Application of Derivatives
Let $\mathrm{g}(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x) > 0, x \in(0,1)$. If $\mathrm{g}$ is decreasing in the interval $(0, a)$ and increasing in the interval $(\alpha, 1)$, then $\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$ is equal to :
The slope of tangent at any point (x, y) on a curve $y=y(x)$ is ${{{x^2} + {y^2}} \over {2xy}},x > 0$. If $y(2) = 0$, then a value of $y(8)$ is :
A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm$^2$) is equal to :
The sum of the absolute maximum and minimum values of the function $f(x)=\left|x^{2}-5 x+6\right|-3 x+2$ in the interval $[-1,3]$ is equal to :
A wire of length $20 \mathrm{~m}$ is to be cut into two pieces. A piece of length $l_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $l_{2}$ is made into a circle of area $A_{2}$. If $2 A_{1}+3 A_{2}$ is minimum then $\left(\pi l_{1}\right): l_{2}$ is equal to :
and $g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b$
have a common extreme point, then $a+2 b+7$ is equal to :
The number of points on the curve $y=54 x^{5}-135 x^{4}-70 x^{3}+180 x^{2}+210 x$ at which the normal lines are parallel to $x+90 y+2=0$ is :
Let the function $f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6$ have a maxima for some value of $x < 0$ and a minima for some value of $x > 0$. Then, the set of all values of p is
Let $x=2$ be a local minima of the function $f(x)=2x^4-18x^2+8x+12,x\in(-4,4)$. If M is local maximum value of the function $f$ in ($-4,4)$, then M =
Let $f:(0,1)\to\mathbb{R}$ be a function defined $f(x) = {1 \over {1 - {e^{ - x}}}}$, and $g(x) = \left( {f( - x) - f(x)} \right)$. Consider two statements
(I) g is an increasing function in (0, 1)
(II) g is one-one in (0, 1)
Then,
If the maximum and the minimum perimeters of such triangles are obtained at
$t=\alpha$ and $t=\beta$ respectively, then $6 \alpha+21 \beta$ is equal to ___________.
Explanation:
We want to find the maximum and minimum perimeters of such triangles, which occur at $t=\alpha$ and $t=\beta$, respectively.
To find the minimum perimeter, we use a geometric approach. Reflect point $B$ over the line $y=4$ to get $B'(0,8)$. The line segment $AB'$ intersects the line $y=4$ (which is the $y$-coordinate of point $C$) at the point which gives the minimum perimeter.
The slope of $AB'$ is :
$m_{AB'} = \frac{8 - 1}{0 - 2} = \frac{-7}{2}$
The equation of the line $AB'$ is then $y - 1 = m_{AB'}(x - 2)$, or $7x + 2y = 16$.
Solving this equation for $y = 4$ yields $x = \frac{8}{7}$. So, the minimum perimeter is achieved at point $C\left(\frac{8}{7}, 4\right)$, so $\beta = \frac{8}{7}$.
For the maximum perimeter, we notice that it will be achieved when point $C$ is either at $(0,4)$ or at $(4,4)$, since these are the furthest points from $A(2,1)$ within the range of $t$. By calculating the perimeters at these points, we find that the maximum perimeter is achieved at $\alpha = 4$.
Finally, we calculate $6\alpha + 21\beta = 6 \cdot 4 + 21 \cdot \frac{8}{7} = 24 + 24 = 48$.
Therefore, $6\alpha + 21\beta = 48$.
Let the quadratic curve passing through the point $(-1,0)$ and touching the line $y=x$ at $(1,1)$ be $y=f(x)$. Then the $x$-intercept of the normal to the curve at the point $(\alpha, \alpha+1)$ in the first quadrant is __________.
