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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Vertex

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

2.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $(1),$
$1 = 3^{2} - 4 c$
$\Rightarrow c = 2$

Suggest Corrections

Similar questions

Q. If least value of

f (x) = x^{2} + b x + c

- \frac{1}{4}

and maximum value of

g (x) = - x^{2} + b x + 2

occurs at

\frac{3}{2}

, then

c

is equal to

Q. If least value of

f (x) = x^{2} + b x + c

- \frac{1}{4}

and maximum value of

g (x) = - x^{2} + b x + 2

occurs at

\frac{3}{2}

, then

c

is equal to

Q. Let

f (x) = x^{2} + b_{1} x + c_{1}, g (x) = x^{2} + b_{2} x + c_{2}

. Let the real roots of

f (x) = 0

α, β

and real roots of

g (x) = 0

α + h, β + h

. The least value of

f (x)

- \frac{1}{4}

. The least value of

g (x)

occurs at

x = - \frac{7}{2}

The least value of

g (x)

Q. Let for

a \neq a_{1} \neq 0,

f (x) = a x^{2} + b x + c,

g (x) = a_{1} x^{2} + b_{1} x + c_{1}

and

p (x) = f (x) - g (x),

. If

p (x) = 0

only for

x = - 1

and

p (- 2) = 2

, then the value of

p (2)

is :

Q. If the polynomial

f (x) = a x^{3} + b x - c

is divisible by the polynomial

g (x) = x^{2} + b x + c,

then

a b =

Join BYJU'S Learning Program

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 $(1),$
$1 = 3^{2} - 4 c$
$\Rightarrow c = 2$