Match List-I with List-II
| List-I | List-II | ||
| Function \( y= \) | \( \left(\displaystyle\frac{dy}{dx}\right) \) at \( x=0 \) | ||
| (A) | \( x+e^x \) | (I) | 1 |
| (B) | \( \tan^{-1}x-\displaystyle\frac{x}{2} \) | (II) | \( \displaystyle\frac{1}{2} \) |
| (C) | \( \tan^{-1}x \) | (III) | 0 |
| (D) | \( x+\cos^{-1}x \) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
Each function is differentiated and x = 0 is put in.
(A) \( x+e^x \)
This matches (IV) 2.
\( \displaystyle \frac{dy}{dx}=1+e^x \)
\( \text{at } x=0:\ 1+1=2 \)
(B) \( \tan^{-1}x-\frac{x}{2} \)
This matches (II) \( \frac{1}{2} \).
\( \displaystyle \frac{dy}{dx}=\frac{1}{1+x^2}-\frac{1}{2} \)
\( \displaystyle \text{at } x=0:\ 1-\frac{1}{2}=\frac{1}{2} \)
(C) \( \tan^{-1}x \)
This matches (I) 1.
\( \displaystyle \frac{dy}{dx}=\frac{1}{1+x^2} \)
\( \text{at } x=0:\ 1 \)
(D) \( x+\cos^{-1}x \)
This matches (III) 0.
\( \displaystyle \frac{dy}{dx}=1-\frac{1}{\sqrt{1-x^2}} \)
\( \text{at } x=0:\ 1-1=0 \)
So the matching is (A)-(IV), (B)-(II), (C)-(I), (D)-(III).
Tip: Remember the minus sign in \( \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}} \); with a plus sign (D) would give 2 instead of 0.