Let f k x = 1 k sin k x + cos k x where x ∈ R and k ≥1 . Then f 4 x - f 6 x equals

f 4 x = 1 4 sin 4 x + cos 4 x = 1 4 sin 2 x + cos 2 x 2 - 2 sin 2 x cos 2 x = 1 4 1 - 2 sin 2 x cos 2 x f 6 x = 1 6 sin 6 x + cos 6 x = 1 6 sin 2 x + cos 2 x 3 - 3 sin 2 x cos 2 x sin 2 x + cos 2 x = 1 6 1 - 3 sin 2 x cos 2 x 1 f 4 x - f 6 x = 1 4 - 1 6 = 1 1 2

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Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

MathematicsTrigonometric Ratios & IdentitiesAddition, Multiplication, Subtraction formula based,Multiple/Submultiple anglesHard2 minPYQ_2014

MathematicsHardsingle choice

Let $f_{k} (x) = \frac{1}{k} (\sin^{k} x + \cos^{k} x)$ where $x \in R$ and $k \geq1$ . Then $f_{4} (x) - f_{6} (x)$ equals

Options:

Answer:

Solution:

$f_{4} (x) = \frac{1}{4} (\sin^{4} x + \cos^{4} x)$

$= \frac{1}{4} [{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x]$

$= \frac{1}{4} [1 - 2 \sin^{2} x \cos^{2} x]$

$f_{6} (x) = \frac{1}{6} [\sin^{6} x + \cos^{6} x]$

$= \frac{1}{6} [{(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)]$

$= \frac{1}{6} [1 - 3 \sin^{2} x \cos^{2} x (1)]$

$f_{4} (x) - f_{6} (x) = \frac{1}{4} - \frac{1}{6}$

$= \frac{1}{12}$

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Addition, Multiplication, Subtraction formula based,Multiple/Submultiple angles

⏱ 2mℹ️ Source: PYQ_2014

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