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Functions question

2022 · Shift 1 · Q32
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Functions question

2022 · Shift 1 · Q32

JEE AdvancedMathematicsFunctionsMultiple correct+4 / −2
Let ∣M∣|M|∣M∣ denote the determinant of a square matrix MMM. Let g:[0,π2]→Rg:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}g:[0,2π​]→R be the function defined by g(θ)=f(θ)−1+f(π2−θ)−1g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}g(θ)=f(θ)−1​+f(2π​−θ)−1​ where f(θ)=12∣1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1∣+∣sin⁡πcos⁡(θ+π4)tan⁡(θ−π4)sin⁡(θ−π4)−cos⁡π2log⁡e(4π)cot⁡(θ+π4)log⁡e(π4)tan⁡π∣.f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} \sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi \end{array}\right| .f(θ)=21​​1−sinθ−1​sinθ1−sinθ​1sinθ1​​+​sinπsin(θ−4π​)cot(θ+4π​)​cos(θ+4π​)−cos2π​loge​(4π​)​tan(θ−4π​)loge​(π4​)tanπ​​. Let p(x)p(x)p(x) be a quadratic polynomial whose roots are the maximum and minimum values of the function g(θ)g(\theta)g(θ), and p(2)=2−2p(2)=2-\sqrt{2}p(2)=2−2​. Then, which of the following is/are TRUE ?
  1. A
    p(3+24)<0p\left(\frac{3+\sqrt{2}}{4}\right)\lt 0p(43+2​​)<0
  2. B
    p(1+324)>0p\left(\frac{1+3 \sqrt{2}}{4}\right)\gt 0p(41+32​​)>0
  3. C
    p(52−14)>0p\left(\frac{5 \sqrt{2}-1}{4}\right)\gt 0p(452​−1​)>0
  4. D
    p(5−24)<0p\left(\frac{5-\sqrt{2}}{4}\right)\lt 0p(45−2​​)<0
View written solutionFree

Correct answer: A, C

  1. Compute the first determinant in f(θ)f(\theta)f(θ)

Let s=sin⁡θs=\sin\thetas=sinθ. Then D1=∣1s1−s1s−1−s1∣.D_1=\begin{vmatrix}1&s&1\\-s&1&s\\-1&-s&1\end{vmatrix}.D1​=​1−s−1​s1−s​1s1​​.

Expanding along the first row, D1=1∣1s−s1∣−s∣−ss−11∣+1∣−s1−1−s∣.D_1=1\begin{vmatrix}1&s\\-s&1\end{vmatrix}-s\begin{vmatrix}-s&s\\-1&1\end{vmatrix}+1\begin{vmatrix}-s&1\\-1&-s\end{vmatrix}.D1​=1​1−s​s1​​−s​−s−1​s1​​+1​−s−1​1−s​​.

Now, ∣1s−s1∣=1+s2,\begin{vmatrix}1&s\\-s&1\end{vmatrix}=1+s^2,​1−s​s1​​=1+s2, ∣−ss−11∣=−s+s=0,\begin{vmatrix}-s&s\\-1&1\end{vmatrix}=-s+s=0,​−s−1​s1​​=−s+s=0, ∣−s1−1−s∣=s2+1.\begin{vmatrix}-s&1\\-1&-s\end{vmatrix}=s^2+1.​−s−1​1−s​​=s2+1.

Hence, D1=(1+s2)+0+(1+s2)=2(1+s2).D_1=(1+s^2)+0+(1+s^2)=2(1+s^2).D1​=(1+s2)+0+(1+s2)=2(1+s2). Therefore, 12D1=1+sin⁡2θ.\frac12 D_1=1+\sin^2\theta.21​D1​=1+sin2θ.

