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(a) To show f(x) is not differentiable at x=1:
f(x) = { x², if x≥1; x, if x<1 }
Left-hand derivative (LHD) at x=1:
LHD = lim (h->0) [f(1-h) - f(1)] / -h = lim (h->0) [(1-h) - 1²] / -h = lim (h->0) [1-h-1] / -h = lim (h->0) [-h] / -h = 1
Right-hand derivative (RHD) at x=1:
RHD = lim (h->0) [f(1+h) - f(1)] / h = lim (h->0) [(1+h)² - 1²] / h = lim (h->0) [1 + 2h + h² - 1] / h = lim (h->0) [2h + h²] / h = lim (h->0) [2 + h] = 2
Since LHD ≠ RHD, f(x) is not differentiable at x=1.
(b) To find λ if f(x) is continuous at x=0:
f(x) = { (sin²λx)/x², if x≠0; 1, if x=0 }
For f(x) to be continuous at x=0, lim (x->0) f(x) = f(0)
lim (x->0) (sin²λx)/x² = 1
lim (x->0) (sin λx / x)² = 1
lim (x->0) (λ * sin λx / λx)² = 1
λ² * lim (x->0) (sin λx / λx)² = 1
Since lim (x->0) (sin λx / λx) = 1, we have λ² * 1² = 1
λ² = 1
λ = ±1
Correct Answer: (a) LHD ≠ RHD, hence not differentiable. (b) λ = ±1
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