Application of Derivatives

$\begin{aligned} & f(t)=\frac{|t+1|}{t^2}, t<0 \\ & \text { critical point } t+1=0 \Rightarrow t=-1\end{aligned}$

$f(t)= \begin{cases}-\left(\frac{t+1}{t^2}\right), & t \leq-1 \\ \frac{t+1}{t^2}, & -1 < t < 0\end{cases}$

$f^{\prime}(t)= \begin{cases}\frac{t+2}{t^3}, & t < -1 \\ -\left(\frac{t+2}{t^3}\right), & -1 < t <0\end{cases}$

If $t<-1, f^{\prime}(t)=\left(\frac{t+2}{t^3}\right)$

roots is $t=-2$

for $t \in(-2,-1), f^{\prime}(t)<0$, decreasing interval

$t \in(-\infty,-2), f^{\prime}(t)>0$, Increasing interval

If $t \in(-1,0), f^{\prime}(t)=-\left(\frac{t+2}{t^3}\right)$

$f^{\prime}(t)>0$

so $t \in(-1,0)$ is increasing interval.

It is given that $(2 \alpha, \alpha)$ is the largest interval in which $f(t)$ is decreasing matches with $t \in (-2,-1)$.

so, $\alpha=-1$.

now, $g(x)=2 \log _e(x-2)+\alpha x^2+4 x-\alpha, x>2$

substitute $\alpha=-1$

$ g(x)=2 \log _e(x-2)-x^2+4 x+1, x>2 $

$\Rightarrow $ $g^{\prime}(x)=2 \cdot \frac{1}{x-2}-2 x+4$

$=\frac{2-2 x(x-2)+4(x-2)}{(x-2)}=\frac{-2 x^2+8 x-6}{x-2}$

$ =\frac{-2\left(x^2-4 x+3\right)}{x-2}=\frac{-2(x-3)(x-1)}{(x-2)} $

roots of $g^{\prime}(x)$ for $x>2$ is 3

$ g^{\prime}\left(3^{-}\right)=\text {positive, } g^{\prime}\left(3^{+}\right)=- \text {ve } $

so sign of $g^{\prime}(x)$ changing positive to negative about $x=3$. so $x=3$ is point of local maxima calculating local maximum value

$ g(3)=2 \log _e(3-2)-(3)^2+4(3)+1 $

$ \begin{aligned} & =2 \times \log _e 1-9+12+1 \\\\ & =2 \times 0+4=4 \end{aligned} $

The local maximum value is 4 .

For equal roots, discriminant is 0 .

$ 4\left(f^{\prime}(x)\right)^2-4(f(x))\left(f^{\prime \prime}(x)\right)=0 $

Let $y=f(x)$

$ \begin{aligned} & \left(y^{\prime}\right)^2=y y^{\prime \prime} \\ & \Rightarrow \frac{y^{\prime}}{y}=\frac{y^{\prime \prime}}{y^{\prime}} \end{aligned} $

$ \begin{array}{r} \Rightarrow \ln |y|=\ln \left|y^{\prime}\right|+\ln K \\ \ln (1)=\ln |2|+\ln (K) \Rightarrow \\ \ln K=-\ln 2 \end{array} $

$ \begin{aligned} & \Rightarrow \quad \ln |y|=\ln \left|y^{\prime}\right|-\ln 2 \\ & \Rightarrow \quad \ln \left|\frac{y}{y^{\prime}}\right|=-\ln 2=\ln \left(\frac{1}{2}\right) \\ & \Rightarrow \quad 2 y=y^{\prime} \\ & \Rightarrow \quad 2=\left|\frac{y^{\prime}}{y}\right| \Rightarrow 2 x=\ln |y|+\ln \lambda \\ & \quad 0=\ln |1|+\ln \lambda \\ & \Rightarrow \quad \ln \lambda=0 \\ & \Rightarrow \quad \ln |y|=2 x \\ & y=e^{2 x} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & f^{(\ln x-x)}=e^{2(\ln x-x)}=e^{2\left(\ln x^2-2 x\right)} \\ & =\frac{x^2}{e^{2 x}} \end{aligned}\\ &\text { The function } \frac{x^2}{e^{2 x}} \text { is increasing in }(0,1)\\ &\Rightarrow \alpha+\beta=1 \end{aligned} $

The maximum area of such a parallelogram $A F D E$, with one vertex fixed at $A$ and the other three points lying on the sides of triangle $A B C$, is half the area of triangle $A B C$.

