In (A)-(C), [z] denotes the floor of z. Match List I to List II. List I: (A) If f ( x ) = sin − 1 x and lim x → ( 1/2 ) + f ( 3 x − 4 x 3 ) = a − 3 lim x → ( 1/2 ) + f ( x ) , find [a]. (B) If f (

Question

In (A)-(C), [z] denotes the floor of z. Match List I to List II. List I: (A) If $f (x) = sin^{- 1} x$ and $lim_{x \to (1/2)^{+}}$ $f (3 x - 4 x^{3}) = a - 3 lim_{x \to (1/2)^{+}} f (x)$ , find [a]. (B) If $f (x) = tan^{- 1} ((3 x - x^{3}) / (1 - 3 x^{2}))$ and $lim [h \to 0] (f (a + 3 h) - f (a)) / (3 h) = 3/ (1 + a^{2})$ for $- 1/3 < a < 1/ 3$ , find $[lim [h \to 0] (f (1/2 + 6 h) - f (1/2)) / (6 h)]$ . (C) If $cos^{- 1} (4 x^{3} - 3 x) = a + b cos^{- 1} x$ for $- 1 < x < - 1/2$ , find $[a + b + 2]$ . (D) If $f (x) = cos^{- 1} (4 x^{3} - 3 x), a = lim_{x \to (1/2)^{+}} f^{'} (x), b = lim_{x \to (1/2)^{-}} f^{'} (x)$ , find $a + b - 3$ . List II: (P) 2; (Q) 3; (R) 4; (S) -2; (T) -3.

Options:

Answer:

A

Solution:

A: $sin^{- 1} (3 x - 4 x^{3}) = π - 3 sin^{- 1} x$ to the right of $1/2$ , hence $a = π$ and $[a] = 3$ . B: $f = 3 tan^{- 1} x$ in the given interval, so $f^{'} (1/2) = 12/5$ , whose floor is 2 . $C : cos^{- 1} (4 x^{3} - 3 x) = 3 cos^{- 1} x - 2 π$ on $(- 1, - 1/2)$ , so $[a + b + 2] = [- 2 π + 5] = - 2$ . $D : f^{'} (1/ 2^{+}) = - 23$ and $f^{'} (1/ 2^{-}) = 23$ , hence $a + b - 3 = - 3$ .

TestHub · Accepted Answer

A: sin − 1 ( 3 x − 4 x 3 ) = π − 3 sin − 1 x to the right of 1/2 , hence a = π and [ a ] = 3 . B: f = 3 tan − 1 x in the given interval, so f ′ ( 1/2 ) = 12/5 , whose floor is 2 . C : cos − 1 ( 4 x 3 − 3 x ) = 3 cos − 1 x − 2 π on ( − 1 , − 1/2 ) , so [ a + b + 2 ] = [ − 2 π + 5 ] = − 2 . D : f ′ ( 1/ 2 + ) = − 2 3 ​ and f ′ ( 1/ 2 − ) = 2 3 ​ , hence a + b − 3 = − 3 .

Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Options:

Doubts & Discussion