Search any question & find its solution
Question:
Answered & Verified by Expert
The number of integral values of $K$, for which the equation $7 \cos x+5 \sin x=2 K+1$ has a solution, is
Options:
Solution:
1162 Upvotes
Verified Answer
The correct answer is:
8
$\begin{array}{l}
-\sqrt{7^{2}+5^{2}} \leq(7 \cos x+5 \sin x) \leq \sqrt{7^{2}+5^{2}} \\
\Rightarrow-\sqrt{74} \leq(2 K+1) \leq \sqrt{74} \\
\Rightarrow-8.6 \leq(2 K+1) \leq 8.6 \\
\Rightarrow-9.6 \leq 2 K \leq 7.6 \\
\Rightarrow-4.8 \leq K \leq 3.8
\end{array}$
So, integral values of $\mathrm{K}$ are $-4,-3,-2,-1,0,1,2,3$ (eight values)
-\sqrt{7^{2}+5^{2}} \leq(7 \cos x+5 \sin x) \leq \sqrt{7^{2}+5^{2}} \\
\Rightarrow-\sqrt{74} \leq(2 K+1) \leq \sqrt{74} \\
\Rightarrow-8.6 \leq(2 K+1) \leq 8.6 \\
\Rightarrow-9.6 \leq 2 K \leq 7.6 \\
\Rightarrow-4.8 \leq K \leq 3.8
\end{array}$
So, integral values of $\mathrm{K}$ are $-4,-3,-2,-1,0,1,2,3$ (eight values)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.