GATE CSE 2014 | Set 1 | Question: 6

Question

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 $f'(\theta)\neq  0$

• A. I only
• B. II only
• C. Both I and II
• D. Neither I nor II

Comments

Can you share the new link @vikas. Its showing page not found.

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@Hirak, They menctioned derivative along with openinterval ,So wrt rolles theorem, it just satisfies.

And in rolles theorem, atleast one x needs to satisfy f'(x)=0, but with other x values it can be non zero.

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$f(\pi/6) = 0$ (because 1st and 2nd rows are equal)

$f(\pi/3) = 0$ (because 1st and 3rd rows are equal)

By Rolle's Theorem Statement $I$ is True.

check with an intermediate value $f(\pi/4) \neq 0$ $\neq f(\pi/6)$ We can see that $f(x)$ is not a constant function 

$f′(θ)$ cannot be equal to 0 everywhere

so Statement $II$ is also True.

Answer 1

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.
 

Comments

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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@Aashish Lakhchaura but 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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@Ayush Upadhyaya

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..!

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@Ayush Upadhyaya sir  plz explain for statement one also. 

Thanks

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f(theta) after calculating determinant is not continuous on (30, 60) due to tan function. this violates condition 1 of rolle's theorem, so how can rolle's theorem be applied?

from graph we can say there exists some theta between (30,60) such that f'(theta) = 0 and for some other theta f'(theta) != 0 thus both statement 1 and 2 asked in question are true.

but we need theoretical way since during exam we don't get graphing tools

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@Akriti sood @PEKKA

sum of continuous function is also continuous. 

$\text{Proof:} $  https://math.stackexchange.com/questions/2750160/prove-that-the-sum-of-two-continuous-functions-is-continuous

 

Answer 2

Comments

U didn't even calculated correct determinant

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how f(π/6​)=f(π/3​)=0?
do we need to solve the determinant?

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yes

Answer 3

Question
Let $f(\theta)$ be defined as:



$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta \\ \sin\frac{\pi}{6} & \cos\frac{\pi}{6} & \tan\frac{\pi}{6} \\ \sin\frac{\pi}{3} & \cos\frac{\pi}{3} & \tan\frac{\pi}{3} \end{vmatrix}, \quad \theta \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right]$$

Consider the following statements:

I. There exists $\theta \in (\frac{\pi}{6}, \frac{\pi}{3})$ such that $f'(\theta) = 0$.

II. There exists $\theta \in (\frac{\pi}{6}, \frac{\pi}{3})$ such that $f'(\theta) \neq 0$.

Best Approach: Rolle's Theorem
This problem can be solved efficiently by utilizing the properties of determinants and Rolle's Theorem, avoiding tedious differentiation or expansion.

Key Observation: Determinant Properties
A determinant is zero if any two rows are identical.

If we plug in $\theta = \frac{\pi}{6}$, Row 1 becomes identical to Row 2. Thus, $f(\frac{\pi}{6}) = 0$.

If we plug in $\theta = \frac{\pi}{3}$, Row 1 becomes identical to Row 3. Thus, $f(\frac{\pi}{3}) = 0$.

Step 1: Evaluating Statement I
According to Rolle's Theorem, if a function $f(x)$ is:

Continuous on $[a, b]$

Differentiable on $(a, b)$

And $f(a) = f(b)$

Then there must exist at least one $c \in (a, b)$ such that $f'(c) = 0$.

Here, $f(\frac{\pi}{6}) = f(\frac{\pi}{3}) = 0$.

Since $f(\theta)$ is a combination of trigonometric functions, it is continuous and differentiable in the given interval.

Conclusion: There exists $\theta \in (\frac{\pi}{6}, \frac{\pi}{3})$ such that $f'(\theta) = 0$.

👉 Statement I is TRUE.

Step 2: Evaluating Statement II
Statement II asks if there is any point where the derivative is non-zero.

If $f'(\theta) = 0$ for all $\theta \in (\frac{\pi}{6}, \frac{\pi}{3})$, then $f(\theta)$ must be a constant function.

We know $f(\frac{\pi}{6}) = 0$. If we pick an intermediate value like $\theta = \frac{\pi}{4}$:



$$f\left(\frac{\pi}{4}\right) = \begin{vmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 1 \\ 1/2 & \sqrt{3}/2 & 1/\sqrt{3} \\ \sqrt{3}/2 & 1/2 & \sqrt{3} \end{vmatrix}$$

The rows are linearly independent, so $f(\frac{\pi}{4}) \neq 0$.

Since the function is not constant, its derivative cannot be zero everywhere.

👉 Statement II is TRUE.

Final Answer
Both statements I and II are correct.

Correct Option: (C)

Comments

Extending in 1st point : To make the applicability of Rolle's theorem more clearer :  The function is a linear combination of \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\):  \(\sin\theta\) and \(\cos\theta\) are continuous and differentiable for all real numbers. \(\tan\theta\) is continuous and differentiable everywhere except at odd multiples of \(\frac{\pi }{2}\) (like \(90^{\circ }\)), where it has vertical asymptotes. Since here the interval is \(\left[\frac{\pi}{6}, \frac{\pi}{3}\right]\) (which translates to \([30^\circ, 60^\circ]\)), it avoids \(90^{\circ }\) entirely. Therefore, the function is continuous and differentiable in the given range.

Answer 4

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.

Comments

Very good question. Checks many concepts.

Answer 5

Here is the graph of f'(x).

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 ?

Answer 6

Solutions are already given , just for fun see the graph.  

Comments

Graph is not showing @Kshitij Sharma. Upload new graph.