Explanation:
The curve passes through $(-1,0)$
$0=a-b+c \Rightarrow a+c=b$ ..........(i)
The curve also passes through $(1,1)$
$ \begin{gathered} a+b+c=1 .........(ii)\\\\ 2 b=1 \Rightarrow b=\frac{1}{2} \end{gathered} $
$ f^{\prime}(x)=2 a x+b $
Slope tangent of curve $=f^{\prime}(x)$ at $(1,1)=2 a+b$
Slope of line $y=x$ is 1
$ \begin{aligned} &\therefore 2 a+b =1 \\\\ &2 a+\frac{1}{2} =1 \Rightarrow a=\frac{1}{4} \\\\ &c =1-a-b \Rightarrow c=\frac{1}{4} \end{aligned} $
$ \begin{aligned} & f(x)=\frac{1}{4} x^2+\frac{1}{2} x+\frac{1}{4} = \frac{(x+1)^2}{4} \\\\ & f^{\prime}(x)=\frac{2(x+1)}{4}=\frac{x+1}{2} \end{aligned} $
$ \begin{aligned} & f(x) \text { passes through }(\alpha, \alpha+1) \\\\ & \begin{array}{l} \therefore \alpha+1=\frac{(\alpha+1)^2}{4} \\\\ \Rightarrow \alpha+1=4 \text { or } \alpha=3 \end{array} \end{aligned} $
Equation of normal at $(3,4)$ is
$ y-4=-\frac{1}{2}(x-3) $
For $x$, intercept, $y=0$
$ x-3=8 \text { or } x=11 $
Hence, the required $x$ intercept is 11 .
If $a_{\alpha}$ is the greatest term in the sequence $\alpha_{n}=\frac{n^{3}}{n^{4}+147}, n=1,2,3, \ldots$, then $\alpha$ is equal to _____________.
Explanation:
For maxima/minima, put $\frac{d y}{d x}=0$
$ \begin{aligned} & \Rightarrow 441 x^2-x^6=0 \Rightarrow x^4=441 \\\\ & \Rightarrow x= \pm \sqrt{21}, \pm \sqrt{21} i \end{aligned} $
Now, by descrates rule on number line we have
Since sign changes from negative to positive at 0 .
$\therefore$ Maximum value of is at $x=\sqrt{21}=4.58$
Now, $4<4.5<5$
$ \begin{aligned} & \therefore \text { yat } x=4=\frac{64}{403}=0.159 \\\\ & y \text { at } x=5=\frac{125}{772}=0.162 \end{aligned} $
So, $y$ is maximum at $x=5$
$ \therefore \alpha=5 $
Let a curve $y=f(x), x \in(0, \infty)$ pass through the points $P\left(1, \frac{3}{2}\right)$ and $Q\left(a, \frac{1}{2}\right)$. If the tangent at any point $R(b, f(b))$ to the given curve cuts the $\mathrm{y}$-axis at the point $S(0, c)$ such that $b c=3$, then $(P Q)^{2}$ is equal to __________.
Explanation:
$ y-f(b)=f^{\prime}(b)(x-b) $
which passes through $S(0, c)$
$ \begin{aligned} & \therefore c-f(b)=f^{\prime}(b)(0-b) \\\\ & b f^{\prime}(b)-f(b)=-c \\\\ & \Rightarrow b f^{\prime}(b)-f(b)=\frac{-3}{b} (\because b c=3) \\\\ & \Rightarrow \frac{b f^{\prime}(b)-f(b)}{b^2}=\frac{-3}{b^3} \\\\ & \Rightarrow d\left(\frac{f(b)}{b}\right)=\frac{-3}{b^3} \\\\ & \Rightarrow \frac{f(b)}{b}=\frac{3}{2 b^2}+c \end{aligned} $
which passes through $P\left(1, \frac{3}{2}\right)$
$ \begin{aligned} & \Rightarrow \frac{3 / 2}{1}=\frac{3}{2}+c \\\\ & \Rightarrow c=0 \\\\ & \therefore f(b)=\frac{3}{2 b^2} \times b \\\\ & \Rightarrow f(b)=\frac{3}{2 b} \end{aligned} $
$\because$ It passes through $Q\left(a, \frac{1}{2}\right)$
$ \begin{aligned} & \therefore \frac{1}{2}=\frac{3}{2 a} \\\\ & \Rightarrow a=3 \\\\ & \therefore P \equiv\left(1, \frac{3}{2}\right) \text { and } Q \equiv\left(3, \frac{1}{2}\right)\\\\ & \therefore (P Q)^2=(3-1)^2+\left(\frac{1}{2}-\frac{3}{2}\right)^2=4+1=5 \end{aligned} $
The number of points, where the curve $y=x^{5}-20 x^{3}+50 x+2$ crosses the $\mathrm{x}$-axis, is ____________.
