Match List I with List II
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LIST I |
LIST II |
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A. Slope of the tangent to curve $x^3 - 2x$ at $x=2$ |
I. -81 |
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B. Slope of line passing through the points (0, 2) and (5,-6) |
II. 10 |
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C. Point at which the tangent to the curve $y=\sqrt{4x-3}$ has its slope $\frac{2}{3}$ |
III. $-\frac{8}{5}$ |
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D. Slope of normal to the curve $y =\frac{x-2}{x-1}$ at x = 10 |
IV. (3,3) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
A. $y=x^3 - 2x$
$y'=3x^2-2$
Slope at $x = 2$
$y'|_{x=2}=3×2^2-2=10$
B. points $P(0, 2)$ and $Q(5,-6)$
Slope → $\frac{-6-2}{5-0}=-\frac{8}{5}$
C. $y=\sqrt{4x-3}$ for $y'=\frac{2}{3}$
$y'=\frac{1}{2}\frac{4}{\sqrt{4x-3}}⇒\frac{2}{3}=\frac{4}{2\sqrt{4x-3}}$
so $\sqrt{4x-3}=3=y$
$⇒4x-3=9=4x=12⇒x=3$ $⇒(3,3)$
D. $y =\frac{x-2}{x-1}$, so $y'=-\frac{(x-2)+(x-1)}{(x-1)^2}$
$y'=\frac{+1}{(x-1)^2}$ (slope of tangent)
slope of normal = $-(x-1)^2$ at $x = 10$
Slope of normal = $-(10-1)^2=-81$