JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 21)
The number of integral values of 'k' for which the equation $$3\sin x + 4\cos x = k + 1$$ has a solution, k$$\in$$R is ___________.
Answer
11
Explanation
We know,
$$ - \sqrt {{a^2} + {b^2}} \le a\cos x + b\sin x \le \sqrt {{a^2} + {b^2}} $$
$$ \therefore $$ $$ - \sqrt {{3^2} + {4^2}} \le 3\cos x + 4\sin x \le \sqrt {{3^2} + {4^2}} $$
$$ - 5 \le k + 1 \le 5$$
$$ - 6 \le k \le 4$$
$$ \therefore $$ Set of integers = $$ - 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$$ = Total 11 intergers.
$$ - \sqrt {{a^2} + {b^2}} \le a\cos x + b\sin x \le \sqrt {{a^2} + {b^2}} $$
$$ \therefore $$ $$ - \sqrt {{3^2} + {4^2}} \le 3\cos x + 4\sin x \le \sqrt {{3^2} + {4^2}} $$
$$ - 5 \le k + 1 \le 5$$
$$ - 6 \le k \le 4$$
$$ \therefore $$ Set of integers = $$ - 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$$ = Total 11 intergers.
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