Using the determinant formula for area of triangle with vertices $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ :

Area $\triangle A B C$

$=\frac{1}{2}\left|x_1\left(y_2-y_3^2\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Substitute the coordinates:

$\begin{aligned} & =\frac{1}{2}|4(1-(-3))+1((-3)-(-2))+9((-2)-1)| \\ & =\frac{1}{2}|4(4)+1(-1)+9(-3)| \\ & =\frac{1}{2}|16-1-27|=\frac{1}{2}|-12|=\frac{12}{2}=6 \end{aligned}$

Maximum area of parallelogram $A F D E$ $=\frac{1}{2} \times$ area of triangle $=\frac{1}{2} \times 6=3$

$\begin{aligned} & 5 x^3-15 x-a=0 \\ & f(x)=5 x^3-15 x \\ & f'(x)=15 x^2-15=15(x-1)(x+1) \end{aligned}$

$\begin{aligned} &\mathrm{a} \in(-10,10)\\ &\alpha=-10, \beta=10\\ &\beta-2 \alpha=10+20=30 \end{aligned}$

$\begin{aligned} & f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\ & f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\ & f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\ & 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\ & \Rightarrow\left[-4 p^2+24 p-32,4 p^2-24 p+32\right]+(4-2 p) \\ & {\left[-4 p^2+22 p-28,4 p^2-26 p+36\right]} \\ & {[(p-2)(-4 p+14),(p-2)(4 p-18)]} \\ & \Rightarrow(p-2)[(-4 p+14), 4 p-18] \Rightarrow p \in\left(\frac{7}{2}, \frac{9}{2}\right) \\ & \Rightarrow a=\frac{7}{2}, b=\frac{9}{2} \\ & \Rightarrow 16 a b=4 \times 63=252 \end{aligned}$

$\begin{aligned} & f(x)=1+x\left(\lambda^2-x^2\right) \\ & f(x)=-x^3+\left(\lambda^2 x+1\right) \\ & f^{\prime}(x)=-3 x^2+\lambda^2 \\ & x= \pm \frac{\lambda}{\sqrt{3}} \end{aligned}$

$-\frac{\lambda}{\sqrt{3}}$ should satisfy the given condition

$\begin{aligned} & \frac{x^2+x+2}{x^2+5 x+6}<6 \\ & \frac{1}{(x+2)(x+3)}<0 \end{aligned}$

$\begin{aligned} &\begin{aligned} & x \in(-3,-2) \\ & -3<-\frac{\lambda}{\sqrt{3}}<-2 \\ & -3 \sqrt{3}<-\lambda<-2 \sqrt{3} \\ & 2 \sqrt{3}<\lambda<3 \sqrt{3} \\ & \left.\begin{array}{l} \alpha=2 \sqrt{3} \\ \beta=2 \sqrt{3} \end{array}\right\} \\ & (2 \sqrt{3})^2+(3 \sqrt{2})^2 \\ & 12+27 \\ \end{aligned}\\ &39 \end{aligned}$

$\begin{aligned} & y^2=2 x \\ & a=\left(\frac{1}{2}\right) \end{aligned}$

$\begin{aligned} & A(t)=2 t \times\left(24-\frac{t^2}{2}\right) \\ & A=48 t-t^3 \end{aligned}$

$\begin{aligned} & \frac{d A}{d t}=48-3 t^2 \\ & 48-3 t^2=0 \\ & 3 t^2=48 \\ & t^2=16 \\ & t= \pm 4 \end{aligned}$

$\begin{aligned} & A(4)=48 \times 4-4^3 \\ & =192-64 \\ & A(4)=128 \end{aligned}$

$\begin{aligned} & \text { Let } y=\sqrt{8 x-x^2-12} \Rightarrow(x-4)^2+y^2=2^2 \\ & \Rightarrow d=(y-4)^2+(x-7)^2 \end{aligned}$

