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Let the function

$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) & \tan(\frac{\pi}{6}) & \\ \sin(\frac{\pi}{3}) & \cos(\frac{\pi}{3}) & \tan(\frac{\pi}{3}) \end{vmatrix}$$

where

$\theta \in \left[ \frac{\pi}{6},\frac{\pi}{3} \right]$ and $f'(\theta)$     denote the derivative of $f$ with respect to $\theta$. Which of the following statements is/are TRUE?

1. There exists $\theta \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta) = 0$
2. There exists $\theta \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta)\neq 0$
1. I only
2. II only
3. Both I and II
4. Neither I nor II

Thanks. @Sambhrant Maurya understood :)

In both the options we are provided with open interval..so how can someone put the value π/6,π/3 in place of theta in the determinant?

So option 2 is only satisfying..!

Since the previous link posted by @vikas is not working . I have posted a new link for finding derivative of determinant.

Learn Derivative of Determinant in 3 minutes. (toppr.com)

We need to solve this by Rolle's theorem. To apply Rolle's theorem following $3$ conditions should be satisfied:

1. $f(x)$ should be continuous in interval $[a, b],$
2. $f(x)$ should be differentiable in interval $(a, b),$ and
3. $f(a) = f(b)$

If these $3$ conditions are satisfied simultaneously then, there exists at least one $'x'$ such that $f '(x) = 0$

For the given question, it satisfies all the three conditions, so we can apply Rolle's theorem, i.e, there exists at least one $\theta$ that gives $f '(\theta) = 0$

Also, the given function is also not a constant function, i.e., for some $\theta,$ $f '(\theta) ≠ 0$

There is small typo at 8 line from bottom

@VIDYADHAR SHELKE 1-It's not Mod.It's determinant.

@Aashish Lakhchaura

In both the options we are provided with open interval..so how can someone put the value π/6,π/3 in place of theta in the determinant?

So option 2 is only satisfying..!

Solution:

Show me one such point where f’($\theta$) is 0

there is some calculation mistake in  while calculating the determinant.

Here is the graph of f'(x).

Its f(X)= 1.21 *sinX - 0.366 cosX + 0.5 tanX (If f(X) is determinant as shown in ques. )

can we put directly f(a)=pi/6 and f(b)=pi/3 in given function

by solving this f(a)=f(b) (which is zero)

by this it is proved that given function is continuous as well as differentiable (Rolle's Theorem)

but why ii point is also true ?

One major reason of confusion here is that people are differentiating the determinant in a wrong manner. The differentiation of a determinant is done in the following manner:

1. Select a Row
2. Differentiate that Row keeping others constant

$\frac{d}{dt}\begin{vmatrix} a_{11}(t) & a_{12}(t) & a_{13}(t) \\ a_{21}(t) & a_{22}(t) & a_{23}(t) \\ a_{31}(t) & a_{32}(t) & a_{33}(t) \end{vmatrix}=\begin{vmatrix} a'_{11}(t) & a'_{12}(t) & a'_{13}(t) \\ a_{21}(t) & a_{22}(t) & a_{23}(t) \\ a_{31}(t) & a_{32}(t) & a_{33}(t) \end{vmatrix}+\begin{vmatrix} a_{11}(t) & a_{12}(t) & a_{13}(t) \\ a'_{21}(t) & a'_{22}(t) & a'_{23}(t) \\ a_{31}(t) & a_{32}(t) & a_{33}(t) \end{vmatrix}+\begin{vmatrix} a_{11}(t) & a_{12}(t) & a_{13}(t) \\ a_{21}(t) & a_{22}(t) & a_{23}(t) \\ a'_{31}(t) & a'_{32}(t) & a'_{33}(t) \end{vmatrix}.$

So the differentiation of

$f(\theta) = \begin{vmatrix} sin(\theta) & cos(\theta) & tan(\theta)\\ sin(\frac{\pi}{6}) & cos(\frac{\pi}{6}) & tan(\frac{\pi}{6})\\ sin(\frac{\pi}{3}) & cos(\frac{\pi}{3}) & tan(\frac{\pi}{3})\\ \end{vmatrix}$

represented as $f’(\theta)$ would not have 2 rows zero and would not zero out.

### 1 comment

Very good question. Checks many concepts.