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For $0 < x < \frac{\pi}{6}$ , all values of

$ \tan^2 3x \cos^2 x - 4\tan 3x \sin 2x + 16\sin^2 x$ lie in interval

A. $(0 , \frac{121}{36})$

B.$(1,\frac{121}{9})$

C.$(-1,0)$

D.None of above

Now i completed square and write as $(\tan 3x \cos x - 4\sin x)^2$. How do i proceed? Thanks

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    You missed minus sign in the square. – 2017-02-19
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    @VikrantDesai Thanks – 2017-02-19

1 Answers 1

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The answer is $D$ since you have a perfect square and hence the value has to be positive. Indeed, let $f(x)=(\tan 3x \cos x - 4\sin x)^2$.

For $x\to 0$ the value of $f(x)$ tends to $0$. This leaves you with possibilities: A) or D).

Now remark that $f(x)$ is continuous in the interval you are considering and that for $x\to\frac{\pi}{6}$ the function $f(x)\to\infty$

Hence the only option is D).

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    But sir ,my textbook states answer to be a – 2017-02-19
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    Did you forget to add some informations in the statement? Otherwise I think that the answer of the textbook is wrong. – 2017-02-19
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    i added everything – 2017-02-19
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    Is it possible that there are some parenthesis? I mean that $\tan(3x)\cos x$ is in fact $\tan(3x\cos x)$ ? Because if this is the case then the answer is A) – 2017-02-19