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Exercise 7.10 · Q7

Q.The slope of the line normal to the curve f(x)=2cos4xf(x)=2\\cos4x at x=dfracpi12x=\\dfrac{\\pi}{12} is

(1) −43-4\sqrt3
(2) −4-4
(3) 312\dfrac{\sqrt3}{12}
(4) 434\sqrt3
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Differentiate to find the tangent's slope, then take the negative reciprocal for the normal.

Step 1. Differentiate. f(x)=2cos⁡4x⇒f′(x)=−8sin⁡4xf(x)=2\cos4x\Rightarrow f'(x)=-8\sin4x.

Step 2. Evaluate at x=π/12x=\pi/12. 4x=π/34x=\pi/3, sin⁡(π/3)=32\sin(\pi/3)=\dfrac{\sqrt3}{2}.

f′ ⁣(π12)=−8(32)=−43.f'\!\left(\frac{\pi}{12}\right)=-8\left(\frac{\sqrt3}{2}\right)=-4\sqrt3. …

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