SATHEE: Trigonometrical Equations 3 Question 7

Trigonometrical Equations 3 Question 7

8. The smallest positive root of the equation $\tan x - x = 0$ lies in

(1987, 2M)

(a) $0, \frac{π}{2}$

(b) $\frac{π}{2}, π$

(d) $\frac{3 π}{2}, 2 π$

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Answer:

Correct Answer: 8. (c)

Solution:

Let $f (x) = \tan x - x$

We know, for $0 < x < \frac{π}{2}$

$\Rightarrow \tan x > x$

$∴ f (x) = \tan x - x$ has no root in $(0, π / 2)$

For $π / 2 < x < π, \tan x$ is negative.

$\begin{aligned} ∴ f (x) = \tan x - x < 0 \\ So, f (x) = 0 has no root in \frac{π}{2}, π . \\ For \frac{3 π}{2} < x < 2 π, \tan x is negative. \\ ∴ f (x) = \tan x - x < 0 \end{aligned}$

So, $f (x) = 0$ has no root in $\frac{3 π}{2}, 2 π$ .

We have, $f (π) = 0 - π < 0$

and $f \frac{3 π}{2} = \tan \frac{3 π}{2} - \frac{3 π}{2} > 0$

$∴ f (x) = 0$ has at least one root between $π$ and $\frac{3 π}{2}$ .

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Trigonometrical Equations 3 Question 7

8. The smallest positive root of the equation tan⁡x−x=0 lies in

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8. The smallest positive root of the equation $\tan x - x = 0$ lies in