$\begin{aligned} \Rightarrow & M=P A^2=16+25=41 \\ & m=P Q^2=(\sqrt{16+9}-2)^2=9 \\ \Rightarrow & M^2-m^2=1681-81=1600 \end{aligned}$

$\begin{aligned} & \therefore \mathrm{m}=3 \\ & \mathrm{f}^{\prime}(\mathrm{x})=2^{\mathrm{x}} \ln 2-2 \mathrm{x}=0 \\ & 2^{\mathrm{x}} \ln 2=2 \mathrm{x} \end{aligned}$

$\begin{aligned} & \therefore \mathrm{n}=2 \\ & \Rightarrow \mathrm{m}+\mathrm{n}=5 \end{aligned}$

$\begin{aligned} & f(x)-f(y) \geq \ln x-\ln y+x-y \\ & \frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1 \end{aligned}$

Let $x>y$

$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1\quad\text{.... (1)}$

Let $x< y$

$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1 \quad\text{.... (2)}$

$\begin{aligned} & \mathrm{f}^1\left(\mathrm{x}^{-}\right)=\mathrm{f}^1\left(\mathrm{x}^{+}\right) \\ & \mathrm{f}^1(\mathrm{x})=\frac{1}{\mathrm{x}}+1 \\ & \mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}^2}\right)=\mathrm{x}^2+1 \end{aligned}$

$\begin{aligned} & \sum_{x=1}^{20}\left(x^2+1\right)=\sum_{x-1}^{20} x^2+20 \\ & =\frac{20 \times 21 \times 41}{6}+20 \\ & =2890 \end{aligned}$

We have a triangle with vertices $A(2,1)$, $B(0,0)$, and $C(t, 4)$, where $t$ belongs to the interval $[0,4]$.

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 be $f(x)=a x^2+b x+c$

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 .

$ \begin{aligned} & \text { Let } y=\frac{x^3}{x^4+147} \\\\ & \Rightarrow \frac{d y}{d x}=\frac{\left(x^4+147\right) \times 3 x^2-x^3\left(4 x^3\right)}{\left(x^4+147\right)^2} \\\\ & =\frac{3 x^6+441 x^2-4 x^6}{\left(x^4+147\right)^2}=\frac{441 x^2-x^6}{\left(x^4+147\right)^2} \end{aligned} $

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 $

Equation of tangent at $R(b, f(b))$

$ 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} $

Given equation of curve

$ \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 $

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 $1 + b \ne 0$

$b = - 3,a = 7$ $\therefore$ $b \ne - 1$

$\therefore$ $a + b = 04$

Slope of tangent to curve $y=5 x^{2}+2 x-25$

$ =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

$\therefore \quad a=3, b=21$

$\therefore|2 a+9 b|=195$

$v = {1 \over 3}\pi {r^2}h$ ..... (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$

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.

$\therefore$ $N = 1$.

$\therefore$ $M + N = 2$.

$\delta '(x) = {{4{x^2} - 1} \over x}$ 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 ${x \over 9} + {y \over {36}} = 1$

$f(x) = |5x - 7| + [{x^2} + 2x]$

$ = |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 $4 + 11 = 15$

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$

${{{y_1} - 4} \over {{x_1} - 6}} = - {1 \over {4{x_1} + 1}}$

$ \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$.

$f(x) = \left| {(x - 1)(x + 1)(x - 3)} \right| + (x - 3)$

$f(x) = \left\{ {\matrix{ {(x - 3)({x^2})} & {3 \le x \le 4} \cr {(x - 3)(2 - {x^2})} & {1 \le x < 3} \cr {(x - 3)({x^2})} & {0 < x < 1} \cr } } \right.$

$f'(x) = \left\{ {\matrix{ {3{x^2} - 6x} & {3 < x < 4} \cr { - 3{x^2} + 6x + 2} & {1 < x < 3} \cr {3{x^2} - 6x} & {0 < x < 1} \cr } } \right.$

$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.

