Question

If the curves $y=b{x}^2+c$ and $y=x^3+ax$ have a common tangent at $(-1,0)$ then the possible value of $a^4+b^4+c^4$ is

• A. $12$
• B. $20$
• C. $3$
• D. $2$

Answer 1

The correct option is C $3$

Let $y_1=b{x}^2+c$ & $y_2=x^3+ax$

as both the curves have common tangent at (-1,0)

So,

$\frac{d{y}_1}{dx}{|}_{(-1,0)}=\frac{d{y}_2}{dx}{|}_{(-1,0)}$

$(2bx+0){|}_{(-1,0)}=(3x^2+a){|}_{(-1,0)}$

$-2b=3+a\Rightarrow a+2b+3=0...\left(i\right)$

As (-1, 0) lying on both the curves, so it will satisfy

both the curves.

$\therefore y_1$ at $(-1,0)$ & $y_2$ at (-1, 0)

$0=b+c...\left(ii\right)$

$=0-1-a$

$\Rightarrow a=-1...\left(iii\right)$

From $e{q}^n$ (i) & (ii)

$a=-1$ & $a+2b+3=0$

$-1+3+2b=0$

$b=-1$

From $e{q}^n$ (ii)

$b+c=0$

$-b=+c$

$C=1$

$\therefore a^4+b^4+c^4=(-1{)}^4+(-2{)}^4+(1{)}^4$

$[a^4+b^4+c^4=3]$