JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: If $1, lo g_{10} (4^{x} - 2)$ and $lo g_{10} (4^{x} + \frac{18}{5})$ are in arithmetic progression for a real number $x$ , then the value of the determinant $210 (x - 1) 01 x^{2} 1 x$ is equal to :

Q: Wait, why do we reject the negative root $y = -2$?

Great catch! Remember that $y = 4^x$. Since $4$ is a positive base, $4^x$ will always be strictly greater than zero for any real number $x$. A negative value is physically impossible in this context, so we discard it to keep our domain valid.

Q: Why did we convert $1$ into $\log_{10} 10$?

To combine logarithmic terms using the product rule $\log A + \log B = \log(AB)$, all terms must be expressed as logarithms with the same base. Converting $1$ to $\log_{10} 10$ is the standard 'key' to unlocking the equation.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: When you see three terms in an Arithmetic Progression, the middle term is the arithmetic mean of the outer two. Visualize this as a balance:

2 b = a + c

. Don't let the logarithmic expressions intimidate you; they are just numbers waiting to be simplified.

Hint 2

Hint 2: Your toolkit for logarithms is essential here. Remember that

n lo g_{10} a = lo g_{10} a^{n}

and that the constant

1

can be written as

lo g_{10} 10

. This allows you to combine the right-hand side into a single logarithm, setting the stage to remove the logs entirely.

Hint 3

Hint 3: Once you equate the arguments, you will have a quadratic equation in terms of

4^{x}

. Let

y = 4^{x}

to make it look friendly. Solve for

y

, but remember the physical constraint:

4^{x}

must always be positive. Once you find

x

, the determinant is just a simple arithmetic calculation.

Visualized Solution

Terms in Arithmetic Progression

Given terms: $1, lo g_{10} (4^{x} - 2), lo g_{10} (4^{x} + \frac{18}{5})$
Condition for $a, b, c$ in A.P.: $2 b = a + c$

Applying the A.P. Formula

Substitute $a = 1$ , $b = lo g_{10} (4^{x} - 2)$ , $c = lo g_{10} (4^{x} + \frac{18}{5})$
$2 lo g_{10} (4^{x} - 2) = 1 + lo g_{10} (4^{x} + \frac{18}{5})$

Applying Logarithmic Power Rule

Use the property $n lo g a = lo g a^{n}$ on the left side.
$lo g_{10} (4^{x} - 2)^{2} = 1 + lo g_{10} (4^{x} + \frac{18}{5})$

Combining Terms on the Right Side

Rewrite $1$ as $lo g_{10} 10$ to match bases.
$lo g_{10} (4^{x} - 2)^{2} = lo g_{10} 10 + lo g_{10} (4^{x} + \frac{18}{5})$
Apply $lo g a + lo g b = lo g (ab)$ :
$lo g_{10} (4^{x} - 2)^{2} = lo g_{10} [10 (4^{x} + \frac{18}{5})]$

Equating the Logarithmic Arguments

Since $lo g_{10} A = lo g_{10} B$ , we can equate $A = B$ .
$(4^{x} - 2)^{2} = 10 (4^{x} + \frac{18}{5})$

Expanding and Rearranging

Expand LHS: $(4^{x})^{2} - 4 (4^{x}) + 4$
Expand RHS: $10 (4^{x}) + 36$
Equate and rearrange: $(4^{x})^{2} - 14 (4^{x}) - 32 = 0$

Solving for $4^{x}$

Let $y = 4^{x}$ . The equation becomes $y^{2} - 14 y - 32 = 0$ .
Factorize: $(y - 16) (y + 2) = 0$
Roots: $y = 16$ or $y = - 2$
Since $4^{x} > 0$ , we reject $y = - 2$ . Thus, $4^{x} = 16$ .

Finding the Value of $x$

We have $4^{x} = 16$ .
Rewrite $16$ as a power of $4$ : $4^{x} = 4^{2}$ .
Comparing exponents, we get $x = 2$ .

Setting up the Determinant

We need to find the value of $Δ = 210 (x - 1) 01 x^{2} 1 x$
We can substitute $x = 2$ directly or expand first. Let's expand first for algebraic practice.

Expanding the Determinant

Expand along the first column for simplicity (since it has a zero):
$Δ = 2 (0 \cdot x - 1 \cdot 1) - 1 ((x - 1) \cdot x - x^{2} \cdot 1) + 0$
$Δ = 2 (- 1) - (x^{2} - x - x^{2})$

Simplifying and Substituting

Simplify the expression: $Δ = - 2 - (x^{2} - x - x^{2})$
Notice that $x^{2}$ cancels out: $Δ = - 2 - (- x) = x - 2$
Substitute $x = 2$ : $Δ = 2 - 2 = 0$
The final value of the determinant is $0$ .

00:00 / 00:00

The Sigma Insight: Properties of Determinants

Analyzing the Arithmetic Bridge

We are given three terms:

1

lo g_{10} (4^{x} - 2)

, and

lo g_{10} (4^{x} + \frac{18}{5})

. The problem states that these terms are in an Arithmetic Progression (A.P.).

