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

The correct option is B 9f(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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Rolle's Theorem

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$a = 9$

Suggest Corrections

Similar questions

Q. The least value of '

a

' for which equation

\frac{4}{sin x} + \frac{1}{1 - sin x} = a

has at least one solution in the interval

[0, \frac{π}{2}]

Q. The least value of

a

for which the equation

\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)

Q. The least value of a for which the expression

\frac{4}{sin x} + \frac{1}{1 - sin x} = a^{2}

has at least one solution on the interval

(0, π / 2)

Q. The least value of a for which the expression

f (x) = \frac{4}{sin x} + \frac{1}{1 - sin x} = a^{2}

has at least one solution on the interval

(0, π / 2)

Q. The least value of 'a' for which the equation,

\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a

has atleast one solution on the interval

(0, π / 2)

is:

Join BYJU'S Learning Program

The least value of 'a' for which 4sin x+11−sin x=a has at least one solution in the interval (0,π/2) is

The correct option is A 9f(x)=4sin x+11−sin x=af′(x)=−4 cos xsin2 x+cos x(1−sin x)2=cos x(1(1−sin x)2−4sin2 x)∴f′(x)=0⇒1(1−sin x)2−4sin2 x=0 as cos x≠0 in(0,π2) This gives sin x=23 Substituting Eq. (i), we get a=9

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