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Exercise 3.12 · Q3

Q.The maximum value of 4sin⁡2x+3cos⁡2x+sin⁡x2+cos⁡x24\sin^2x+3\cos^2x+\sin\dfrac x2+\cos\dfrac x2 is

(1) 4+24+\sqrt2
(2) 3+23+\sqrt2
(3) 99
(4) 44
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✓ Free question

Both pieces, 4sin⁡2x+3cos⁡2x=3+sin⁡2x4\sin^2x+3\cos^2x=3+\sin^2x and sin⁡(x/2)+cos⁡(x/2)\sin(x/2)+\cos(x/2), hit their own maximum at the same x=90∘x=90^\circ — so the maxima simply add.

Step 1. 4sin⁡2x+3cos⁡2x=3sin⁡2x+3cos⁡2x+sin⁡2x=3+sin⁡2x4\sin^2x+3\cos^2x=3\sin^2x+3\cos^2x+\sin^2x=3+\sin^2x, whose maximum is 3+1=43+1=4, attained whenever sin⁡2x=1\sin^2x=1, i.e. x=90∘,270∘,…x=90^\circ,270^\circ,\ldots

Step 2. sin⁡x2+cos⁡x2=2sin⁡(x2+45∘)\sin\tfrac x2+\cos\tfrac x2=\sqrt2\sin\left(\tfrac x2+45^\circ\right), whose maximum is 2\sqrt2, attained when x2+45∘=90∘\tfrac x2+45^\circ=90^\circ, i.e. x=90∘x=90^\circ (plus multiples of 720∘720^\circ).

Step 3. Both individual maxima occur at x=90∘x=90^\circ simultaneously: 4sin⁡2(90∘)+3cos⁡2(90∘)=4(1)+3(0)=44\sin^2(90^\circ)+3\cos^2(90^\circ)=4(1)+3(0)=4, and sin⁡45∘+cos⁡45∘=12+12=2\sin45^\circ+\cos45^\circ=\tfrac1{\sqrt2}+\tfrac1{\sqrt2}=\sqrt2.

Step 4. So the maximum of the sum is 4+24+\sqrt2, matching option (1).

✓Final answer

Option (1): 4+24+\sqrt2.

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