Embibe Experts Solutions for Chapter: Application of Derivatives, Exercise 1: JEE Main - 10th April 2015

The range of the function $f\left(x\right)=\sqrt{3-x}+\sqrt{2+x}$ is

A wire of length $20 \mathrm{m}$ is to be cut into two pieces. A piece of length ${\ell }_1$ is bent to make a square of area $A_1$ and the other piece of length ${\ell }_2$ is made into a circle of area $A_2$. If $2 A_1+3 A_2$ is minimum then $\left(\pi {\ell }_1\right):{\ell }_2$ is equal to:

Let $a_1,a_2,a_3,\dots \dots$. be an A.P. If $a_7=3$, the product $\left(a_1{a}_4\right)$ is minimum and the sum of its first $n$ terms is zero then $n!-4a_{n\left(n+2\right)}$ is equal to

The absolute minimum value, of the function $f\left(x\right)=\left|x^2-x+1\right|+\left[x^2-x+1\right]$, where $\left[t\right]$ denotes the greatest integer function, in the interval $\left[-1,2\right]$, is

Let $f:R-\left\{2,6\right\}\to R$ be real valued function defined as $f\left(x\right)=\frac{x+2x+1}{x^2-8x+12}$. Then range of $f$ is

Let $f\left(x\right)=2x+{\tan }^{-1}x$ and $g\left(x\right)={\log }_e\left(\sqrt{1+x^2}+x\right), x\in \left[0,3\right]$. Then

The area enclosed by the closed curve $C$ given by the differential equation $\frac{dy}{dx}+\frac{x+a}{y-2}=0,y\left(1\right)=0$ is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is

The sum of the abosolute maximum and minimum values of the function $f\left(x\right)=\left|x^2-5x+6\right|-3x+2$ in the interval $\left[-1,3\right]$ is equal to :