By the definition of an A.P., the middle term is the arithmetic mean of the outer terms. This gives us the following relationship:

2 lo g_{10} (4^{x} - 2) = 1 + lo g_{10} (4^{x} + \frac{18}{5})

Unlocking the Logarithms

To solve for

x

, we apply logarithmic properties. First, we use the power rule

n lo g a = lo g a^{n}

on the left side and express

1

lo g_{10} 10

on the right side.

Applying the product rule

lo g A + lo g B = lo g (A B)

, the equation becomes:

lo g_{10} (4^{x} - 2)^{2} = lo g_{10} [10 (4^{x} + \frac{18}{5})]

Equating the arguments of the logarithms, we obtain the algebraic equation:

(4^{x} - 2)^{2} = 10 (4^{x} + \frac{18}{5})

The Quadratic Reveal

Let

y = 4^{x}

. Substituting this into our equation, we get

(y - 2)^{2} = 10 y + 36

, which simplifies to

y^{2} - 4 y + 4 = 10 y + 36

Rearranging the terms into a standard quadratic form yields:

y^{2} - 14 y - 32 = 0

Factoring the quadratic, we find

(y - 16) (y + 2) = 0

, resulting in

y = 16

y = - 2

. Since

4^{x}

must be positive, we discard

y = - 2

Thus,

4^{x} = 16

, which implies

4^{x} = 4^{2}

, and we conclude that $x = 2$ .

The Final Determinant

With

x = 2

determined, we substitute this value into the given matrix to evaluate the determinant

Δ

Δ = 210 (2 - 1) 01 2^{2} 12 = 210101412

Expanding the determinant along the first column:

Δ = 20112 - 11142 + 01041

Δ = 2 (0 - 1) - 1 (2 - 4) + 0 = - 2 + 2 = 0

The final value of the determinant is $0$ .

Common Traps & FAQs

Wait, why do we reject the negative root

y = - 2

Great catch! Remember that

y = 4^{x}

. Since

4

is a positive base,

4^{x}

will always be strictly greater than zero for any real number

x

. A negative value is physically impossible in this context, so we discard it to keep our domain valid.

Can I expand the determinant before solving for

x

Absolutely! In fact, it is often a great strategy. Expanding the determinant symbolically gives you an expression in terms of

x

. You will find that the

x^{2}

terms cancel out beautifully, leaving you with a simple linear expression. It is a great way to double-check your work!

Why did we convert

1

into

lo g_{10} 10

To combine logarithmic terms using the product rule

lo g A + lo g B = lo g (A B)

, all terms must be expressed as logarithms with the same base. Converting

1

lo g_{10} 10

is the standard 'key' to unlocking the equation.

Is there a shortcut for the determinant calculation?

Yes! If you look closely at the determinant, you can perform row or column operations to create more zeros. However, since we already know

x = 2

, direct substitution is very fast and less prone to algebraic errors.

What if I made a mistake in the quadratic expansion?

It happens to the best of us! Always check your signs. When you expand

(4^{x} - 2)^{2}

, ensure you get

(4^{x})^{2} - 4 (4^{x}) + 4

. If you miss that middle term, the entire quadratic will shift, and you won't get a clean integer for

x

Animated Solution for Mathematics - Matrices and Determinants: If $1, lo g_{10} (4^{x} - 2)$ and $lo g_{10} (4^{x} + \frac{18}{5})$ are in arithmetic progression for a real number $x$ , then the value of the determinant $210 (x - 1) 01 x^{2} 1 x$ is equal to :

Enter Numerical Value:

Visualized Solution

Terms in Arithmetic Progression

Applying the A.P. Formula

Applying Logarithmic Power Rule

Combining Terms on the Right Side

Equating the Logarithmic Arguments

Expanding and Rearranging

Solving for $4^{x}$

Finding the Value of $x$

Setting up the Determinant

Expanding the Determinant

Simplifying and Substituting

The Sigma Insight: Properties of Determinants

Analyzing the Arithmetic Bridge

Unlocking the Logarithms

The Quadratic Reveal

The Final Determinant

Similar Questions

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If $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ are in G.P., then the determinant $Δ = lo g a_{n} lo g a_{n + 3} lo g a_{n + 6} lo g a_{n + 1} lo g a_{n + 4} lo g a_{n + 7} lo g a_{n + 2} lo g a_{n + 5} lo g a_{n + 8}$ is equal to

Let $a > 0, d > 0$ . Find the value of the determinant $\frac{1}{a} \frac{1}{( a + d )} \frac{1}{( a + 2 d )} \frac{1}{a ( a + d )} \frac{1}{( a + d ) ( a + 2 d )} \frac{1}{( a + 2 d ) ( a + 3 d )} \frac{1}{( a + d ) ( a + 2 d )} \frac{1}{( a + 2 d ) ( a + 3 d )} \frac{1}{( a + 3 d ) ( a + 4 d )}$ .

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Enter Numerical Value:

Visualized Solution

Terms in Arithmetic Progression

Applying the A.P. Formula

Applying Logarithmic Power Rule

Combining Terms on the Right Side

Equating the Logarithmic Arguments

Expanding and Rearranging

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4x

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Setting up the Determinant

Expanding the Determinant

Simplifying and Substituting

The Sigma Insight: Properties of Determinants

Analyzing the Arithmetic Bridge

Unlocking the Logarithms

The Quadratic Reveal

The Final Determinant

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