IIT JEE 1983 Maths - 'Adapted' - FIB to Multi-Correct Q10

Calculus Level 5

The larger of $\cos (\ln \theta)$ and $\ln (\cos \theta)$ is $\text{\_\_\_}$ . Which of the following can fill the blank correctly?

(A) $\cos(\ln\theta)$ in the interval $(0,\ e^{-\pi/2})$
(B) $\cos(\ln\theta)$ in the interval $(e^{-\pi/2},\ 1)$
(C) $\ln(\cos\theta)$ in the interval $(1,\ \pi/2)$
(D) $\ln(\cos\theta)$ in the interval $(\pi/2,\ e^{\pi/2})$

Enter your answer as a 4 digit string of 1s and 9s - 1 for correct option, 9 for wrong. Eg. 1199 indicates A and B are correct, C and D are incorrect. None, one or all may also be correct.

In case you are preparing for IIT JEE, you may want to try IIT JEE 1983 Mathematics Archives .

The answer is 9199.

1 solution

Ajinkya Shivashankar
Feb 3, 2017

$f=\ln(\cos(\theta))$ is always negative (or 0). $g=\cos(\ln(\theta))$ is between (-1,1).

f is a periodic function and defined only where $\cos(\theta) \geq 0$ .

For A: Interval begins from 0 so we check values near 0.

As $\theta$ gets close to 0 , g oscillates between its extreme values, while f near 0 is almost 0. Thus they intersect each other a lot and no function is clearly larger. Thus A is incorrect.

For B: We know that f is always -ve. In this range , g is always positive. Hence this is correct.

For C and D: Again, In this range, g is always positive. Hence this is wrong.

Thus answer is 9199 .

Note: We cannot compare $\ln(-something)$ with a real number. So the answer for D is False and even if it was the other way round (i.e if it is asked if g is more than f) it will still be False.

IIT JEE 1983 Maths - 'Adapted' - FIB to Multi-Correct Q10

In case you are preparing for IIT JEE, you may want to try IIT JEE 1983 Mathematics Archives .

The answer is 9199.

1 solution

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