JEE Main Application of Derivatives Practice Questions With Solutions

Let $f:\to \mathbb{R}\to (0,\mathrm{\infty })$f :→ \mathbb{R} \rightarrow \left(\right. 0 , \infty \left.\right) be strictly increasing function such that $\underset{x\to \mathrm{\infty }}{lim}\frac{f(7x)}{f(x)}=1$\underset{x \rightarrow \infty}{lim} \frac{f \left(\right. 7 x \left.\right)}{f \left(\right. x \left.\right)} = 1. Then, the value of $\underset{x\to \mathrm{\infty }}{lim}[\frac{f(5x)}{f(x)}-1]$\underset{x \rightarrow \infty}{lim} \left[\right. \frac{f \left(\right. 5 x \left.\right)}{f \left(\right. x \left.\right)} - 1 \left]\right. is equal to

If the function $f:(-\mathrm{\infty },-1]\to (a,b]$f : \left(\right. - \infty , - 1 \left]\right. \rightarrow \left(\right. a , b \left]\right. defined by $f(x)=e^{x^3-3x+1}$f \left(\right. x \left.\right) = e^{x^{3} - 3 x + 1} is one - one and onto, then the distance of the point $P(2b+4,a+2)$P \left(\right. 2 b + 4 , a + 2 \left.\right) from the line $x+e^{-3}y=4$x + e^{- 3} y = 4 is :

$\text{ If }f(x)=\left|\begin{array}{ccc}{x}^3& 2x^2+1& 1+3x\\ 3x^2+2& 2x& x^3+6\\ x^3-x& 4& x^2-2\end{array}\right|\text{ for all }x\in \mathbb{R}\text{, then }2f(0)+f^{\prime }(0)\text{ is equal to }$\textrm{ }\text{If}\textrm{ } f \left(\right. x \left.\right) = \left|\right. x^{3} & 2 x^{2} + 1 & 1 + 3 x \\ 3 x^{2} + 2 & 2 x & x^{3} + 6 \\ x^{3} - x & 4 & x^{2} - 2 \left|\right. \textrm{ }\text{for all}\textrm{ } x \in \mathbb{R} ,\text{ then}\textrm{ } 2 f \left(\right. 0 \left.\right) + f^{′} \left(\right. 0 \left.\right) \textrm{ }\text{is equal to}\textrm{ }

Let $f(x)=(x+3{)}^2(x-2{)}^3,x\in [-4,4]$f \left(\right. x \left.\right) = \left(\right. x + 3 \left.\right)^{2} \left(\right. x - 2 \left.\right)^{3} , x \in \left[\right. - 4 , 4 \left]\right.. If $M$M and $m$m are the maximum and minimum values of $f$f, respectively in $[-4,4]$\left[\right. - 4 , 4 \left]\right., then the value of $M-m$M - m is

The maximum area of a triangle whose one vertex is at $(0,0)$\left(\right. 0 , 0 \left.\right) and the other two vertices lie on the curve $y=-2x^2+54$y = - 2 x^{2} + 54 at points $(x,y)$\left(\right. x , y \left.\right) and $(-x,y)$\left(\right. - x , y \left.\right), where $y>0$y > 0, is :

The function $f(x)=\frac{x}{x^2-6x-16},x\in \mathbb{R}-\{-2,8\}$f \left(\right. x \left.\right) = \frac{x}{x^{2} - 6 x - 16} , x \in \mathbb{R} - \left{\right. - 2 , 8 \left.\right}

The function $f(x)=2x+3(x{)}^{\frac{2}{3}},x\in \mathbb{R}$f \left(\right. x \left.\right) = 2 x + 3 \left(\right. x \left.\right)^{\frac{2}{3}} , x \in \mathbb{R}, has

Consider the function $f:[\frac{1}{2},1]\to \mathbb{R}$f : \left[\right. \frac{1}{2} , 1 \left]\right. \rightarrow \mathbb{R} defined by $f(x)=4\sqrt{2}{x}^3-3\sqrt{2}x-1$f \left(\right. x \left.\right) = 4 \sqrt{2} x^{3} - 3 \sqrt{2} x - 1. Consider the statements

(I) The curve $y=f(x)$y = f \left(\right. x \left.\right) intersects the $x$x-axis exactly at one point.

(II) The curve $y=f(x)$y = f \left(\right. x \left.\right) intersects the $x$x-axis at $x=\cos \frac{\pi }{12}$x = cos ⁡ \frac{\pi}{12}.

Then

Let $g(x)=3f\left(\frac{x}{3}\right)+f(3-x)$g \left(\right. x \left.\right) = 3 f \left(\right. \frac{x}{3} \left.\right) + f \left(\right. 3 - x \left.\right) and $f^{\prime \prime }(x)>0$f^{′ ′} \left(\right. x \left.\right) > 0 for all $x\in (0,3)$x \in \left(\right. 0 , 3 \left.\right). If $g$g is decreasing in $(0,\alpha )$\left(\right. 0 , \alpha \left.\right) and increasing in $(\alpha ,3)$\left(\right. \alpha , 3 \left.\right), then $8\alpha$8 \alpha is :