Let f(x) = ax³ + bx² + cx + d

f'(x) = 3ax² + 2bx + c $\Rightarrow$ f''(x) = 6ax + 2b

f'(x) has local minima at x = $-$1, so

$\because$ f''($-$1) = 0 $\Rightarrow$ $-$6a + 2b = 0 $\Rightarrow$ b = 3a ..... (i)

f(x) has local minima at x = 1

f'(1) = 0

$\Rightarrow$ 3a + 6a + c = 0

$\Rightarrow$ c = $-$9a ..... (ii)

f(1) = $-$10

$\Rightarrow$ $-$5a + d = $-$10 ..... (iii)

f($-$1) = 6

$\Rightarrow$ 11a + d = 6 ..... (iv)

Solving Eqs. (iii) and (iv)

a = 1, d = $-$5

From Eqs. (i) and (ii),

b = 3, c = $-$9

$\therefore$ f(x) = x³ + 3x² $-$ 9x $-$ 5

So, f(3) = 27 + 27 $-$ 27 $-$ 5 = 22

f(x) = x² + ax + 1

f'(x) = 2x + a

when f(x) is increasing on [1, 2]

2x + a $\ge$ 0 $\forall$ x$\in$[1, 2]

a $\ge$ $-$2x $\forall$ x$\in$[1, 2]

R = $-$4

when f(x) is decreasing on [1, 2]

2x + a $\le$ 0 $\forall$ x$\in$[1, 2]

a $\le$ $-$2 $\forall$ x$\in$[1, 2]

S = $-$2

|R $-$ S| = | $-$4 + 2 | = 2

3x⁴ + 4x³ $-$ 12x² + 4 = 0

So, let f(x) = 3x⁴ + 4x³ $-$ 12x² + 4

$\therefore$ f'(x) = 12x(x² + x $-$ 2)

= 12x (x + 2) (x $-$ 1)

$ \therefore $ f'(x) = 12x³ + 12x² – 24x = 12x(x + 2) (x – 1)

Points of extrema are at x = 0, –2, 1

f(0) = 4

f(–2) = –28

f(1) = –1

So, 4 Real Roots

Let x + y = 36

x is perimeter of square and y is perimeter of circle side of square = x/4

radius of circle = ${y \over {2\pi }}$

Sum Areas = ${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$

$ = {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$

For min Area :

$x = {{144} \over {\pi + 4}}$

$\Rightarrow$ Radius = y = 36 $-$ ${{144} \over {\pi + 4}}$

$\Rightarrow$ k = ${{36\pi } \over {\pi + 4}}$

$\left( {{4 \over \pi } + 1} \right)k$ = 36

$f(x) = a{x^2} + bx + c$

$f'(x) = 2ax + b,$

$f''(x) = 2a$

Given, $f''( - 1) = {1 \over 2}$

$ \Rightarrow a = {1 \over 4}$

$f'( - 1) = 1 \Rightarrow b - 2a = 1$

$ \Rightarrow b = {3 \over 2}$

$f( - 1) = a - b + c = 2$

$ \Rightarrow c = {{13} \over 4}$

Now, $f(x) = {1 \over 4}({x^2} + 6x + 13),x \in [ - 1,1]$

$f'(x) = {1 \over 4}(2x + 6) = 0$

$ \Rightarrow x = - 3 \notin [ - 1,1]$

$f(1) = 5,f( - 1) = 2$

$f(x) \le 5$

So, $\alpha$_minimum = 5

Let the equation of normal is Y $-$ y = $-$${1 \over m}(X - x)$, where, m = ${{dy} \over {dx}}$

As it passes through (a, b)

$b - y = - {1 \over m}(a - x) = - {{dx} \over {dy}}(a - x)$

$ \Rightarrow (b - y)dy = (x - a)dx$

by $ - {{{y^2}} \over 2} = {{{x^2}} \over 2} - ax + c$ ..... (i)

It passes through (3, $-$3) & (4, $-$2$\sqrt 2 $)

$ \therefore $ $ - 3b - {9 \over 2} = {9 \over 2} - 3a + c$

$ \Rightarrow - 6b - 9 = 9 - 6a + 2c$

$ \Rightarrow 6a - 6b - 2c = 18$

$ \Rightarrow 3a - 3b - c = 9$ .... (ii)

Also,

$ - 2\sqrt 2 b - 4 = 8 - 4a + c$

$4a - 2\sqrt 2 b - c = 12$ .... (iii)

