If A,B,C,D are the slopes of the tangents to the curves y=6+x−x2 at (2,4), y=sinx at x=π4,y=tanx at x=π3 and y2=10x at (10,10) respectively then the ascending order is

y=6+x−x2 dydx=1−2x⇒slope=−3=A y=sinx dydx=cosx⇒slope=12=B y=tanx dydx=sec2x⇒slope=4=C y2=10x dydx=102y=5y⇒slope=12=D ascending is A, D, B, C

If A,B,C,D are the slopes of the tangents to the curves y=6+x−x2 at (2,4), y=sinx at x=π4,y=tanx at x=π3 and y2=10x at (10,10) respectively then the ascending order is

y=6+x−x2 dydx=1−2x⇒slope=−3=A y=sinx dydx=cosx⇒slope=12=B y=tanx dydx=sec2x⇒slope=4=C y2=10x dydx=102y=5y⇒slope=12=D ascending is A, D, B, C

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If A,B,C,D are the slopes of the tangents to the curves $y = 6 + x - x^{2}$ at $(2, 4),$ $y = \sin x$ at $x = \frac{π}{4}, y = \tan x$ at $x = \frac{π}{3}$ and $y^{2} = 10 x$ at $(10, 10)$ respectively then the ascending order is

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A, B, C, D

B, C, D, A

A, D, B, C

C, D, B, A

answer is C.

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Detailed Solution

$y = 6 + x - x^{2}$

$\frac{d y}{d x} = 1 - 2 x \Rightarrow slope = - 3 = A$

$y = \sin x$

$\frac{d y}{d x} = \cos x \Rightarrow slope = \frac{1}{\sqrt{2}} = B$

$y = \tan x$

$\frac{d y}{d x} = \sec^{2} x \Rightarrow slope = 4 = C$

$y^{2} = 10 x$

$\frac{d y}{d x} = \frac{10}{2 y} = \frac{5}{y} \Rightarrow slope = \frac{1}{2} = D$

ascending is A, D, B, C

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If A,B,C,D are the slopes of the tangents to the curves y=6+x−x2 at (2,4), y=sinx at x=π4,y=tanx at x=π3 and y2=10x at (10,10) respectively then the ascending order is

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If A,B,C,D are the slopes of the tangents to the curves $y = 6 + x - x^{2}$ at $(2, 4),$ $y = \sin x$ at $x = \frac{π}{4}, y = \tan x$ at $x = \frac{π}{3}$ and $y^{2} = 10 x$ at $(10, 10)$ respectively then the ascending order is