The possible value(s) of k for which Lim x → ∞ π 8 x 3 c o t − 1 ∣ k x ∣ + k 2 x 6 s i n x 3 1 − 3 k x 3 2 x 3 − ( t a n − 1 x ) 3 = 2 1 is false, is

Question

The possible value(s) of k for which Lim x → ∞ ​ π 8 ​ x 3 c o t − 1 ∣ k x ∣ + k 2 x 6 s i n x 3 1 ​ − 3 k x 3 2 x 3 − ( t a n − 1 x ) 3 ​ = 2 1 ​ is false, is

TestHub · Accepted Answer

To find the limit, divide numerator and denominator by $x^{3}$ .

For $k = 0$ , the limit is $2/ (\frac{8}{π} \cdot \frac{π}{2}) = 1/2$ .

For $k \neq = 0$ , the limit is $2/ (k^{2} - 3 k) = 1/2$ , which gives $k^{2} - 3 k - 4 = 0$ , so $k = 4, - 1$ .

The question asks for the value(s) of $k$ for which the statement is FALSE.

The statement is true for $k = 0, 4, - 1$ . Thus, the statement is false for values not in this set.

The options are not properly formatted, but assuming they refer to values of $k$ , the correct option would be a value not equal to $0, 4, - 1$ .

Mathematics - Limits Question with Solution | TestHub

Options:

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JEE Main - 1 year (12th class)

JEE Advanced (2028)