Match the following. List-I List-II (a) If ( tan − 1 3 x ) 2 − 4 tan − 1 3 x − 5 = 0 , then x = (p) 2 (b) If { tan − 1 ( 3 x + 2 ) } 2 + 2 tan − 1 ( 3 x + 2 ) = 0 , then x = (q) − 3 tan 1 (c) If 3

Question

Match the following. List-I List-II (a) If ( tan − 1 3 x ​ ) 2 − 4 tan − 1 3 x ​ − 5 = 0 , then x = (p) 2 (b) If { tan − 1 ( 3 x + 2 ) } 2 + 2 tan − 1 ( 3 x + 2 ) = 0 , then x = (q) − 3 tan 1 (c) If 3 ( tan − 1 x ) 2 − 4 π tan − 1 x + π 2 = 0 , then x = (r) 1 (d) Given that 0 ≤ x ≤ 2 1 ​ , the value of tan [ sin − 1 ( 2 ​ x ​ + 2 1 − x 2 ​ ​ ) − sin − 1 x ] is (s) 3 ​ (t) − 3 2 ​

TestHub · Accepted Answer

(a) tan − 1 3 x ​ = 2 4 ± 16 + 20 ​ ​ = 5 (rejected) or -1, so 3 x ​ = − tan 1 , x = − 3 tan 1 (q). (b) tan − 1 ( 3 x + 2 ) = − 2 or 0 ; 3 x + 2 = 0 ⇒ x = − 3 2 ​ (t). (c) tan − 1 x = 6 4 π ± 16 π 2 − 12 π 2 ​ ​ = π (rejected) or 3 π ​ , so x = 3 ​ (s). (d) Put x = sin y , 0 ≤ y ≤ 6 π ​ : sin − 1 ( 2 ​ 1 ​ sin y + 2 ​ c o s y ​ ) = sin − 1 ( sin ( y + 4 π ​ ) ) = y + 4 π ​ , so the value is tan 4 π ​ = 1 (r).

List-I	List-II
(a) If $(tan^{- 1} \frac{x}{3})^{2} - 4 tan^{- 1} \frac{x}{3} - 5 = 0$ , then $x =$	(p) 2
(b) If ${tan^{- 1} (3 x + 2)}^{2} + 2 tan^{- 1} (3 x + 2) = 0$ , then $x =$	(q) $- 3 tan 1$
(c) If $3 (tan^{- 1} x)^{2} - 4 π tan^{- 1} x + π^{2} = 0$ , then $x =$	(r) 1
(d) Given that $0 \leq x \leq \frac{1}{2}$ , the value of $tan [sin^{- 1} (\frac{x}{2} + \frac{1 - x ^{2}}{2}) - sin^{- 1} x]$ is	(s) $3$
	(t) $- \frac{2}{3}$

Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

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