JEE Main 2010 — Calculus

Let $f: R \rightarrow R$ be a continuous function defined by $f(x)=\frac{1}{e^x+2 e^{-x}}$. Statement-1: $f(c)=\frac{1}{3}$, for some $c \in R$. Statement-2: $0 < f(x) \leq \frac{1}{2 \sqrt{2}}$, for all $x \in R$

Let $f:(-1,1) \rightarrow R$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. Let $g(x)=[f(2 f(x)+2)]^2$. Then $g^{\prime}(0)=$

Let $p(x)$ be a function defined on $R$ such that $p^{\prime}(x)=p^{\prime}(1-x)$, for all $x \in[0,1], p(0)=1$ and $p(1)=41$. Then $\int_0^1 p(x) d x$ equals

Let $f: R \rightarrow R$ be a positive increasing function with $\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1$. Then $\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=$

Let $f: R \rightarrow R$ be defined by $f(x)=\left\{\begin{array}{ll}k-2 x, & \text { if } x \leq-1 \\ 2 x+3, & \text { if } x>-1\end{array}\right.$. If $f$ has a local minimum at $x=-1$, then a possible value of $\mathrm{k}$ is

Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \frac{\pi}{2}$ is

The area bounded by the curves $y=\cos x$ and $y=\sin x$ between the ordinates $x=0$ and $x=\frac{3 \pi}{2}$ is