Question

Let the number 2, b, c be in A.P. and $A\ =v\ \left[ \begin{matrix} 1 & 1 & 1 \\\ 2 & b & c \\\ 4 & {{b}^{2}} & {{c}^{2}} \\\ \end{matrix} \right]$. If $\det \left( A \right)\in \left[ 2,16 \right]$, then c lies in the interval:
A) $[2,3)$
B) $\left( 2+{{2}^{\dfrac{3}{4}}},4 \right)$
C) $\left[ 3,2+{{2}^{\dfrac{3}{4}}} \right]$
D) $\left[ 4,6 \right]$

Answer 1

Hint: Use the fact that 2, b, c is in A.P. the substitute $b\ =\ \dfrac{2+c}{2}$ and hence find the determinant. After finding the determinant use the fact it belongs to $\left[ 2,16 \right]$ and hence, find $\left[ 4,6 \right]$.
Complete step-by-step answer:
In the question it is given that 2, b, c is in A.P. and a matrix A is given which is $\ \left[ \begin{matrix} 1 & 1 & 1 \\\ 2 & b & c \\\ 4 & {{b}^{2}} & {{c}^{2}} \\\ \end{matrix} \right]$. Further it is said that if the range of $\det \left( A \right)\in \left[ 2,16 \right]$, then we have to find the interval of c.
As 2, b, c is in A.P. so we can say that,
$b\ =\ \dfrac{2+c}{2}$.
Now, we will substitute the b in term of c in matrix A so we get,
$A\ =\ \left[ \begin{matrix} 1 & 1 & 1 \\\ 2 & \dfrac{2+c}{2} & c \\\ 4 & {{\left( \dfrac{2+c}{2} \right)}^{2}} & {{c}^{2}} \\\ \end{matrix} \right]$
Now, we will take det(A) so,
$\det \left( A \right)\ =\ 1\left( \left( \dfrac{2+c}{2} \right){{c}^{2}}-c{{\left( \dfrac{2+c}{2} \right)}^{2}} \right)-1\left( 2{{c}^{2}}-4c \right)+1\left( 2{\left( \dfrac{2+c}{2} \right)^2}-4\left( \dfrac{2+c}{2} \right) \right)$
On further expanding we get,
$\det \left( A \right)\ =\ \dfrac{2{{c}^{2}}+{{c}^{3}}}{2}-\dfrac{c}{4}\left( 4+4c+{{c}^{2}} \right)-2{{c}^{2}}+4c+\dfrac{\left( 4+4c+{{c}^{2}} \right)}{2}-4-2c$
$=\ \dfrac{4{{c}^{2}}+2{{c}^{3}}-4c-4{{c}^{2}}-{{c}^{3}}-8{{c}^{2}}+16c+8+8c+2{{c}^{2}}}{4}+\dfrac{-16-8c}{4}$
On simplification we get,
$\det \left( A \right)\ =\ \dfrac{{{c}^{3}}-6{{c}^{2}}+12c-8}{4}$
Now at first we will factorize the given numerator which is ${{c}^{3}}-6{{c}^{2}}+12c-8$ as ${{\left( c \right)}^{3}}+3\times \left( -2 \right)\times {{\left( c \right)}^{2}}+3\times {{\left( -2 \right)}^{2}}\times c+{{\left( -2 \right)}^{3}}$ . Now we know that the formula of ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ is ${{\left( a+b \right)}^{3}}$ . So, we can factorize and write it as ${{\left( c-2 \right)}^{3}}$ .
Now, we can write $\left( {{c}^{3}}-6{{c}^{2}}+12c-8 \right)$ as ${{\left( c-2 \right)}^{3}}$
$\det \left( A \right)\ =\ {{\left( c-2 \right)}^{3}}$
Now, as we know det(A) lie between 2 and 16 so,
$2\le \dfrac{{{\left( c-2 \right)}^{3}}}{4}\le 16$
On multiplying by 4 all the sides we get,
$8\le {{\left( c-2 \right)}^{3}}\le 64$
Taking cube root in all the sides we get
$2\le \left( c-2 \right)\le 4$
So, $$$$
$2+2\le c-2+2\le 4+2$
Hence, $4\le c\le 6$
The correct option is ‘D’.
Note: Students should be cautious while finding out the determinant because there is a high chance of making calculation mistakes. Also, while doing inequality they should know that taking cube root doesn't change signs but square root can.