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We have,
f(x) = x2 - 3x + 4
Now,
f(2x + 1) = (2x + 1)2 - 3(2x + 1) + 4
= 4x2 + 1 + 4x - 6x - 3 + 4
= 4x2 - 2x + 2
It is given that
f(x) = f(2x + 1)
⟹ x2 — 3x + 4 = 4x2 - 2x + 2
⟹ 0 = 4x2 - x2 - 2x + 3x + 2 - 4
⟹ 3x2 + x - 2 = 0
⟹ 3x2 + 3x - 2x - 2 = 0
⟹ 3x (x + 1) - 2 (x + 1) = 0
⟹ (x +1) (3x -2) = 0
⟹ x + 1 = 0 or 3x - 2 = 0
⟹ x = -1 or x = 2/3
f(x) = (x – a)2 (x – b)2
f(a + b) = (a + b – a)2 (a + b – b)2
= b2a2
⟹ f(a + 6) = a2b2
Hence, proved
∴ f[f(x)] = x Hence, proved
Adding equation (i) and equation (ii), we get
∴ f(tan θ) = sin 2θ Hence, proved.
f(x) = (a - xn)1/n, a > 0
f{f(x)} = f(a - xn)1/n
= [a-{(a - xn)1/n}n]1/n
= [a - (a - xn)]1/n
= [a - a + xn]1/n
= (xn)1/n
= (x)n×1/n
= x
∴ f{f(x)} = x Hence, proved.
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