3.5k views

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
edited | 3.5k views
+3
Someone explain it please.. I m getting both (i) and (ii).. since if we take derivative f(x) then the 2nd and the 3rd row is will be 0.. So determinant is 0 which is independent of theta..!!
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How to solve quickly :

Put pi/3 into the determinant (without solving it), row 1 and row 3 become same, f(pi/3) = 0

Similarly by putting theta = pi/6, f(pi/6) = 0

Then apply roll's theorem, making I true

Also calculate the determinant on some different value, it comes out to be nonzero, making II true
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But on evaluating, f(Theta)  is coming out 0.which is constant function.
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@Kirankhare See the link above provided by vikas. You're wrongly differentiating the determinant.

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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..!

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 $\theta,$ $f '(\theta) ≠ 0$

selected
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how can you say that f(x) is continous and differentiable??

secondly,how did you prove the second stateentt as true i.e f'(theta) !=0

plss explain
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"In that given interval" it is continuous as well as differentiable. You should read the definitions of continuity and differentiabilty and check whether it satisfies those properties in that interval.

Also, above function is not constant, it depends on theta, there must be definitely "some theta" where it is not equal to zero. If it would have been constant, then it's derivative i.e. F'(theta) will always be zero.
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i am not able to prove how it is continue and differentiable..can u pls help??not good in continuity and differentiabilty
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some pls explain how it is continuous and differentiable.. here
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@ Akriti,

Trignometric functions are differentiable in their domain also if function is differentiable then it is continuos for sure.

Hope it helps.
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you mean sinx ,cos x,tanx will be differantiable at any value??
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For this question in this range it is continious and differentiable too
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@Akriti simple wy to check them is by vissualizing there graph in the specific domain.
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Sorry, the question may sound dumb, but how to do differentiation of a determinant? Please explain the solution in more detail
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just substitute the value of theta as pi/6 in the function .You will get determinant to be zero .

Similarly for the value pi/3 determinant is zero
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nice explanation saurabh sharma
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Can you elaborate second point.i am not able to get it.
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π/6 is not in the given range. The question asks for f'(theta) = 0 for (π/6,π/3) and not for [π/6,π/3]​​

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$f(\theta)$ turns out to be this

$sin(\theta)\begin{vmatrix} \frac{\sqrt 3}{2} & \frac{1}{\sqrt 3} \\ \frac{1}{2}&\sqrt 3 \end{vmatrix}-cos(\theta)\begin{vmatrix} \frac{1}{2} &\frac{1}{\sqrt 3} \\ \frac{\sqrt 3}{2} &\sqrt 3 \end{vmatrix}+tan(\theta)\begin{vmatrix} \frac{1}{2} &\frac{\sqrt 3}{2} \\ \frac{\sqrt 3}{2} &\frac{1}{2} \end{vmatrix}$

Now, i can rewrite the above equation as

$f(\theta)=asin(\theta)-bcos(\theta)+ctan(\theta)$

for some non-zero constants $a,b,c$

$f'(\theta)=acos(\theta)=bsin(\theta)+csec^2(\theta)$

now at $\theta=\frac{\pi}{4} \in (\frac{\pi}{6},\frac{\pi}{3})$, $f'(\theta)\neq0$

So, II clause is true.

Please let me know if somewhere my this claim for II is incorrect.
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why you mod value took constant a,b,c ...is it not affect on ans?

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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..!

Here is the graph of f'(x). 0

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 ?

Solution: 1