in Calculus edited by
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39 votes
39 votes

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
in Calculus edited by
10.2k views

4 Comments

Thanks. @Sambhrant Maurya understood :) 

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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..!

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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)

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4 Answers

49 votes
49 votes
Best answer

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 $\theta,$ $f '(\theta) ≠ 0$

So, answer is C.
 

selected by

4 Comments

There is small typo at 8 line from bottom
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@VIDYADHAR SHELKE 1-It's not Mod.It's determinant.

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@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..!

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33 votes
33 votes

Solution:

3 Comments

Show me one such point where f’($\theta$) is 0
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there is some calculation mistake in  while calculating the determinant.
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3 votes
3 votes

Here is the graph of f'(x).

2 Comments

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

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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 ?
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2 votes
2 votes

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.
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Answer:

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