Correct Answer - Option 4 : f has no extreme value
Concept:
\(r = {\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_{\left( {a,b} \right)}},\;s = {\left( {\frac{{{\partial ^2}f}}{{\partial x\partial y}}} \right)_{\left( {a,b} \right)}},\;t = \;{\left( {\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right)_{\left( {a,b} \right)}}\)
- If f will be maximum or minimum, if rt – s2 is +ve → (> 0), s2 – rt is –ve → (<0)
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Based on r → -ve ⇒ f is maximum at (a, b)
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Based on r → +ve ⇒ f is minimum at (a, b)
- If rt – s2 is –ve, (a, b) saddle point i.e. neither maximum nor minimum
Calculation:
Given:
\(f\left( {x,y} \right) = \;{y^2} - {x^2}\)
Differentiating with respect to fx,
-2x = 0 ⇒ x = 0
Differentiating with respect to fy,
2y = 0 ⇒ y = 0
Again differentiating for second time, we get
r = fxx = -2
s = fyy = -2
Now, fxy = 0 = s
\(rt - {s^2} = \; - 4\;\left( {negative} \right)\)
∴ Neither maximum nor minimum