Given that f(x) = x² + 2x – 5 … (1)
f '(x) = 2x + 2
At stationary points, f'(x) = 0
\(\Rightarrow\)  2x + 2 = 0 \(\Rightarrow\)  x = –1
f(x) has stationary value at x = –1
When x = –1, from (1)
f(–1) = (–1)² + 2(–1) – 5 = – 6
Stationary value of f (x) is – 6
Hence stationary point is (–1,–6)

(i) p = (a - bx)²
\({dp\over dx}=2(a-bx)(-b)=-2b(a-bx)\)
Elasticity of demand (\(\eta _d\)) \(={-p\over x}.{dx\over dp}\)
\(\Rightarrow -{(a-bx)^2\over x}({-1\over 2b(a-bx)})={a-bx\over x(2b)}\)
When \(\eta_d=1\)
\(\frac{a-b x}{2 b x} =1 \)
\(a-b x =2 b x \)
\(3 b x =a \)
\(x =\frac{a}{3 b} \text { units }\)
(ii) \(p=a-b x^2\)
\(\frac{d p}{d x}=-2 b x\)
\(\eta_d\) Elasticity of demand \(=-\frac{5}{x} \frac{d x}{d p}\)
\(=-\frac{a-b x^2}{x} \cdot \frac{1}{-2 b x}\)
\(=\frac{a-b x^2}{2 b x^2}\)
When \(\eta_d=1\)
\(\frac{a-b x^2}{2 b x^2}=1\) 
\(a-b x^2=2 b x^2\)
\(3 b x^2=a \Rightarrow x^2=\frac{a}{33}\)
\(N=\sqrt{\frac{a}{3 b}}\)units

Given C(x) \(=\frac { 1 }{ 10 } { x }^{ 3 }-4{ x }^{ 2 }-20x+7\) 
(i)  AC =  Avarge Cost = C/x \(=\frac { { x }^{ 2 } }{ 10 } -4{ x }-20+\frac { 7 }{ x } \)
(ii) AVC = Avarge Variable Cost  = f(x)/x 
\(=\frac { { x }^{ 2 } }{ 10 } -4{ x }-20\) 
(iii) AFC = Avarge fixed Cost = \(\frac { k }{ x } =\frac { 7 }{ x } \)
(iv) MC = Marginal Cost = \(\frac { dc }{ dx } \)
= \(\frac { 3x ^ 2 }{ 10 } { }-{8x }-20\)
(v) MAC = Marginal Average Cost  = \(\frac { d (AC) }{ dx }\)
= \(\frac { x }{ 5 } -4-\frac { 7 }{ { x }^{ 2 } } \)

y = x³ + 19
\({dy\over dx}=3x^2 \ \ \ ...(1)\)
\({dy\over dx}=27\ \ \ ...(2)\) [Given]
From (1) and (2), we get
3x² = 27 ⇒ x = ±3

x = 2p²-5p+1
\({dx\over dp}=4p-5\)
Elasticity of supply: \(η_s={P\over x}.{dx\over dp}\)
\(={p\over 2p^2-5p+1}.(4p-5)\) \(={4p^2-5p\over 2p^2-5p+1}\)