The value y for which the equation $4 sin x + 3 cos x = y^{2} - 6 y + 14$ has a real solution is/are:

3

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5

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- 3

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N o n e

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Solution

The correct option is A $3$
As $- 5 \leq 4 sin x + 3 cos x \leq 5$
$y^{2} - 6 y + 14 = - 5$ or $y^{2} - 6 y + 14 = 5$
$y^{2} - 6 y + 19 = 0$ or $y^{2} - 6 y + 9 = 0$
For $y^{2} - 6 y + 19 = 0$ solution is imaginary
And for $y^{2} - 6 y + 9 = 0 \Rightarrow {(y - 3)}^{2} = 0 \Rightarrow y = 3$

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