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Q.(i) Use differentials to approximate √36.6.

(3)
(ii) Consider the curve y = 3x.
(a) Find the slope of the tangent at x = 4.
(1)
(b) Find the equation of tangent at x = 4. (2)
Kerala DhseKerala DHSE Plus Two Board 2025Subjective· 6mImportance★★★★★
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Use dy≈f′(x) dxdy\approx f'(x)\,dx to approximate 36.6\sqrt{36.6} near the perfect square 3636. For the given straight line y=3xy=3x, the slope is the constant coefficient of xx, and the tangent at any point coincides with the line itself.

(i) Let y=f(x)=xy=f(x)=\sqrt{x}. Take x=36x=36 (a nearby perfect square) and dx=36.6−36=0.6dx=36.6-36=0.6.

f′(x)=12x⇒dy=1236×0.6=0.612=0.05f'(x) = \frac{1}{2\sqrt{x}} \Rightarrow dy = \frac{1}{2\sqrt{36}}\times0.6 = \frac{0.6}{12} = 0.05

36.6≈36+dy=6+0.05=6.05\sqrt{36.6} \approx \sqrt{36}+dy = 6+0.05 = 6.05

(ii) Curve: y=3xy=3x (a straight line through the origin with constant slope 3).

(a) Since y=3xy=3x, dydx=3\dfrac{dy}{dx}=3 for every xx, in particular at x=4x=4. So the slope of the tangent at x=4x=4 is 33.

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