The function sin x-asin x-b has no maxima or minima if

The correct options are A b a=nπ, n∈ I B b a=(2n+1)nπ, n∈ I C b a=2nπ, n∈ I Given : f( x ) =sin (x+a) #160; sin (x+b) #160; #160; ...

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Standard XII

Mathematics

Integration as Antiderivative

The function ...

Question

The function $\frac{sin (x + a)}{sin (x + b)}$ has no maxima or minima if

b - a = n π, n \in I

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b - a = (2 n + 1) n π, n \in I

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b - a = 2 n π, n \in I

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None of the above.

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Solution

The correct options are
A $b - a = n π, n \in I$
B $b - a = (2 n + 1) n π, n \in I$
C $b - a = 2 n π, n \in I$
Given : $f (x) = \frac{sin (x + a)}{sin (x + b)}$

$f^{^{'}} (x) = \frac{sin (x + b) cos (x + a) - sin (x + a) cos (x + b)}{{sin}^{2} (x + b)}$ $= \frac{sin (b - a)}{{sin}^{2} (x + b)}$

For no maxima and minima condition, $sin (b - a) \neq 0$
$\Rightarrow b - a \neq n π$
Therefore, options A,B and C are correct.

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The function sin(x+a)sin(x+b) has no maxima or minima if

The correct options are A b−a=nπ,n∈I B b−a=(2n+1)nπ,n∈I C b−a=2nπ,n∈IGiven : f(x)=sin(x+a)sin(x+b)f′(x)=sin(x+b)cos(x+a)−sin(x+a)cos(x+b)sin2(x+b) =sin(b−a)sin2(x+b)For no maxima and minima condition, sin(b−a)≠0⇒b−a≠nπTherefore, options A,B and C are correct.

The function $\frac{sin (x + a)}{sin (x + b)}$ has no maxima or minima if