Let fx-4-y2 x2-2 x y-1 and gy be the minimum value of fx- If y - then the maximum value of gy is

F(x)=(4+y^2) x^2+(2 y) x+1 Here the coefficient of x^2 is 4+y^2 nbsp;which is always positive. nbsp; Thus y=f(x) is a quadratic curve opened upwards. The m ...

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Standard XII

Mathematics

Location of Roots

Let fx=4+y2 x...

Question

Let $f (x) = (4 + y^{2}) x^{2} + 2 x y + 1$ and $g (y)$ be the minimum value of $f (x)$ . If $y \in R,$ then the maximum value of $g (y)$ is

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Solution

The correct option is A $1$
$f (x) = (4 + y^{2}) x^{2} + (2 y) x + 1$
Here the coefficient of $x^{2}$ is $4 + y^{2}$ which is always positive.
Thus $y = f (x)$ is a quadratic curve opened upwards.

The minimum value of the curve $y = a x^{2} + b x + c, a > 0$ is $\frac{4 a c - b^{2}}{4 a}$
Minimum value of $f (x) = (4 + y^{2}) x^{2} + (2 y) x + 1$ is
$g (y) = \frac{4 (4 + y^{2}) - 4 y^{2}}{4 (4 + y^{2})} = \frac{4}{4 + y^{2}}$

So, $g (y) = \frac{4}{4 + y^{2}} \in (0, 1], \forall y \in R$
Hence, the maximum value of $g (y)$ is $1.$

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Let f(x)=(4+y2)x2+2xy+1 and g(y) be the minimum value of f(x). If y∈R, then the maximum value of g(y) is

Let $f (x) = (4 + y^{2}) x^{2} + 2 x y + 1$ and $g (y)$ be the minimum value of $f (x)$ . If $y \in R,$ then the maximum value of $g (y)$ is