The number of integral values of k for which | vedclass

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(A) The given equation is $3 \sin x + 4 \cos x = k + 1$.We know that the range of the function $f(x) = a \sin x + b \cos x$ is $[-\sqrt{a^2 + b^2}, \s

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$11$

Vedclass · Answer

(A) The given equation is $3 \sin x + 4 \cos x = k + 1$.We know that the range of the function $f(x) = a \sin x + b \cos x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 3$ and $b = 4$,so the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.For the equation to have a solution,the value $k + 1$ must lie within this range:$-5 \le k + 1 \le 5$Subtracting $1$ from all sides,we get:$-6 \le k \le 4$The integral values of $k$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4\}$.The total number of such integral values is $11$.

The number of integral values of $k$ for which the equation $3 \sin x + 4 \cos x = k + 1$ has a solution,where $k \in R$,is:

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