Answer

Verified

412.5k+ views

**Hint:**In this particular question use the concept that if something is positive than it is always greater than zero, later on in the solution use the concept that arithmetic mean is always greater than or equal to geometric mean so use these concepts to reach the solution of the question.

__Complete step-by-step answer__:Given determinant

$\left| {\begin{array}{*{20}{c}}

a&1&1 \\

1&b&1 \\

1&1&c

\end{array}} \right|$

Now we have to find out the value of determinant if the determinant is positive.

So as the determinant is positive so the value of determinant is always greater than zero.

$ \Rightarrow \left| {\begin{array}{*{20}{c}}

a&1&1 \\

1&b&1 \\

1&1&c

\end{array}} \right| > 0$

Now expand the determinant we have,

\[ \Rightarrow \left[ {a\left| {\begin{array}{*{20}{c}}

b&1 \\

1&c

\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}

1&1 \\

1&c

\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}

1&b \\

1&1

\end{array}} \right|} \right] > 0\]

Now expand the mini determinants we have,

\[ \Rightarrow \left[ {a\left( {bc - 1} \right) - 1\left( {c - 1} \right) + 1\left( {1 - b} \right)} \right] > 0\]

\[ \Rightarrow \left[ {a\left( {bc - 1} \right) - 1\left( {c - 1} \right) + 1\left( {1 - b} \right)} \right] > 0\]

Now simplify we have,

\[ \Rightarrow \left[ {abc - a - c + 1 + 1 - b} \right] > 0\]

\[ \Rightarrow abc - a - b - c + 2 > 0\]

\[ \Rightarrow abc + 2 > a + b + c\]................. (1)

Now consider three numbers a, b, and c.

Arithmetic mean (A.M) of the numbers is $ = \dfrac{{a + b + c}}{3}$

And the Geometric mean (G.M) of the numbers is $\sqrt[3]{{abc}}$

Now as we know that

$A.M \geqslant G.M$, so use this property

$\therefore \dfrac{{a + b + c}}{3} \geqslant \sqrt[3]{{abc}}$

$ \Rightarrow a + b + c \geqslant 3\sqrt[3]{{abc}}$............... (2)

Now from equation (1) and (2) we have,

\[ \Rightarrow abc + 2 > 3\sqrt[3]{{abc}}\]

Now let abc = ${x^3}$.................. (3), so substitute this in the above equation we have,

\[ \Rightarrow {x^3} + 2 > 3x\]

\[ \Rightarrow {x^3} - 3x + 2 > 0\]................... (4)

Now first solve the cubic equation, \[{x^3} - 3x + 2\]

$ \Rightarrow {x^3} - 3x + 2 = 0$

So solve this by hit and trial method, so substitute x = 1 in the above equation we have,

$ \Rightarrow {1^3} - 3 + 2 = 0$

$ \Rightarrow 0 = 0$

So 1 is the root of the equation, so (x - 1) is the factor of the cubic equation.

Now substitute x = -2 in the above equation we have,

$ \Rightarrow {\left( { - 2} \right)^3} - 3\left( { - 2} \right) + 2 = 0$

$ \Rightarrow - 8 + 6 + 2 = 0$

$ \Rightarrow 0 = 0$

So -2 is the root of the equation, so (x + 2) is the factor of the cubic equation.

Let the other factor be (x - a).

Therefore, (x - a) (x + 2) (x - 1) = \[{x^3} - 3x + 2\]

$ \Rightarrow \left( {x - a} \right)\left( {{x^2} + x - 2} \right) = {x^3} - 3x + 2$

$ \Rightarrow \left( {{x^3} + \left( {1 - a} \right){x^2} - \left( {2 + a} \right)x + 2a} \right) = {x^3} - 3x + 2$

Now on comparing the constant term we have,

$ \Rightarrow 2a = 2$

Therefore, a = 1

So from equation (4) we have,

\[ \Rightarrow {\left( {x - 1} \right)^2}\left( {x + 2} \right) > 0\]

Now as we know that the square term is always positive for any value of x.

But (x + 2) cannot be positive for any value of x.

So for the determinant to be positive (x + 2) must be greater than zero.

Therefore, x > - 2

Now take cube on both sides we have,

$ \Rightarrow {x^3} > {\left( { - 2} \right)^3}$

$ \Rightarrow {x^3} > - 8$

Now from equation (3) we have,

$ \Rightarrow abc > - 8$

**Hence option (b) is the correct answer.**

**Note**: Whenever we face such types of questions the key concept we have to remember is that always recall the arithmetic mean of three numbers (a, b, and c) is $\dfrac{{a + b + c}}{3}$ and the geometric mean of three numbers (a, b, and c) is $\sqrt[3]{{abc}}$, and always recall that the quickly solution of a cubic equation if the solution is an integer value is by hit and trial method as applied above.

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Name 10 Living and Non living things class 9 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail