Question

If [5 sinx] + [cosx] + 6 = 0 then range of f(x) = sinx + $\sqrt{3}$cos x, corresponding to the solution set of the given equation, is (where [.] denotes the greatest integer function)

• A. [-2, 0]
• B. (−$\sqrt{3}$, 0|
• C. [−2, $\sqrt{3}$)
• D. |−2, −$\sqrt{3}$)

Answer 1

Answer: (D)

|−2, −$\sqrt{3}$)

Answer 2

[5 sinx] + [cosx] = -6

⇒ [5 sinx] = −5, (cosx] = −1

⇒ −5 ≤ 5 sin x < −4, -1 ≤ cos x < 0

⇒ −1 < sin x < $- \frac { 4 } { 5 }$, −1 ≤ cos θ < cos x < 0

$\pi + \sin ^ { - 1 } \left( \frac { 4 } { 5 } \right) < x < \frac { 3 \pi } { 2 }$

Now f(x) = sinx + $\sqrt { 3 }$ cosx = $2 \sin \left( x + \frac { \pi } { 6 } \right)$

we have, $\pi + \frac { \pi } { 6 } + \sin ^ { - 1 } \left( \frac { 4 } { 5 } \right) < x + \frac { \pi } { 6 } < \frac { 3 \pi } { 2 } + \frac { \pi } { 6 }$

⇒ $- 1 \leq \sin \left( x + \frac { \pi } { 6 } \right) < - \frac { \sqrt { 3 } } { 2 }$