CUET Preparation Today
Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(IV), (B)-(I), (C)-(II), (D)-(III) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
(A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Use: \(\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C\), \(\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C\) Use: \(\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right)+C\), \(\int \frac{dx}{\sqrt{x^2-a^2}}=\log\!\left|x+\sqrt{x^2-a^2}\right|+C\) (A) \(\int \frac{dx}{x^2-16}\ \to\ \frac{1}{8}\log\left|\frac{x-4}{x+4}\right|\) ⇒ (III) (B) \(\int \frac{dx}{x^2+16}\ \to\ \frac{1}{4}\tan^{-1}\!\left(\frac{x}{4}\right)\) ⇒ (IV) (C) \(\int \frac{dx}{16-x^2}\ \to\ \frac{1}{8}\log\left|\frac{4+x}{4-x}\right|\) ⇒ (I) (D) \(\int \frac{dx}{\sqrt{x^2-16}}\ \to\ \log\!\left|x+\sqrt{x^2-16}\right|\) ⇒ (II) Matching: A–III, B–IV, C–I, D–II |
