some definite integral or it is a new function whose derivative is the original function. By looking at the question, we can clearly observe that we are required to find the integral i.e. the numerical values which \[K\] can take (which can’t be in the form of a fraction or decimal). Hence, in order to find the required number of possible integral values, we had found the range in which \[K\] lies. This means that the largest and the smallest possible values in which the value of \[K\] lies. Hence, counting all the integrals lying in that range helps us to find the required answer.