Also, $a - 2\sqrt 2 \,b = 3$ .... (iv) (given)

$(ii) - (iii) \Rightarrow - a + \left( {2\sqrt 2 - 3} \right)b = - 3$ ... (v)

$(iv) + (v) \Rightarrow b = 0,a = 3$

$ \therefore $ ${a^2} + {b^2} + ab = 9$

Let, $f(x) = 2{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 10x + 10$

$ \Rightarrow f'(x) = 10({x^4} + 2{x^3} + 3{x^2} + 2x + 1)$

$ = 10\left( {{x^2} + {1 \over {{x^2}}} + 2\left( {x + {1 \over x}} \right) + 3} \right)$

$ = 10\left( {{{\left( {x + {1 \over x}} \right)}^2} + 2\left( {x + {1 \over x}} \right) + 1} \right)$

$ = 10{\left( {\left( {x + {1 \over x}} \right) + 1} \right)^2} > 0;\forall x \in R$

$ \therefore $ f(x) is strictly increasing function. Since, it is an odd degree polynomial it will have exactly one real root.

Now, by observation.

$f( - 1) = 3 > 0$

$f( - 2) = - 64 + 80 - 80 + 40 - 20 + 10$

$ = - 34 < 0$

$ \Rightarrow f(x)$ has at least one root in $( - 2, - 1) \equiv (a,a + 1)$

$ \Rightarrow a = - 2$

$ \Rightarrow $ |a| = - 2

$x = {y^4}$ and $xy = k$

for intersection ${y^5} = k$ ..... (1)

Also $x = {y^4}$

$ \Rightarrow 1 = 4{y^3}{{dy} \over {dx}} \Rightarrow {{dy} \over {dx}} = {1 \over {4{y^3}}}$

for $xy = k \Rightarrow x = {k \over y}$

$ \Rightarrow 1 = - {k \over {{y^2}}}.{{dy} \over {dx}}$

$ \Rightarrow {{dy} \over {dx}} = {{ - {y^2}} \over k}$

$ \because $ Curve cut orthogonally

$ \Rightarrow {1 \over {4{y^3}}} \times \left( {{{ - {y^2}} \over k}} \right) = - 1$

$ \Rightarrow y = {1 \over {4k}}$

$ \therefore $ from (1), ${y^5} = k$

$ \Rightarrow {1 \over {{{(4k)}^5}}} = k$

$ \Rightarrow 4 = {(4k)^6}$

$f(x) = {x^6} + a{x^5} + b{x^4} + {x^3}$

$\therefore$ $f'(x) = 6{x^5} + 5a{x^4} + 4b{x^3} + 3{x^2}$

Roots 1 & $-$1

$ \therefore $ $6 + 5z + 4b + 3 = 0$ & $ - 6 + 5a - 4b + 3 = 0$ solving

$a = - {3 \over 5}$

$b = - {3 \over 2}$

$ \therefore $ $f(x) = {x^6} - {3 \over 5}{x^5} - {3 \over 2}{x^4} + {x^3}$

$ \therefore $ $5.f(2) = 5\left[ {64 - {{96} \over 5} - 24 + 8} \right] = 144$

$f(x) = {4 \over {\sin x}} + {1 \over {1 - \sin x}}$

Let sinx = t $ \because $ $x \in \left( {0,{\pi \over 2}} \right) \Rightarrow 0 < t < 1$

$f(t) = {4 \over t} + {1 \over {1 - t}}$

$f'(t) = {{ - 4} \over {{t^2}}} + {1 \over {{{(1 - t)}^2}}}$

$ = {{{t^2} - 4{{(1 - t)}^2}} \over {{t^2}{{(1 - t)}^2}}}$

$ = {{(t - 2(1 - t))(t + 2(1 - t))} \over {{t^2}{{(1 - t)}^2}}}$

$ = {{(3t - 2)(2 - t)} \over {{t^2}{{(1 - t)}^2}}}$

${f_{\min }}$ at $t = {2 \over 3}$

${\alpha _{\min }} = f\left( {{2 \over 3}} \right) = {4 \over {{2 \over 3}}} + {1 \over {1 - {2 \over 3}}}$

$ = 6 + 3$

$ = 9$

y = x² – 3x + 2

$ \Rightarrow $ y = (x – 1)(x – 2)

At x-axis y = 0
$ \Rightarrow $ x = 1, 2

So this curve intersects the x-axis at A(1, 0) and B(2, 0).

