For m- n - N- Let An - 16 cosn- - sinn- and Bm - 16 cosm- - sinm- and log 102 - 0-3010If m is the minimum value of B 6 16- then the value of log 2 m - is equal to

The correct option is A 0 B(m)=16(cos ^ m θ #160; +sin ^ m θ #160; ) B(6) 16 means 16(cos ^ 6 θ #160; +sin ^ 2 θ #160; ) 16 m ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

For m, n ∈ ...

Question

For m, $n \in N$ ,
Let A(n) = $16 ({cos}^{n} θ - {sin}^{n} θ)$ and B(m) = $16 ({cos}^{m} θ + {sin}^{m} θ)$ and ${log}_{10} 2 = 0.3010$
If m is the minimum value of $\frac{B (6)}{16}$ , then the value of ${log}_{2} (m)$ , is equal to

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Solution

The correct option is A 0
$B (m) = 16 (cos^{m} θ + sin^{m} θ)$
$\frac{B (6)}{16}$ means
$\frac{16 (cos^{6} θ + sin^{2} θ)}{16} m = 6 {log}_{2} (6) \approx 0$

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Similar questions

Q. For m,

n \in N

,
Let A(n) =

16 ({cos}^{n} θ - {sin}^{n} θ)

and B(m) =

16 ({cos}^{m} θ + {sin}^{m} θ)

and

{log}_{10} 2 = 0.3010

If A(6) =

a cos 2 θ + b cos 6 θ

, then value of (a+b) is equal to

Q. For m,

n \in N

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Let A(n) =

16 ({cos}^{n} θ - {sin}^{n} θ)

and B(m) =

16 ({cos}^{m} θ + {sin}^{m} θ)

and

{log}_{10} 2 = 0.3010

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θ \in [0, 2 π]

such that

A (4) = B (3)

is equal to

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where

m, n \in N

, then

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Q. If

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Q. If sum of maximum and minimum value of

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For m, n∈N, Let A(n) = 16(cosnθ−sinnθ) and B(m) = 16(cosmθ+sinmθ) and log102=0.3010If m is the minimum value of B(6)16, then the value of log2(m), is equal to

The correct option is A 0B(m)=16(cosmθ+sinmθ)B(6)16 means16(cos6θ+sin2θ)16m=6log2(6)≈0

For m, $n \in N$ ,
Let A(n) = $16 ({cos}^{n} θ - {sin}^{n} θ)$ and B(m) = $16 ({cos}^{m} θ + {sin}^{m} θ)$ and ${log}_{10} 2 = 0.3010$
If m is the minimum value of $\frac{B (6)}{16}$ , then the value of ${log}_{2} (m)$ , is equal to

The correct option is A 0
$B (m) = 16 (cos^{m} θ + sin^{m} θ)$
$\frac{B (6)}{16}$ means
$\frac{16 (cos^{6} θ + sin^{2} θ)}{16} m = 6 {log}_{2} (6) \approx 0$