Question

The values of $\lambda$ for which the equation

$(x^2+3x+5{)}^2-(\lambda -3\left)\right(x^2+3x+4\left)\right(x^2+3x+5)+(\lambda -4\left)\right(x^2+3x+4{)}^2=0$

has at least one real root are

• A. $(5,\frac{39}{7})$
• B. $[5,\frac{39}{7}]$
• C. $(5,\frac{39}{7}]$
• D. $[5,\frac{39}{7})$

Answer 1

Answer: (B)

$[5,\frac{39}{7}]$

Given $(x^2+3x+5{)}^2-(\lambda -3\left)\right(x^2+3x+4\left)\right(x^2+3x+5)+(\lambda -4\left)\right(x^2+3x+4{)}^2=0$

for $x^2+3x+4$ $D=3^2-4\left(1\right)\left(4\right)\Rightarrow D=9-16=-7$

as $D<0$ $x^2+3x+4>0$

So,

$Dividingthegivenequationby{(x^2+3x+4)}^{2}itbecomes{y}^2-(\lambda -3)y+(\lambda -4)=0....\left(1\right)where,y=\frac{x^2+3x+5}{x^2+3x+4}=1+\frac{1}{x^2+3x+4}=1+\frac{1}{{(x+\frac{3}{2})}^2+\frac{7}{4}}\Rightarrow y>1and(atx=-\frac{3}{2})willbe\le 1+\frac{1}{7/4}\Rightarrow y\le \frac{11}{7}Nowsolvingforequation\left(1\right)y=\frac{(\lambda -3)\pm \sqrt{{(\lambda -3)}^2-4(\lambda -4)}}{2}=[\alpha ,\beta ]\therefore atleastoneof[\alpha ,\beta ]\in [1,\frac{11}{7}]⟶\left[realvalue\right]\therefore {(\lambda -3)}^2>4(\lambda -4)forrealvalueofy.\Rightarrow {\lambda }^2-10\lambda +25>0\Rightarrow {(\lambda -5)}^2>0\Rightarrow \lambda >5Again\lambda -4\le \frac{11}{7}\Rightarrow \lambda \le \frac{39}{7}\therefore 5<\lambda \le \frac{39}{7}AnsOptionC.$