Explanation:
$ \begin{aligned} & y=x^5-20 x^3+50 x+2 \\\\ & \Rightarrow \frac{d y}{d x}=5 x^4-60 x^2+50 \end{aligned} $
On putting $\frac{d y}{d x}=0$
$ \begin{array}{ll} \Rightarrow & 5\left(x^4-12 x^2+10\right)=0 \\\\ \Rightarrow & x^2=\frac{12 \pm \sqrt{144-40}}{2}=6 \pm \sqrt{26} \\\\ \Rightarrow & x^2=6-\sqrt{26}, 6+\sqrt{26} \\\\ \Rightarrow & x^2=6-5.10,6+5.10 \\\\ \Rightarrow & x^2=09,11.1 \\\\ \Rightarrow & x= \pm \sqrt{0.9}, \pm \sqrt{11.1} \\\\ \Rightarrow & x=-0.95,0.95,-3.33,3.33 \end{array} $
Now,
$ \begin{aligned} y(0) & =2(+\mathrm{ve}) \Rightarrow y(1)=+\mathrm{ve} \\\\ y(2) & =-\mathrm{ve} \Rightarrow y(3.3)=-\mathrm{ve} \\\\ y(-1) & =-\mathrm{ve} \Rightarrow y(-2)=+\mathrm{ve} \\\\ y(-3.3) & =-\mathrm{ve} \end{aligned} $
$ \because \text { Required number of points }=5 $
If the equation of the normal to the curve $y = {{x - a} \over {(x + b)(x - 2)}}$ at the point (1, $-$3) is $x - 4y = 13$, then the value of $a + b$ is equal to ___________.
Explanation:
Given curve : $y = {{x - a} \over {(x + b)(x - 2)}}$ at $(1, - 3)$
$\therefore$ $ - 3 = {{1 - a} \over {(1 + b)( - 1)}} \Rightarrow 3 + 3b = 1 - a$
$\beta \Rightarrow a + 3b + 2 = 0$
$y = {{x - a} \over {(x + b)(x - 2)}}$
${{dy} \over {dx}} = {{(x + b)(x - 2) - (x - a)[(x + b) + (x - 2)]} \over {{{[(x + b)(x - 2)]}^2}}}$
at $(1, - 3)\,{m_T} = {{ - (1 + b) - (1 - a)(b)} \over {{{(1 + b)}^2}}} = - 4$
$\therefore$ $1 + b + b - ab = 4{(1 + b)^2}$
$ \Rightarrow 1 + 2b + b(3b + 2) = 4{b^2} + 4 + 8b$
$ \Rightarrow {b^2} + 4b + 3 = 0$
$(b + 1)(b + 3) = 0$
$b = - 1,a = 1$ but $1 + b \ne 0$
$b = - 3,a = 7$ $\therefore$ $b \ne - 1$
$\therefore$ $a + b = 04$
Let $f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in \mathrm{R}$. Then which of the following statements are true?
$\mathrm{P}: x=0$ is a point of local minima of $f$
$\mathrm{Q}: x=\sqrt{2}$ is a point of inflection of $f$
$R: f^{\prime}$ is increasing for $x>\sqrt{2}$
The function $f(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}$, is :
If the minimum value of $f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0$, is 14 , then the value of $\alpha$ is equal to :
If the maximum value of $a$, for which the function $f_{a}(x)=\tan ^{-1} 2 x-3 a x+7$ is non-decreasing in $\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$, is $\bar{a}$, then $f_{\bar{a}}\left(\frac{\pi}{8}\right)$ is equal to :
If the absolute maximum value of the function $f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$ in the interval $[-3,0]$ is $f(\alpha)$, then :
The curve $y(x)=a x^{3}+b x^{2}+c x+5$ touches the $x$-axis at the point $\mathrm{P}(-2,0)$ and cuts the $y$-axis at the point $Q$, where $y^{\prime}$ is equal to 3 . Then the local maximum value of $y(x)$ is:
If xy4 attains maximum value at the point (x, y) on the line passing through the points (50 + $\alpha$, 0) and (0, 50 + $\alpha$), $\alpha$ > 0, then (x, y) also lies on the line :
Let $f(x) = 4{x^3} - 11{x^2} + 8x - 5,\,x \in R$. Then f :
Let f : R $\to$ R be a function defined by f(x) = (x $-$ 3)n1 (x $-$ 5)n2, n1, n2 $\in$ N. Then, which of the following is NOT true?
A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :
The number of real solutions of
${x^7} + 5{x^3} + 3x + 1 = 0$ is equal to ____________.
Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :
The sum of the absolute minimum and the absolute maximum values of the
function f(x) = |3x $-$ x2 + 2| $-$ x in the interval [$-$1, 2] is :
Let S be the set of all the natural numbers, for which the line ${x \over a} + {y \over b} = 2$ is a tangent to the curve ${\left( {{x \over a}} \right)^n} + {\left( {{y \over b}} \right)^n} = 2$ at the point (a, b), ab $\ne$ 0. Then :
Let $f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10$, $x \in [ - 1,1]$. If [a, b] is the range of the function f, then 4a $-$ b is equal to :
Water is being filled at the rate of 1 cm3 / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm2 / sec) at which the wet conical surface area of the vessel increases is
If the angle made by the tangent at the point (x0, y0) on the curve $x = 12(t + \sin t\cos t)$, $y = 12{(1 + \sin t)^2}$, $0 < t < {\pi \over 2}$, with the positive x-axis is ${\pi \over 3}$, then y0 is equal to:
The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by ${{{x^2}} \over {xy - {x^2}{y^2} - 1}}$. If the curve passes through the point (1, 1), then e . y(e) is equal to
Let $\lambda$$^ * $ be the largest value of $\lambda$ for which the function ${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36x + 48$ is increasing for all x $\in$ R. Then ${f_{{\lambda ^ * }}}(1) + {f_{{\lambda ^ * }}}( - 1)$ is equal to :
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :
For the function
$f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1$, which one of the following is NOT correct?
If the tangent at the point (x1, y1) on the curve $y = {x^3} + 3{x^2} + 5$ passes through the origin, then (x1, y1) does NOT lie on the curve :
The sum of absolute maximum and absolute minimum values of the function $f(x) = |2{x^2} + 3x - 2| + \sin x\cos x$ in the interval [0, 1] is :
Let $\lambda x - 2y = \mu $ be a tangent to the hyperbola ${a^2}{x^2} - {y^2} = {b^2}$. Then ${\left( {{\lambda \over a}} \right)^2} - {\left( {{\mu \over b}} \right)^2}$ is equal to :
If the tangent to the curve $y=x^{3}-x^{2}+x$ at the point $(a, b)$ is also tangent to the curve $y = 5{x^2} + 2x - 25$ at the point (2, $-$1), then $|2a + 9b|$ is equal to __________.
Explanation:
$ =m=\left(\frac{d y}{d x}\right)_{\mathrm{at}(2,-1)}=22 $
$\therefore \quad$ Equation of tangent $: y+1=22(x-2)$
$\therefore \quad y=22 x-45$.
Slope of tangent to $y=x^{3}-x^{2}+x$ at point $(a, b)$
$ =3 a^{2}-2 a+1 $
$3 a^{2}-2 a+1=22$
$3 a^{2}-2 a-21=0$
$\therefore \quad a=3$ or $-\frac{7}{3}$
Also $b=a^{3}-a^{2}+a$
Then $(a, b)=(3,21)$ or $\left(-\frac{7}{3},-\frac{151}{9}\right)$.
$\left(-\frac{7}{3},-\frac{151}{9}\right)$ does not satisfy the equation of tangent
$\therefore \quad a=3, b=21$
$\therefore|2 a+9 b|=195$
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $\tan ^{-1} \frac{3}{4}$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ______________.
Explanation:
$v = {1 \over 3}\pi {r^2}h$ ..... (i)
And $\tan \theta = {3 \over 4} = {r \over h}$ ...... (ii)
i.e. if $h = 4,\,r = 3$
$v = {1 \over 3}\pi {r^2}\left( {{{4r} \over 3}} \right)$
${{dv} \over {dt}} = {{4\pi } \over 9}3{r^2}{{dr} \over {dt}} \Rightarrow 6 = {{4\pi } \over 3}(9){{dr} \over {dt}}$
$ \Rightarrow {{dr} \over {dt}} = {1 \over {2\pi }}$
Curved area $ = \pi r\sqrt {{r^2} + {h^2}} $
$ = \pi r\sqrt {{r^2} + {{16{r^2}} \over 9}} $
$ = {5 \over 3}\pi {r^2}$
${{dA} \over {dt}} = {{10} \over 3}\pi r{{dr} \over {dt}}$
$ = {{10} \over 3}\pi \,.\,3\,.\,{1 \over {2\pi }}$
$ = 5$
Let $M$ and $N$ be the number of points on the curve $y^{5}-9 x y+2 x=0$, where the tangents to the curve are parallel to $x$-axis and $y$-axis, respectively. Then the value of $M+N$ equals ___________.
