Question

Let $f\left(x\right)=\{\begin{array}{cc}5,& x=0\\ x^2+5x+a,& 0<x<2\\ bx+c,& 2\le x\le 3\end{array}$
If $f\left(x\right)$ satisfies all the conditions of Lagrange's mean value theorem in $[0,3]$ and $P(\lambda ,f(\lambda \left)\right)$ is the point on the curve f(x) in [0, 3], where tangent is parallel to the chord joining the end points, then the correct relation is/are

• A. $a+2b+4\lambda =\frac{89}{3}$
• B. $b+c+6\lambda =18$
• C. $a+2b+3c+4\lambda =\frac{94}{3}$
• D. $2a+3\lambda =9$

Answer 1

Answer: (B), (C)

$a+2b+3c+4\lambda =\frac{94}{3}$

$\because f\left(x\right)$ satisfies LMVT on interval [0, 3]
$\therefore f\left(x\right)$ is continuous at x = 0, 2 and 3 and f(x) is differentiable at x = 2

$\therefore 5=a$ ...(i)
and $4+10+a=2b+c$
$\therefore 2b+c=19$ ...(ii)
and $2x+5=b\text{at}x=2$
$b=9\Rightarrow c=1$ ...(iii)

$\because f^{\prime }\left(\lambda \right)=\frac{f\left(3\right)-f\left(0\right)}{3-0}$

$f^{\prime }\left(\lambda \right)=\frac{28-5}{3}$

$2\lambda +5=\frac{23}{3}$

$\therefore \lambda =\frac{4}{3}$

$\therefore a+2b+3c+4\lambda =5+18+3+\frac{16}{3}$

$=\frac{94}{3}$