Sin3 x-sin3 600-x-sin3 1200-x- -m-4-m-4- m-

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Graphs of Basic Inverse Trigonometric Functions

sin3 x-sin3 6...

Question

${sin}^{3} x - {sin}^{3} (60^{0} - x) + {sin}^{3} (120^{0} + x) \in$
$[- \frac{m}{4}, \frac{m}{4}] \Rightarrow m =$

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Solution

The correct option is B 3
Let $f (x) = s i n 3 x - s i n 3 (60^{o} + x) + s i n (120^{o} + x)$
$= s i n 3 x - s i n (π + 3 x) + s i n (2 π + 3 x)$
$= s i n 3 x + s i n 3 x + s i n 3 x$
$= 3 s i n 3 x$
range of $f (x) ϵ [- 3, 3]$
range of $s i n x ϵ [- 1, 1]$

Suggest Corrections

Similar questions

{sin}^{3} x + {sin}^{3} (120^{0} + x) + {sin}^{3} (240^{0} + x) \in

[- \frac{p}{4}, \frac{p}{4}] \supset p =

{sin}^{3} x + {sin}^{3} (60^{0} - x) - {sin}^{3} (60^{0} + x) \in

Expand:
(1) $(a + 2) (a - 1)$
(2) $(m - 4) (m + 6)$
(3) $(p + 8) (p - 3)$
(4) $(13 + x) (13 - x)$
(5) $(3 x + 4 y) (3 x + 5 y)$
(6) $(9 x - 5 t) (9 x + 3 t)$
(7) $(m + \frac{2}{3}) (m - \frac{7}{3})$
(8) $(x + \frac{1}{x}) (x - \frac{1}{x})$
(9) $(\frac{1}{y} + 4) (\frac{1}{y} - 9)$

Q. Solve using formula.
(1) x² + 6x + 5 = 0
(2) x² – 3x – 2 = 0
(3) 3m² + 2m – 7 = 0
(4) 5m² – 4m – 2 = 0
(5)

y^{2} + \frac{1}{3} y = 2

(6) 5x² + 13x + 8 = 0

Q. Simplify:

(1)

{(18)}^{\frac{1}{3}} \times {(18)}^{\frac{1}{2}}

(2)

x^{\frac{2}{3}} \times x^{\frac{1}{3}}

(3)

{(y^{\frac{2}{3}})}^{\frac{3}{4}}

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m^{\frac{14}{5}} \div m^{\frac{4}{5}}

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sin3x−sin3(600−x)+sin3(1200+x)∈[−m4,m4]⇒m=

The correct option is B 3Let f(x)=sin3x−sin3(60o+x)+sin(120o+x)=sin3x−sin(π+3x)+sin(2π+3x)=sin3x+sin3x+sin3x=3sin3xrange of f(x)ϵ[−3,3]range of sinxϵ[−1,1]

${sin}^{3} x - {sin}^{3} (60^{0} - x) + {sin}^{3} (120^{0} + x) \in$
$[- \frac{m}{4}, \frac{m}{4}] \Rightarrow m =$

The correct option is B 3
Let $f (x) = s i n 3 x - s i n 3 (60^{o} + x) + s i n (120^{o} + x)$
$= s i n 3 x - s i n (π + 3 x) + s i n (2 π + 3 x)$
$= s i n 3 x + s i n 3 x + s i n 3 x$
$= 3 s i n 3 x$
range of $f (x) ϵ [- 3, 3]$
range of $s i n x ϵ [- 1, 1]$