If least value of fx-x2-bx-c be -14 and maximum value of gx-x2-bx-2 occurs at 32- then c is equal to

F(x)=x^2+bx+c Least value of f(x)= D4= 14 ⇒ D=b^2 4c=1 ⋯(1) g(x)= x^2+bx+2 Maximum value of g(x) is at x= b 2=32 ⇒ b=3 From (1), 1=3^2 4c ⇒ c=2 ...

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Standard XIII

Mathematics

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If least valu...

Question

If least value of $f (x) = x^{2} + b x + c$ be $- \frac{1}{4}$ and maximum value of $g (x) = - x^{2} + b x + 2$ occurs at $\frac{3}{2}$ , then $c$ is equal to

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Solution

$f (x) = x^{2} + b x + c$
Least value of $f (x) = - \frac{D}{4} = - \frac{1}{4}$
$\Rightarrow D = b^{2} - 4 c = 1 \dots (1)$

$g (x) = - x^{2} + b x + 2$
Maximum value of $g (x)$ is at $x = \frac{- b}{- 2} = \frac{3}{2}$
$\Rightarrow b = 3$
From $(1),$
$1 = 3^{2} - 4 c$
$\Rightarrow c = 2$

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If least value of f(x)=x2+bx+c be −14 and maximum value of g(x)=−x2+bx+2 occurs at 32, then c is equal to

f(x)=x2+bx+c Least value of f(x)=−D4=−14 ⇒D=b2−4c=1 ⋯(1) g(x)=−x2+bx+2 Maximum value of g(x) is at x=−b−2=32 ⇒b=3 From (1), 1=32−4c ⇒c=2

If least value of $f (x) = x^{2} + b x + c$ be $- \frac{1}{4}$ and maximum value of $g (x) = - x^{2} + b x + 2$ occurs at $\frac{3}{2}$ , then $c$ is equal to

$f (x) = x^{2} + b x + c$
Least value of $f (x) = - \frac{D}{4} = - \frac{1}{4}$
$\Rightarrow D = b^{2} - 4 c = 1 \dots (1)$

$g (x) = - x^{2} + b x + 2$
Maximum value of $g (x)$ is at $x = \frac{- b}{- 2} = \frac{3}{2}$
$\Rightarrow b = 3$
From $(1),$
$1 = 3^{2} - 4 c$
$\Rightarrow c = 2$