Explanation:
Here equation of curve is
${y^5} - 9xy + 2x = 0$ ...... (i)
On differentiating : $5{y^4}{{dy} \over {dx}} - 9y - 9x{{dy} \over {dx}} + 2 = 0$
$\therefore$ ${{dy} \over {dx}} = {{9y - 2} \over {5{y^4} - 9x}}$
When tangents are parallel to x-axis then $9y - 2 = 0$
$\therefore$ $M = 1$.
For tangent perpendicular to x-axis
$5{y^4} - 9x = 0$ ...... (ii)
From equation (i) and (ii) we get only one point.
$\therefore$ $N = 1$.
$\therefore$ $M + N = 2$.
Let the function $f(x)=2 x^{2}-\log _{\mathrm{e}} x, x>0$, be decreasing in $(0, \mathrm{a})$ and increasing in $(\mathrm{a}, 4)$. A tangent to the parabola $y^{2}=4 a x$ at a point $\mathrm{P}$ on it passes through the point $(8 \mathrm{a}, 8 \mathrm{a}-1)$ but does not pass through the point $\left(-\frac{1}{a}, 0\right)$. If the equation of the normal at $P$ is : $\frac{x}{\alpha}+\frac{y}{\beta}=1$, then $\alpha+\beta$ is equal to ________________.
Explanation:
$\delta '(x) = {{4{x^2} - 1} \over x}$ so f(x) is decreasing in $\left( {0,{1 \over 2}} \right)$ and increasing in $\left( {{1 \over 2},\infty } \right) \Rightarrow a = {1 \over 2}$
Tangent at ${y^2} = 2x \Rightarrow y = ,x + {1 \over {2m}}$
It is passing through $(4,3)$
$3 = 4m + {1 \over {2m}} \Rightarrow m = {1 \over 2}$ or ${1 \over 4}$
So tangent may be
$y = {1 \over 2}x + 1$ or $y = {1 \over 4}x + 2$
But $y = {1 \over 2}x + 1$ passes through $( - 2,0)$ so rejected.
Equation of normal
$y = - 4x - 2\left( {{1 \over 2}} \right)( - 4) - {1 \over 2}{( - 4)^3}$
or $y = - 4x + 4 + 32$
or ${x \over 9} + {y \over {36}} = 1$
The sum of the maximum and minimum values of the function $f(x)=|5 x-7|+\left[x^{2}+2 x\right]$ in the interval $\left[\frac{5}{4}, 2\right]$, where $[t]$ is the greatest integer $\leq t$, is ______________.
Explanation:
$f(x) = |5x - 7| + [{x^2} + 2x]$
$ = |5x - 7| + [{(x + 1)^2}] - 1$
Critical points of
$f(x) = {7 \over 5},\sqrt 5 - 1,\,\sqrt 6 - 1,\,\sqrt 7 - 1,\,\sqrt 8 - 1,\,2$
$\therefore$ Maximum or minimum value of $f(x)$ occur at critical points or boundary points
$\therefore$ $f\left( {{5 \over 4}} \right) = {3 \over 4} + 4 = {{19} \over 4}$
$f\left( {{7 \over 5}} \right) = 0 + 4 = 4$
as both $|5x - 7|$ and ${x^2} + 2x$ are increasing in nature after $x = {7 \over 5}$
$\therefore$ $f(2) = 3 + 8 = 11$
$\therefore$ $f{\left( {{7 \over 5}} \right)_{\min }} = 4$ and $f{(2)_{\max }} = 11$
Sum is $4 + 11 = 15$
A hostel has 100 students. On a certain day (consider it day zero) it was found that two students are infected with some virus. Assume that the rate at which the virus spreads is directly proportional to the product of the number of infected students and the number of non-infected students. If the number of infected students on 4th day is 30, then number of infected students on 8th day will be __________.
