Exam

JEE Advanced

Year

2009

Paper

Paper 2

Subject

Mathematics

Chapter

Functions

JEE Advanced 2009 · Paper 2 · Mathematics · Functions

Match the statements/expressions given in Column I with the values...

Match the statements/expressions given in Column I with the values given in Column II.

Step-by-step solution

(a) Let $f(x)=x e^{\sin x}-\cos x$ $\overrightarrow{f^{\prime}}(x)=e^{\sin x}+x e^{\sin x} \cos x+\sin x \geq 0$ For interval $x \in\left(0, \frac{\pi}{2}\right)$, $f$ is strictly increasing. $\therefore \quad f(0)=-1$ and $f\left(\frac{\pi}{2}\right)=\frac{\pi}{2} e \Rightarrow$ one solution (b) Since, \left|$\begin{array}{lll}k & 4 & 1 \\ 4 & k & 2 \\ 2 & 2 & 1\end{array}$\right|=0 $\Rightarrow k(k-4)-4(0)+1(8-2 k)=0$ $\Rightarrow \quad k^2-6 k+8=0$ $\Rightarrow \quad k=2,4$ (c) Let $y=|x-1|+|x-2|+|x+1|$ $+|x+2|$ For solutions, $4 k \geq 6 \Rightarrow k \geq 3 / 2$
$\therefore$ Integer values of $k$ are $2,3,4,5$. (d) Given, $\frac{d y}{d x}=y+1$ $\Rightarrow \quad \ln |(y+1)|=x+C$ $\Rightarrow \ln 2=C \Rightarrow \ln |y+1|=x+\ln 2$ Put $x=\ln 2$ $\therefore \quad \ln (y+1)=\ln 2+\ln 2=\ln 4$ $y+1=4 \Rightarrow y=3$