Question

In the given figure vertices of $△ABC$ lie on y = $f\left(x\right)=a{x}^2+bx+c$. (vertex lies on y axis). The $△ABC$ is right angled isosceles triangle whose hypotenuse $AC=8units$, then

• A. $y=f\left(x\right)=\frac{x^2}{4}-4$
• B. minimum value of f(x) is $-4$
• C. Number of integral values of x lying between the roots of f(x) = 0 are $7$
• D. Area of the $△ABC$ is $16$ sq.units

Answer 1

The correct option is B minimum value of f(x) is $-4$

$y=f\left(x\right)=a{x}^2+bx+c$ $f^{\prime }\left(x\right)=2ax+b$

$AC=8units$

$\Rightarrow AO=OC=4units$

Coordinates of $A(-4,0),B(?),C=(4,0)$

In $△ABC$

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

$AB=BC$ ( sides of isosceles triangle)

$(AC{)}^2=2(AB{)}^2$

$64=2(AB{)}^2$

$(AB{)}^2=32$

$AB=\sqrt{32}$

$AB=\sqrt{16\times 2}$

$AB=4\sqrt{2}units$

Now In $△AOB,△AOB={90}^o$

$(AB{)}^2=(OA{)}^2+(OB{)}^2$

$(A\sqrt{2}{)}^2=(4{)}^2+(OB{)}^2$

$32=16+(OB{)}^2$

$(OB{)}^2=16$

$OB=4$

Coordinates of $B$ is $(0,-4)$

$A,B$ and $C$ lies on curve $a{x}^2+bx+c=f\left(x\right)=y$........(1)

$\Rightarrow 16a-4b+c=0$

$\Rightarrow 16a+4b+c=0$

$c=4$

Hence $a=1/4,b=0,c=-4$

put eq (1)

$y=f\left(x\right)=\frac{1}{4}{x}^2-4$ option $\left(A\right)$

Minimize $e\times \Delta t$ at point $(0,-4)$, [ clear from the figure]

minimum value $=y=\frac{1}{4}\left(0\right)-4=y=-4$ option (B)

Area $=1/2\times Base\times Height$

$=1/2\times 8\times 4=16squnits$

$y=f\left(x\right)=\frac{1}{4}{x}^2-4$

$f\left(x\right)=0$

$\Rightarrow \frac{1}{4}{x}^2-4=0$

$\frac{x^2}{4}=4$

$x^2=16$

$x=\pm 4$

Integral value being b/w $-4,4$ are $-3,-2,-1,0,1,2,3$

$=7values$

option- C