If f ( x ) = ∣ ∣ ∣ ∣ s i n x s i n a s i n b c o s x c o s a c o s b t a n x t a n a t a n b ∣ ∣ ∣ ∣ , where 0 < a < b < π 2 then the equation f ′ ( x ) = 0 has in the interval ( a , b ) Options: A . &nbspAtleast one root B . &nbspAtmost one root C . &nbspNo root D . &nbspexactly one root

Answer: Option A : A Here f ( a ) = ∣ ∣ ∣ ∣ s i n a s i n a s i n b c o s a c o s a c o s b t a n a t a n a t a n b ∣ ∣ ∣ ∣ = 0. Also f ( b ) = 0. Moreover, as sin x, cos x and tan x are continuos and differentiable in (a, b) for 0 < a < b < π 2 , therefore f(x) is also continuos and differentiable in [a, b]. Hence, by Rolle's theorem, there exists atleast one real number c in (a, b) such that f ' (c) = 0. Hence (a) is the correct answer.

If f(x)=∣∣ ∣∣sin xsin asin bcos

Question

f (x) = ∣ ∣∣ ∣ \begin{matrix} s i n x & s i n a & s i n b c o s x & c o s a & c o s b t a n x & t a n a & t a n b \end{matrix} ∣ ∣∣ ∣,

where

0 < a < b < \frac{π}{2}

then the equation

f^{'} (x) = 0

has in the interval

(a, b)

Options:

A . Atleast one root

B . Atmost one root

C . No root

D . exactly one root

Answer: Option A
:
A
Here

f (a) = ∣ ∣∣ ∣ \begin{matrix} s i n a & s i n a & s i n b c o s a & c o s a & c o s b t a n a & t a n a & t a n b \end{matrix} ∣ ∣∣ ∣ = 0. Also f (b) = 0.

Moreover, as sin x, cos x and tan x are continuos and differentiable in (a, b) for 0 < a < b <

\frac{π}{2}

, therefore f(x) is also continuos and differentiable in [a, b]. Hence, by Rolle's theorem, there exists atleast one real number c in (a, b) such that f ' (c) = 0.
Hence (a) is the correct answer.

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