Let M=sin4⁡θ-1-sin2⁡θ1+cos2⁡θcos4⁡θ=αI+βM-1, where α=αθ and β=βθ are real number, and I is the 2×2 identity matrix. If α* is the minimum of the set { αθ:θ∈0, 2π} and β* is the minimum of the set βθ:θ∈0, 2π, then the value of α*+β* is

JEE Advanced · Mathematics · 18. Matrices

Let $M = [\begin{matrix} \sin^{4} θ & - 1 - \sin^{2} θ \\ 1 + \cos^{2} θ & \cos^{4} θ \end{matrix}] = α I + β M^{- 1},$ where $α = α (θ)$ and $β = β (θ)$ are real number, and $I$ is the $2 \times 2$ identity matrix. If $α^{}$ is the minimum of the set { $α (θ) : θ \in [0, 2 π)}$ and $β^{}$ is the minimum of the set $\{β (θ) : θ \in [0, 2 π)\},$ then the value of $α^{} + β^{}$ is

A $- \frac{37}{16}$
B $- \frac{31}{16}$
C $- \frac{29}{16}$
D $- \frac{17}{16}$

Verified Solution

Answer & Solution

Correct Answer

(C) $- \frac{29}{16}$

Step-by-step Solution

Detailed explanation

Method I
As given,

M = [\begin{matrix} {s i n}^{4} θ & - 1 - {s i n}^{2} θ \\ 1 + {c o s}^{2} θ & {c o s}^{4} θ \end{matrix}] = α I + β M^{- 1}

a d j (M) = [\begin{matrix} {c o s}^{4} θ & 1 + {s i n}^{2} θ \\ - 1 - {c o s}^{2} θ & {s i n}^{4} θ \end{matrix}]

M^{- 1} = \frac{1}{| M |} [\begin{matrix} {c o s}^{4} θ & 1 + {s i n}^{2} θ \\ - 1 - {c o s}^{2} θ & {s i n}^{4} θ \end{matrix}]

[\begin{matrix} {s i n}^{4} θ & - 1 - {s i n}^{2} θ \\ 1 + {c o s}^{2} θ & {c o s}^{4} θ \end{matrix}] = α [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + \frac{β}{|M|} [\begin{matrix} {c o s}^{4} θ & 1 + {s i n}^{2} θ \\ - 1 - {c o s}^{4} θ & {s i n}^{4} θ \end{matrix}]

Compare element

a_{12}

in L.H.S. and R.H.S.

- 1 - {s i n}^{2} θ = 0 + \frac{β}{|M|} (1 + {s i n}^{2} θ)

β = | M |

...(iii)
Compare element

a_{11}

in LHS. and R.H.S.

{s i n}^{4} θ = α + \frac{β}{|M|} ({c o s}^{4} θ)

From equation (iii)

{s i n}^{4} θ = α - {c o s}^{4} θ

α = {s i n}^{4} θ + {c o s}^{4} θ

...(iv)
Now from equation (iv)

α = {s i n}^{4} θ + {c o s}^{4} θ

α = {({s i n}^{2} θ + {c o s}^{2} θ)}^{2} - 2 {s i n}^{2} θ {c o s}^{2} θ

α = 1 - \frac{{s i n}^{2} 2 θ}{2}

0 \leq {s i n}^{2} 2 θ \leq 1

- \frac{1}{2} \leq \frac{- {s i n}^{2} 2 θ}{2} \leq 0

\frac{1}{2} \leq 1 - \frac{{s i n}^{2} 2 θ}{2} \leq 1

α \in [\frac{1}{2}, 1]

α_{m i n} = \frac{1}{2} = α *

For

β

, consider

|M| = {s i n}^{4} θ {c o s}^{4} θ + {s i n}^{2} θ {c o s}^{2} θ + 2

|M| \{{({s i n}^{2} θ {c o s}^{2} θ + \frac{1}{2})}^{2} + \frac{7}{4}\}

|M| \{{(\frac{{s i n}^{2} 2 θ}{4} + \frac{1}{2})}^{2} + \frac{7}{4}\}

0 \leq {s i n}^{2} 2 θ \leq 1

Hence,

|M| \in [2, \frac{37}{16}]

β = - |M|

β \in [- \frac{37}{16}, - 2]

β_{m i n} = - \frac{37}{16} = β *

α * + β * = \frac{1}{2} - \frac{37}{16} = - \frac{29}{16}

Method II

M = α I + β M^{- 1}

Multiply with

M

both side

M^{2} = α M + β I

M^{2} - α M - B I = 0

...(i)
Equation (i) represents characteristic equation of matrix

‘ M ’

Hence, Trace

(M) = α

|M| = - β

Sum of roots of characteristic equation

=

Trace of matrix
Product of roots of characteristic equation

=

determined of matrix

∴ α = {s i n}^{4} θ + {c o s}^{4} θ

β = - ({s i n}^{4} θ {c o s}^{4} θ + {s i n}^{2} θ {c o s}^{2} θ + 2)

Now proceed as Method 1

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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