JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 9)
Let A = $$\left[ {\matrix{
2 & b & 1 \cr
b & {{b^2} + 1} & b \cr
1 & b & 2 \cr
} } \right]$$ where b > 0.
Then the minimum value of $${{\det \left( A \right)} \over b}$$ is -
Then the minimum value of $${{\det \left( A \right)} \over b}$$ is -
$$\sqrt 3 $$
$$-$$ $$2\sqrt 3 $$
$$ - \sqrt 3 $$
$$2\sqrt 3 $$
Explanation
A = $$\left[ {\matrix{
2 & b & 1 \cr
b & {{b^2} + 1} & b \cr
1 & b & 2 \cr
} } \right]$$ (b > 0)
$$\left| A \right|$$ = 2(2b2 + 2 $$-$$ b2) $$-$$ b(2b $$-$$ b) + 1(b2 $$-$$ b2 $$-$$ 1)
$$\left| A \right|$$ = 2(b2 + 2) $$-$$ b2 $$-$$ 1
$$\left| A \right|$$ = b2 + 3
$${{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3 $$
$$b + {3 \over b} \ge 2\sqrt 3 $$
$$\left| A \right|$$ = 2(2b2 + 2 $$-$$ b2) $$-$$ b(2b $$-$$ b) + 1(b2 $$-$$ b2 $$-$$ 1)
$$\left| A \right|$$ = 2(b2 + 2) $$-$$ b2 $$-$$ 1
$$\left| A \right|$$ = b2 + 3
$${{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3 $$
$$b + {3 \over b} \ge 2\sqrt 3 $$
Comments (0)
