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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Advanced 2008 · Paper 2 · Mathematics · Basics of Mathematics

Match the Statements/Expressions in Column I with the...

Match the Statements/Expressions in Column I with the Statements/Expressions in Column II. Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

Options

  1. A(A) r, (B) q,r, (C) s, (D) p,r
  2. B(A) s, (B) q,s, (C) r, (D) p,q
  3. C(A) s, (B) q,r, (C) s, (D) p,q
  4. D(A) r, (B) q,s, (C) r, (D) p,r

Correct answer

D. (A) r, (B) q,s, (C) r, (D) p,r

Step-by-step solution

(A) Let y=x2+2x+4x+2y=\frac{x^2+2 x+4}{x+2}⇒x2+(2−y)x+(4−2y)=0⇒(2−y)2−4(4−2y)≥0⇒y2+4y−12≥0⇒y≤−6,y≥2\begin{aligned} & \Rightarrow x^2+(2-y) x+(4-2 y)=0 \\ & \Rightarrow \quad(2-y)^2-4(4-2 y) \geq 0 \\ & \Rightarrow \quad y^2+4 y-12 \geq 0 \\ & \Rightarrow \quad y \leq-6, y \geq 2 \\ & \end{aligned} \thereforeMinimumvalueofMinimum value ofyis2.(B)Since,is 2 . (B) Since,(A+B)(A-B)=(A-B)(A+B) ⇒A2−AB+BA−B2=A2+AB−BA−B2⇒AB=BA and  (AB)t=(−1)kAB⇒BtAt=(−1)kAB⇒−BA=(−1)kAB[∵Bt=−B,At=A]⇒BA=(−1)k+1AB⇒(−1)k+1=1\begin{aligned} & \Rightarrow \quad A^2-A B+B A-B^2=A^2+A B-B A-B^2 \\ & \Rightarrow \quad A B=B A \\ & \text{\text{ and } } \quad(A B)^t=(-1)^k A B \\ & \Rightarrow \quad B^t A^t=(-1)^k A B \\ & \Rightarrow \quad-B A=(-1)^k A B \quad\left[\because B^t=-B, A^t=A\right] \\ & \Rightarrow \quad B A=(-1)^{k+1} A B \\ & \Rightarrow \quad(-1)^{k+1}=1 \\ & \end{aligned} \therefore k+1isevenoris even orkisodd.(C)1<2(−k+3−a)<2⇒0<−k+3−a<1is odd. (C) 1 < 2^{\left(-k+3^{-a}\right)} < 2 \Rightarrow 0 < -k+3^{-a} < 1 Given, a=log⁡3log⁡32⇒3a=log⁡32a=\log _3 \log _3 2 \Rightarrow 3^a=\log _3 2⇒3−a=log⁡23∴k<log⁡23<2 and  1+k>log⁡23>1⇒k>0\begin{array}{lc} \Rightarrow & 3^{-a}=\log _2 3 \\ \therefore & k < \log _2 3 < 2 \\ \text{\text{ and } } & 1+k>\log _2 3>1 \Rightarrow k>0 \end{array}FromEqs.(ii)and(iii),From Eqs. (ii) and (iii),0 < k < 2 \Rightarrow k=1 [\because kisaninteger]is an integer] (D) sin⁡θ=cos⁡ϕ⇒cos⁡(π2−θ)=cos⁡ϕ⇒π2−θ=2nπ±ϕ,n∈Z⇒θ±ϕ−π2=−2nπ,n∈Z⇒1π(θ±ϕ−π2)=−2n,n∈Z\begin{array}{rlrl} \text{ (D) } & \sin \theta & =\cos \phi \\ \Rightarrow & \cos \left(\frac{\pi}{2}-\theta\right) & =\cos \phi \\ \Rightarrow & \frac{\pi}{2}-\theta & =2 n \pi \pm \phi, n \in Z \\ \Rightarrow & & \theta \pm \phi-\frac{\pi}{2} & =-2 n \pi, n \in Z \\ \Rightarrow & & \frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right) & =-2 n, n \in Z \end{array}$$
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