  1. Compute the second determinant

The second matrix is

\sin\pi & \cos\left(\theta+\frac\pi4\right) & \tan\left(\theta-\frac\pi4\right)\\ \sin\left(\theta-\frac\pi4\right) & -\cos\frac\pi2 & \ln\left(\frac4\pi\right)\\ \cot\left(\theta+\frac\pi4\right) & \ln\left(\frac\pi4\right) & \tan\pi \end{vmatrix}.$$ Use $$\sin\pi=0,\quad \cos\frac\pi2=0,\quad \tan\pi=0,$$ so this becomes $$D_2=\begin{vmatrix} 0 & a & b\\ c & 0 & d\\ e & -d & 0 \end{vmatrix},$$ where $$a=\cos\left(\theta+\frac\pi4\right),\quad b=\tan\left(\theta-\frac\pi4\right),$$ $$c=\sin\left(\theta-\frac\pi4\right),\quad d=\ln\left(\frac4\pi\right),\quad e=\cot\left(\theta+\frac\pi4\right),$$ and we used $$\ln\left(\frac\pi4\right)=-\ln\left(\frac4\pi\right)=-d.$$ Now expand: $$D_2= -a\begin{vmatrix}c&d\\e&0\end{vmatrix}+b\begin{vmatrix}c&0\\e&-d\end{vmatrix}.$$ So, $$D_2=-a(-de)+b(-cd)=ad(e-bc).$$ Now simplify: $$e=\cot\left(\theta+\frac\pi4\right)=\frac{\cos(\theta+\pi/4)}{\sin(\theta+\pi/4)}=\frac{a}{\sin(\theta+\pi/4)}.$$ Also, $$b=\frac{\sin(\theta-\pi/4)}{\cos(\theta-\pi/4)}=\frac{c}{\cos(\theta-\pi/4)}.$$ Thus $$bc=\frac{\sin^2(\theta-\pi/4)}{\cos(\theta-\pi/4)}.$$ A cleaner route is to use angle-shift identities. Let $$u=\theta+\frac\pi4.$$ Then $$\theta-\frac\pi4=u-\frac\pi2.$$ Hence $$\sin\left(\theta-\frac\pi4\right)=\sin\left(u-\frac\pi2\right)=-\cos u=-a,$$ $$\tan\left(\theta-\frac\pi4\right)=\tan\left(u-\frac\pi2\right)=-\cot u=-e.$$ Therefore, $$c=-a,\quad b=-e.$$ So $$e-bc=e-(-e)(-a)=e(1-a).$$ Hence $$D_2=ad\,e(1-a).$$ But $$ae=\cos u\cdot \cot u=\frac{\cos^2u}{\sin u}.$$ This still looks messy, so let us recompute directly using the substitutions $c=-a, b=-e$ in the determinant: $$D_2=\begin{vmatrix}0&a&-e\\-a&0&d\\e&-d&0\end{vmatrix}.$$ Expand along first row: $$D_2=-a\begin{vmatrix}-a&d\\e&0\end{vmatrix}-e\begin{vmatrix}-a&0\\e&-d\end{vmatrix}.$$ Therefore, $$D_2=-a(-de)-e(ad)=ade-ade=0.$$ So the entire second determinant is identically zero. Hence, $$f(\theta)=1+\sin^2\theta.$$ --- 3. **Compute** $g(\theta)$ Given $$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac\pi2-\theta\right)-1},$$ we have $$f(\theta)-1=\sin^2\theta,$$ so on $\left[0,\frac\pi2\right]$, $$\sqrt{f(\theta)-1}=\sqrt{\sin^2\theta}=\sin\theta.$$ Also, $$f\left(\frac\pi2-\theta\right)-1=\sin^2\left(\frac\pi2-\theta\right)=\cos^2\theta,$$ thus $$\sqrt{f\left(\frac\pi2-\theta\right)-1}=\cos\theta.$$ Therefore, $$g(\theta)=\sin\theta+\cos\theta.$$ For $\theta\in\left[0,\frac\pi2\right]$, - minimum value is $1$ (at $\theta=0$ or $\theta=\pi/2$), - maximum value is $\sqrt2$ (at $\theta=\pi/4$). So the roots of $p(x)$ are $1$ and $\sqrt2$. Hence, $$p(x)=k(x-1)(x-\sqrt2).$$ Given $$p(2)=2-\sqrt2,$$ we get $$k(2-1)(2-\sqrt2)=2-\sqrt2 \implies k=1.$$ Thus, $$p(x)=(x-1)(x-\sqrt2).$$ --- 4. **Check each option** Since $1<\sqrt2$, the sign of $$p(x)=(x-1)(x-\sqrt2)$$ is: - positive for $x<1$ or $x>\sqrt2$, - negative for $1<x<\sqrt2$. ### Option A $$x=\frac{3+\sqrt2}{4}.$$ Check location: $$1<\frac{3+\sqrt2}{4}<\sqrt2$$ because $$4<3+\sqrt2 \iff 1<\sqrt2,$$ and $$3+\sqrt2<4\sqrt2 \iff 3<3\sqrt2 \iff 1<\sqrt2.$$ So $x\in(1,\sqrt2)$, hence $$p\left(\frac{3+\sqrt2}{4}\right)<0.$$ So **A is true**. ### Option B $$x=\frac{1+3\sqrt2}{4}.$$ Check location: $$1<\frac{1+3\sqrt2}{4}<\sqrt2$$ because $$4<1+3\sqrt2 \iff 1<\sqrt2,$$ and $$1+3\sqrt2<4\sqrt2 \iff 1<\sqrt2.$$ Thus $x\in(1,\sqrt2)$, so $$p\left(\frac{1+3\sqrt2}{4}\right)<0,$$ not $>0$. So **B is false**. ### Option C $$x=\frac{5\sqrt2-1}{4}.$$ Check whether $x>\sqrt2$: $$\frac{5\sqrt2-1}{4}>\sqrt2 \iff 5\sqrt2-1>4\sqrt2 \iff \sqrt2>1,$$ true. Hence $x>\sqrt2$, so $$p\left(\frac{5\sqrt2-1}{4}\right)>0.$$ So **C is true**. ### Option D $$x=\frac{5-\sqrt2}{4}.$$ Check location: $$1<\frac{5-\sqrt2}{4}<\sqrt2$$ since $$4<5-\sqrt2 \iff \sqrt2<1,$$ which is false. In fact, $$\frac{5-\sqrt2}{4}<1.$$ Therefore $x<1$, so $$p\left(\frac{5-\sqrt2}{4}\right)>0,$$ not $<0$. So **D is false**. --- 5. **Final answer** The true options are $$\boxed{A,\ C}.$$ This matches the stored correct answer.
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