Match list I with list II
| List - I | List - II | ||
| A. | Slope of tangent to the curve \(y = x^3 - x\) at \(x = 2\) | I. | −2 |
| B. | Slope of tangent to the curve \(y = 3x^3 - 4x\) at \(x = 0\) | II. | 11 |
| C. | Slope of normal to the curve \(y = \sin\theta\) at \(\theta = \displaystyle\frac{\pi}{3}\) | III. | 2 |
| D. | Slope of normal to the curve \(y = \cos\theta\) at \(\theta = \displaystyle\frac{\pi}{6}\) | IV. | −4 |
Choose the correct answer from the option given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-IV, C-I, D-III
The slope of the tangent is the value of the derivative at the point, and the slope of the normal is \(-\frac{1}{\text{slope of tangent}}\).
(A) \(y = x^3 - x\) at x = 2
This matches (II) 11.
\(y' = 3x^2 - 1\)
\(y'(2) = 12 - 1 = 11\)
(B) \(y = 3x^3 - 4x\) at x = 0
This matches (IV) −4.
\(y' = 9x^2 - 4\)
\(y'(0) = -4\)
(C) \(y = \sin\theta\) at \(\theta = \frac{\pi}{3}\)
This matches (I) −2.
\( \displaystyle \frac{dy}{d\theta} = \cos\frac{\pi}{3} = \frac{1}{2} \)
\( \displaystyle \text{slope of normal} = -\frac{1}{1/2} = -2 \)
(D) \(y = \cos\theta\) at \(\theta = \frac{\pi}{6}\)
This matches (III) 2.
\( \displaystyle \frac{dy}{d\theta} = -\sin\frac{\pi}{6} = -\frac{1}{2} \)
\( \displaystyle \text{slope of normal} = -\frac{1}{-1/2} = 2 \)
So the matching is A-II, B-IV, C-I, D-III.