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JEE Main 2020
Matrices & Determinants
Matrices and Determinants
Easy

Question

Let M and m respectively be the maximum and the minimum values of f(x)=1+sin2xcos2x4sin4xsin2x1+cos2x4sin4xsin2xcos2x1+4sin4x,xRf(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\end{array}\right|, x \in R Then M4m4M^4 - m^4 is equal to :

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Solution

Key Concepts and Formulas

This problem primarily involves the following mathematical concepts:

  1. Properties of Determinants:

    • Row/Column Operations: The value of a determinant remains unchanged if we apply the operation RiRi+kRjR_i \rightarrow R_i + k R_j (or CiCi+kCjC_i \rightarrow C_i + k C_j) where RiR_i and RjR_j are rows (or CiC_i and CjC_j are columns). This property is crucial for simplifying determinants by introducing zeros.
    • Expansion of a 3×33 \times 3 Determinant: For a matrix (abcdefghi)\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}, its determinant can be expanded along the first row as a(eifh)b(difg)+c(dheg)a(ei - fh) - b(di - fg) + c(dh - eg). Expanding along any row or column yields the same result.
  2. Trigonometric Identities:

    • The fundamental identity: sin2x+cos2x=1\sin^2 x + \cos^2 x = 1.
  3. Range of Trigonometric Functions:

    • For any real number θ\theta, the sine function satisfies 1sinθ1-1 \le \sin \theta \le 1. This property is used to find the maximum and minimum values of the function.

Step-by-Step Solution

Step 1: Simplify the Determinant using Row Operations

We are given the function f(x)f(x) as a determinant: f(x)=1+sin2xcos2x4sin4xsin2x1+cos2x4sin4xsin2xcos2x1+4sin4xf(x)=\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\ \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x \end{array}\right|

Our goal is to simplify this determinant. A common strategy is to introduce zeros into rows or columns using row/column operations, which significantly eases the expansion process. Observe that the first two columns have similar terms in different rows. We can perform row operations to simplify these terms.

Let's apply the following row operations:

  1. R2R2R1R_2 \rightarrow R_2 - R_1
  2. R3R3R1R_3 \rightarrow R_3 - R_1

Why these operations? These operations are chosen because they will create simple entries (including zeros) in the second and third rows, especially in the first two columns. For example, in the second row, the first element will become sin2x(1+sin2x)=1\sin^2 x - (1+\sin^2 x) = -1, and the second element will become (1+cos2x)cos2x=1(1+\cos^2 x) - \cos^2 x = 1. Similarly for the third row. These operations do not change the value of the determinant.

Let's apply them: For R2R2R1R_2 \rightarrow R_2 - R_1:

  • (R2)1=sin2x(1+sin2x)=1(R_2)_{1} = \sin^2 x - (1+\sin^2 x) = -1
  • (R2)2=(1+cos2x)cos2x=1(R_2)_{2} = (1+\cos^2 x) - \cos^2 x = 1
  • (R2)3=4sin4x4sin4x=0(R_2)_{3} = 4 \sin 4x - 4 \sin 4x = 0

For R3R3R1R_3 \rightarrow R_3 - R_1:

  • (R3)1=sin2x(1+sin2x)=1(R_3)_{1} = \sin^2 x - (1+\sin^2 x) = -1
  • (R3)2=cos2xcos2x=0(R_3)_{2} = \cos^2 x - \cos^2 x = 0
  • (R3)3=(1+4sin4x)4sin4x=1(R_3)_{3} = (1+4 \sin 4x) - 4 \sin 4x = 1

After applying these operations, the determinant becomes: f(x)=1+sin2xcos2x4sin4x110101f(x)=\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right|

Step 2: Expand the Simplified Determinant

Now that we have a simplified determinant with zeros, we can expand it. We can expand along any row or column. Expanding along R1R_1 (as in the original solution) or R2R_2 or R3R_3 (due to the presence of zeros) would be efficient. Let's expand along R1R_1:

f(x)=(1+sin2x)Cofactor(R1,C1)(cos2x)Cofactor(R1,C2)+(4sin4x)Cofactor(R1,C3)f(x) = (1+\sin^2 x) \cdot \text{Cofactor}(R_1, C_1) - (\cos^2 x) \cdot \text{Cofactor}(R_1, C_2) + (4 \sin 4x) \cdot \text{Cofactor}(R_1, C_3)

Where Cofactor(Ri,Cj)=(1)i+j×Minor(Ri,Cj)\text{Cofactor}(R_i, C_j) = (-1)^{i+j} \times \text{Minor}(R_i, C_j).

