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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{n\to\infty}\left[\left(\cos\dfrac{k\pi}{4}\right)^{n}-\left(\cos\dfrac{k\pi}{6}\right)^{n}\right]=0

(where

k

is an integer), then which of the following statement(s) is(are) correct?

k

is neither divisible by

4

nor by

6

.correct

k

is divisible by

12

k

is divisible by

24

.correct

k

is either divisible by

24

k

is neither divisible by

4

nor by

6

.correct

Solution

Step 1: For the limit to be zero, both terms must individually have limits and they must be equal. Step 2:

\cos\dfrac{k\pi}{4}

takes values in

\{1,-1,0,\pm\dfrac{1}{\sqrt{2}}\}

depending on

k\pmod 8

. Similarly

\cos\dfrac{k\pi}{6}

depends on

k\pmod{12}

. Step 3: Case analysis based on book solution: If

k

is a multiple of

24

\cos\dfrac{k\pi}{4}=\cos(6m\pi)=1

and

\cos\dfrac{k\pi}{6}=\cos(4m\pi)=1

. So we get

\lim(1^{n}-1^{n})=0

. Valid. If

k

is a multiple of

12

but not

24

\cos\dfrac{k\pi}{4}=\cos(3m\pi)=(-1)^{m}

, and

\cos\dfrac{k\pi}{6}=\cos(2m\pi)=1

. The limit becomes

\lim((-1)^{n}-1^{n})

, which doesn't go to

0

. Invalid. If

k

is not divisible by

4

6

: Then

\left|\cos\dfrac{k\pi}{4}\right|<1

and

\left|\cos\dfrac{k\pi}{6}\right|<1

, so both terms

\to 0

n\to\infty

. Valid. Step 4: From the book's analysis: For

k

multiple of

24

: limit is

0

. Valid. For

k

multiple of

12

(not

24

): limit involves

(-1)^{n}-1^{n}

, oscillates. Invalid. For

k

multiple of

4

(not

12

\cos\dfrac{k\pi}{4}=\pm 1

\cos\dfrac{k\pi}{6}\neq\pm 1

, so first term oscillates while second goes to

0

. Invalid. For

k

multiple of

6

(not

12

): similar issue. Invalid. For

k

neither divisible by

4

nor

6

: both cosines have absolute value

<1

, both terms

\to 0

. Valid. Step 5: So valid cases are: -

k

divisible by

24

, or -

k

neither divisible by

4

nor

6

. This matches option (4). Option (1) is also correct as a sufficient (subset) condition. Option (3) is correct as another sufficient condition. Correct answers: (1), (3) and (4)

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