JEE Main 2024 — Mathematics Algebra

Given: $f(x)=|\begin{matrix}2{\mathrm{cos}}^{4}x & 2{\mathrm{sin}}^{4}x & 3+{\mathrm{sin}}^{2}2x \\ 3+2{\mathrm{cos}}^{4}x & 2{\mathrm{sin}}^{4}x & {\mathrm{sin}}^{2}2x \\ 2{\mathrm{cos}}^{4}x & 3+2{\mathrm{sin}}^{4}x & {\mathrm{sin}}^{2}2x\end{matrix}|$f(x)=∣2cos4x3+2cos4x2cos4x​2sin4x2sin4x3+2sin4x​3+sin22xsin22xsin22x​∣

Applying ${R}_{2}\rightarrow {R}_{2}-{R}_{1},{R}_{3}\rightarrow {R}_{3}-{R}_{1}$R2​→R2​−R1​,R3​→R3​−R1​

$$\Rightarrow f(x)=|\begin{matrix}2{\mathrm{cos}}^{4}x & 2{\mathrm{sin}}^{4}x & 3+{\mathrm{sin}}^{2}2x \\ 3 & 0 & -3 \\ 0 & 3 & -3\end{matrix}|$$⇒f(x)=∣2cos4x30​2sin4x03​3+sin22x−3−3​∣

$$\Rightarrow f(x)=9|\begin{matrix}2{\mathrm{cos}}^{4}x & 2{\mathrm{sin}}^{4}x & 3+{\mathrm{sin}}^{2}2x \\ 1 & 0 & -1 \\ 0 & 1 & -1\end{matrix}|$$⇒f(x)=9∣2cos4x10​2sin4x01​3+sin22x−1−1​∣

$$\Rightarrow f(x)=9(2{\mathrm{cos}}^{4}x+2{\mathrm{sin}}^{4}x+3+{\mathrm{sin}}^{2}2x)$$⇒f(x)=9(2cos4x+2sin4x+3+sin22x)

Now, differentiating above function we get,

$$\Rightarrow {f}^{'}(x)=9[8{\mathrm{cos}}^{3}x(-\mathrm{sin}x)+8{\mathrm{sin}}^{3}x(\mathrm{cos}x)+2\mathrm{sin}2x(\mathrm{cos}2x)(2)]$$⇒f′(x)=9[8cos3x(−sinx)+8sin3x(cosx)+2sin2x(cos2x)(2)]

$$\Rightarrow {f}^{'}(0)=9[0+0+0]$$⇒f′(0)=9[0+0+0]

$$\Rightarrow {f}^{'}(0)=0$$⇒f′(0)=0