  • The minor for (1+sin2x)(1+\sin^2 x) is 1001=(1)(1)(0)(0)=1\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right| = (1)(1) - (0)(0) = 1.
  • The minor for (cos2x)(\cos^2 x) is 1011=(1)(1)(0)(1)=1\left|\begin{array}{cc} -1 & 0 \\ -1 & 1 \end{array}\right| = (-1)(1) - (0)(-1) = -1.
  • The minor for (4sin4x)(4 \sin 4x) is 1110=(1)(0)(1)(1)=0(1)=1\left|\begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array}\right| = (-1)(0) - (1)(-1) = 0 - (-1) = 1.

Substituting these values into the expansion: f(x)=(1+sin2x)(1)(cos2x)(1)+(4sin4x)(1)f(x) = (1+\sin^2 x)(1) - (\cos^2 x)(-1) + (4 \sin 4x)(1) f(x)=1+sin2x+cos2x+4sin4xf(x) = 1+\sin^2 x + \cos^2 x + 4 \sin 4x

Why this simplification? We use the fundamental trigonometric identity sin2x+cos2x=1\sin^2 x + \cos^2 x = 1. f(x)=1+(sin2x+cos2x)+4sin4xf(x) = 1 + (\sin^2 x + \cos^2 x) + 4 \sin 4x f(x)=1+1+4sin4xf(x) = 1 + 1 + 4 \sin 4x f(x)=2+4sin4xf(x) = 2 + 4 \sin 4x This is the simplified form of the function f(x)f(x).

Step 3: Determine the Maximum (M) and Minimum (m) Values of f(x)f(x)

We have f(x)=2+4sin4xf(x) = 2 + 4 \sin 4x. To find the maximum and minimum values of f(x)f(x), we need to consider the range of the sine function. We know that for any real value θ\theta (in our case, θ=4x\theta = 4x): 1sinθ1-1 \le \sin \theta \le 1 1sin4x1-1 \le \sin 4x \le 1

Now, we build up the expression for f(x)f(x) using this inequality: First, multiply by 4: 44sin4x4-4 \le 4 \sin 4x \le 4 Next, add 2 to all parts of the inequality: 242+4sin4x2+42 - 4 \le 2 + 4 \sin 4x \le 2 + 4 2f(x)6-2 \le f(x) \le 6

From this inequality, we can identify the maximum and minimum values of f(x)f(x):

  • Maximum value, M=6M = 6
  • Minimum value, m=2m = -2

Step 4: Calculate M4m4M^4 - m^4

Finally, we need to calculate the value of M4m4M^4 - m^4: M4m4=(6)4(2)4M^4 - m^4 = (6)^4 - (-2)^4

Calculate the powers:

  • 64=6×6×6×6=36×36=12966^4 = 6 \times 6 \times 6 \times 6 = 36 \times 36 = 1296
  • (2)4=(2)×(2)×(2)×(2)=4×4=16(-2)^4 = (-2) \times (-2) \times (-2) \times (-2) = 4 \times 4 = 16

Substitute these values: M4m4=129616M^4 - m^4 = 1296 - 16 M4m4=1280M^4 - m^4 = 1280

Thus, the value of M4m4M^4 - m^4 is 1280.


Tips for Students & Common Mistakes to Avoid

  1. Prioritize Simplification: Always look for opportunities to simplify determinants using row or column operations before expanding. This reduces the number of terms and the chances of calculation errors. Operations like RiRi+kRjR_i \rightarrow R_i + kR_j (or CiCi+kCjC_i \rightarrow C_i + kC_j) are powerful because they do not change the determinant's value.
  2. Trigonometric Identities are Key: Keep fundamental trigonometric identities like sin2x+cos2x=1\sin^2 x + \cos^2 x = 1 handy. They often appear in these problems to simplify expressions.
  3. Range of Functions: When finding maximum/minimum values, remember the basic ranges of trigonometric functions. For sinx\sin x and cosx\cos x, the range is [1,1][-1, 1]. Be careful when multiplying or adding constants to these inequalities.
  4. Powers of Negative Numbers: Pay attention to signs when calculating powers of negative numbers. An even power of a negative number results in a positive number (e.g., (2)4=16(-2)^4 = 16), while an odd power results in a negative number.
  5. Careful Expansion: When expanding a determinant, remember the alternating signs of the cofactors. If expanding along R1R_1, the signs are + - +.

Summary and Key Takeaway

This problem is an excellent illustration of how different mathematical topics—matrices and determinants, trigonometry, and functions—can be integrated into a single question. The most efficient path to the solution involves:

  1. Strategically applying row/column operations to simplify the determinant.
  2. Using trigonometric identities

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