${{dy} \over {dx}} = 2x - 3$

${\left( {{{dy} \over {dx}}} \right)_{x = 1}} = - 1$ and ${\left( {{{dy} \over {dx}}} \right)_{x = 2}} = 1$

Equation of tangent at A(1, 0) :

y = –1(x –1)

$ \Rightarrow $ x + y = 1

and equation of tangent at B(2, 0):

y = 1(x – 2)

$ \Rightarrow $ x – y = 2

So a = 1 and b = 2

$ \Rightarrow $ ${a \over b}$ = 0.5

Let f(x) = ax³ + bx² + cx + d

Given f(-1) = 10, f(1) = -6

$ \therefore $ -a + b - c + d = 10 ....(i)

and a + b + c + d = -6 ......(ii)

adding (i) + (ii)

2(b + d) = 4

$ \Rightarrow $ b + d = 2 ....(iii)

f'(x) = 3ax² + 2bx + c

Given f'(-1) = 0

$ \Rightarrow $ 3a - 2b + c = 0 .....(iv)

f"(x) = 6ax + 2b

Given f"(1) = 0

$ \therefore $ 6a + 2b = 0 ....(v)

$ \Rightarrow $ b = -3a

adding (iv) + (v), we get

9a + c = 0 ....(vi)

$ \Rightarrow $ $9\left( {{{ - b} \over 3}} \right)$ + c = 0

$ \Rightarrow $ c = 3b

f(x) = ${{{ - b} \over 3}{x^3}}$ + bx² + 3bx + (2 - b)

$ \Rightarrow $ f'(x) = -bx² + 2bx + 3b

= -b(x² - 2x - 3)

At maxima and minima f'(x) = 0

$ \therefore $ (x² - 2x - 3) = 0

$ \Rightarrow $ (x - 3) (x + 1) = 0

x = 3, -1

As a + b + c + d = -6

$ \Rightarrow $ ${{{ - b} \over 3}}$ + b + 3b + 2 - b = -6

$ \Rightarrow $ b = -3

$ \therefore $ f'(x) = 3(x² - 2x - 3)

$ \Rightarrow $ f''(x) = 3(2x - 2)

At x = 3, f''(x) = 3(2.3 - 2) = 12 > 0

$ \therefore $ Minima at x = 3.

Given curve : y² – 3x² + y + 10 = 0

$ \Rightarrow $ 2y${{dy} \over {dx}}$ - 6x + ${{dy} \over {dx}}$ = 0

$ \Rightarrow $ ${{dy} \over {dx}}$ = ${{6x} \over {2y + 1}}$

Let P be (x₁, y₁)

Slope of tangent at P = ${{6{x_1}} \over {2{y_1} + 1}}$

$ \therefore $ Slope of normal at P = $ - {{2{y_1} + 1} \over {6{x_1}}}$

$ \Rightarrow $ Equation of normal (y – y₁) = $ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(x – x₁)

This normal passes through point $\left( {0,{3 \over 2}} \right)$.

$ \therefore $ (${{3 \over 2}}$ – y₁) = $ - \left( {{{2{y_1} + 1} \over {6{x_1}}}} \right)$(0 – x₁)

$ \Rightarrow $ y₁ = 1

Put y₁ = 1 in equation of curve , then we get x₁ = $ \pm $2

$ \Rightarrow $ |m| = slope of tangent = $\left| {{{6{x_1}} \over {2{y_1} + 1}}} \right|$ = ${{12} \over 3}$ = 4

We have,

${y_n} = {1 \over n}(n + 1)(n + 2)...{(n + n)^{1/n}}$ and $\mathop {\lim }\limits_{n \to \infty } {y_n} = L$

$ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}{[(n + 1)(n + 2)(n + 3)...(n + n)]^{1/n}}$

$ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {\left[ {\left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)\left( {1 + {3 \over n}} \right)...\left( {1 + {n \over n}} \right)} \right]^{{1 \over n}}}$

