The equation sin x -k-1-kk-0 is

The correct option is B impossible for real values of k Applying AM GM we get k + 1k≥ 2 so sinx = k + 1k #160;is not possible for real values ...

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Standard XII

Mathematics

Complex Numbers

The equation ...

Question

The equation $sin x = k + \frac{1}{k} (k > 0)$ is

possible for real values of

k

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impossible for real values of

k

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true if

k = 3

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true if

k = 1

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Solution

The correct option is B impossible for real values of $k$
Applying AM-GM we get
$k + \frac{1}{k} \geq 2$
so sinx $= k + \frac{1}{k}$ is not possible for real values of k.

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The equation sinx=k+1k(k>0) is

The correct option is B impossible for real values of kApplying AM-GM we getk+1k≥2so sinx =k+1k is not possible for real values of k.

The equation $sin x = k + \frac{1}{k} (k > 0)$ is

The correct option is B impossible for real values of $k$
Applying AM-GM we get
$k + \frac{1}{k} \geq 2$
so sinx $= k + \frac{1}{k}$ is not possible for real values of k.