The least value of -a- for which 4-sin x - 1-1-sin x - a has at least one solution in the interval 0- -2 is

F(x) = 4/sin x + 1/1 sin x = a f'(x) = 4 cos x/sin^2 x + cos x/(1 sin x)^2 = cos x ( 1/(1 sin x)^2 4/sin^2 x ) ∴ f'(x) = 0 ⇒1/(1 sin x)^2 4/sin^2 x = 0 ...

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Byju's Answer

Standard XI

Mathematics

Local Minima

The least val...

Question

The least value of 'a' for which $\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a$ has at least one solution in the interval $(0, π / 2)$ is

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Solution

The correct option is A
9

$f (x) = \frac{4}{s i n x} + \frac{1}{1 - s i n x} = a f^{'} (x) = - \frac{4 c o s x}{s i n^{2} x} + \frac{c o s x}{(1 - s i n x)^{2}} = c o s x (\frac{1}{(1 - s i n x)^{2}} - \frac{4}{s i n^{2} x}) ∴ f^{'} (x) = 0 \Rightarrow \frac{1}{(1 - s i n x)^{2}} - \frac{4}{s i n^{2} x} = 0 a s c o s x \neq 0 i n (0, \frac{π}{2})$
This gives sin $x = \frac{2}{3}$
Substituting Eq. (i), we get
$a = 9$

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