Explanation:
Total students = 100
At t = 0 (zero day), infected student = 2
Let at t = t day infected student = x
$\therefore$ At t = t day non infected student = (100 $-$ x)
Rate of infection $ = {{dx} \over {dt}}$
Given, ${{dx} \over {dt}} \propto x(100 - x)$
$ \Rightarrow \int\limits_{}^{} {{{dx} \over {x(100 - x)}} = \int\limits_{}^{} {k\,dt} } $
$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {{{100 - x + x} \over {x(100 - x)}}dx = k\,t + c} $
$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {\left( {{1 \over x} + {1 \over {100 - x}}} \right)dx = k\,t + c} $
$ \Rightarrow {1 \over {100}}\left[ {\ln x - \ln (100 - x)} \right] = k\,t + c$
$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} = k\,t + c$ ...... (1)
Given, At, t = 0, x = 2
$\therefore$ ${1 \over {100}}\ln {2 \over {98}} = c$
Putting value of c in equation (1), we get
${1 \over {100}}\ln {x \over {100 - x}} = kt + {1 \over {100}}\ln {2 \over {98}}$
$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} - {1 \over {100}}\ln {2 \over {98}} = kt$
$ \Rightarrow {1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = kt$
Given, At t = 4, x = 30
$\therefore$ ${1 \over {100}}\ln {{30 \times 98} \over {2(70)}} = k \times 4$
$ \Rightarrow k = {1 \over {400}}\ln 21$
$\therefore$ ${1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = t \times {1 \over {400}} \times \ln 21$
Now, when t = 8, then r = ?
${1 \over {100}}\ln {{49x} \over {(100 - x)}} = 8 \times {1 \over {400}} \times \ln 21$
$ \Rightarrow \ln {{49x} \over {(100 - x)}} = 2\ln 21$
$ \Rightarrow {{49x} \over {100 - x}} = {21^2}$
$ \Rightarrow {x \over {100 - x}} = {{21 \times 21} \over {49}}$
$ \Rightarrow {x \over {100 - x}} = 9$
$ \Rightarrow x = 900 - 9x$
$ \Rightarrow 10x = 900$
$ \Rightarrow x = 90$
Let l be a line which is normal to the curve y = 2x2 + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line l and O is origin, then the area of the triangle OPQ is equal to ___________.
Explanation:
${{{y_1} - 4} \over {{x_1} - 6}} = - {1 \over {4{x_1} + 1}}$
$ \Rightarrow {{2x_1^2 + {x_1} - 2} \over {{x_1} - 6}} = - {1 \over {4{x_1} + 1}}$
$ \Rightarrow 6 - {x_1} = 8x_1^3 + 6x_1^2 - 7{x_1} - 2$
$ \Rightarrow 8x_1^3 + 6x_1^2 - 6{x_1} - 8 = 0$
So ${x_1} = 1 \Rightarrow {y_1} = 5$
Area $ = \left| {{1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr 6 & 4 & 1 \cr 1 & 5 & 1 \cr } } \right|} \right| = 13$.
Let $f(x) = |(x - 1)({x^2} - 2x - 3)| + x - 3,\,x \in R$. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ____________.
Explanation:
$f(x) = \left| {(x - 1)(x + 1)(x - 3)} \right| + (x - 3)$
$f(x) = \left\{ {\matrix{ {(x - 3)({x^2})} & {3 \le x \le 4} \cr {(x - 3)(2 - {x^2})} & {1 \le x < 3} \cr {(x - 3)({x^2})} & {0 < x < 1} \cr } } \right.$
$f'(x) = \left\{ {\matrix{ {3{x^2} - 6x} & {3 < x < 4} \cr { - 3{x^2} + 6x + 2} & {1 < x < 3} \cr {3{x^2} - 6x} & {0 < x < 1} \cr } } \right.$
$f'({3^ + }) > 0\,\,\,f'({3^ - }) < 0 \to $ Minimum
$f'({1^ + }) > 0\,\,\,f'({1^ - }) < 0 \to $ Minimum
$x \in (1,3)\,\,f'(x) = 0$ at one point $\to$ Maximum
$x \in (3,4)\,\,f'(x) \ne 0$
$x \in (0,1)\,\,f'(x) \ne 0$
So, 3 points.
Statement 1 : there exists x1, x2 $\in$(2, 4), x1 < x2, such that f'(x1) = $-$1 and f'(x2) = 0.
Statement 2 : there exists x3, x4 $\in$ (2, 4), x3 < x4, such that f is decreasing in (2, x4), increasing in (x4, 4) and $2f'({x_3}) = \sqrt 3 f({x_4})$.
Then
${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$ is :
