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Solution


Q.

Let the maximum and minimum values of (8x–x2–12–4)2+(x-7)2, x∈R be M and m, respectively. Then M2–m2 is equal to __________.          [2024]

Ans.

(1600)

Let y = 8x–x2–12

⇒ y2=8x–x2–12

           =–(x2–8x+16)+4=–(x–4)2+4

⇒(x–4)2+y2=4          ... (i)

which is a circle with centre (4, 0) and radius 2.

Now, (y–4)2+(x–7)2 represent distance of (x, y) from (7, 4)

M = Maxium distance = (2–7)2+(0–4)2=25+16=41

m = Minimum distance = Distance between P and (7, 4)

where P is the intersection of circle with line joining (4, 0) and (7, 4).

Now, equation of line joining (4, 0) and (7, 4) is given by

(y–0)=4–07–4(x–4)

i.e.,  y = 43(x–4)

On substituting the value of y is in (i), we get

(x–4)2+169(x–4)2=4

⇒ (x–4)2[259]=4 ⇒ (x–4)2=3625 ⇒ x–4=±65

⇒ x=65+4=265, y=85         (Neglect negative sign)

So, m=(265–7)2+(85–4)2=8125+14425=22525=9

∴  M2–m2=412–92=1600.