$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\log \left( {1 + {1 \over n}} \right) + \log \left( {1 + {2 \over n}} \right)...\log \left( {1 + {n \over n}} \right)} \right]$

$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {\log \left( {1 + {r \over n}} \right)} $

$ \Rightarrow \log L = \int_0^1 {_{II}^1 \times \mathop {\log }\limits_I } (1 + x)\,dx$

$ \Rightarrow \log L = (x.\log (1 + x)_0^1 - \int_0^1 {\left[ {{d \over {dx}}(\log (1 + x)\int {dx} } \right]} dx$

[by using integration by parts]

$ \Rightarrow \log L = [x\log (1 + x)]_0^1 - \int_0^1 {{x \over {1 + x}}} dx$

$ \Rightarrow \log L = \log 2 - \int_0^1 {\left( {{{x + 1} \over {x + 1}} - {1 \over {x + 1}}} \right)} dx$

$ \Rightarrow \log L = \log 2 - [x]_0^1 + [\log (x + 1)]_0^1$

$ \Rightarrow \log L = \log 2 - 1 + \log 2 - 0$

$ \Rightarrow \log L = \log 4 - \log e = \log {4 \over e}$

$ \Rightarrow L = {4 \over e}$

$ \Rightarrow [L] = \left[ {{4 \over e}} \right] = 1$

Given: The inner volume of cylinder $=\mathrm{Vmm}^3$


Thickness of wall $=2 \mathrm{~mm}$

Thickness of bottom circular disc $=2 \mathrm{~mm}$

Let the inner radius of cylinder $=r \mathrm{~mm}$ and height of the inner cylinder $=h \mathrm{~mm}$.

$ \Rightarrow \mathrm{V}=\pi r^2 h $

Now, volume of the material used $=$ volume of outer cylinder - volume of the inner cylinder + Volume of the circular disc

$ \begin{gathered} \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi(r+2)^2 h-\pi r^2 h+\pi(r+2)^2 2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h\left\{(r+2)^2-r^2\right\}+2 \pi(r+2)^2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h(4 r+4)+2 \pi(r+2)^2 \\\\ \left\{\because(a+b)^2=a^2+b^2+2 a b\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{2 h(r+1)+(r+2)^2\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi r^2}(r+1)+(r+2)^2\right\} \\\\ \left\{\because \mathrm{V}=\pi r^2 h\right\} \end{gathered} $

For $\mathrm{V}_{\mathrm{m}}$ to be minimum, $\frac{d v_m}{d r}=0$

Differentiating the above equation w.r.t.r,

$ \begin{aligned} \quad & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \frac{d}{d r}\left\{\frac{r+1}{r^2}\right\}+\frac{d}{d r}(r+2)^2\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2(1)-(r+1)(2 r)}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2-2 r^2-2 r}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi\left\{\frac{\mathrm{V}}{\pi r^3}(-r-2)+(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi(r+2)\left(-\frac{\mathrm{V}}{\pi r^3}+1\right) \\\\ \because & \frac{d \mathrm{~V}_m}{d r}=0 \\\\ \Rightarrow & 4 \pi(r+2)\left(\frac{-\mathrm{V}}{\pi r^3}+1\right)=0 \end{aligned} $

Also, given that $\mathrm{V}_{\mathrm{m}}$ is minimum at $r=10 \mathrm{~mm}$

$ \begin{array}{ll} \Rightarrow & 4 \pi(10+2)\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & 48 \pi\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & \frac{\mathrm{V}}{10^3 \pi}=1 \\\\ \Rightarrow & \frac{\mathrm{V}}{250 \pi}=4 \end{array} $

Given curve

$ \begin{aligned} & \left(y-x^5\right)^2=x\left(1+x^2\right)^2 \\\\ & \Rightarrow 2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+2 x\left(1+x^2\right) \cdot 2 x \end{aligned} $

Now putting $(1,3)$ in it, we get

$ 2(3-1)\left(\frac{d y}{d x}-5\right)=1\{2(2) 2\}+(1+1)^2 $

$\Rightarrow 4\left(\frac{d y}{d x}-5\right)=8+4 \Rightarrow \frac{d y}{d x}=8$

Thus, the slope at $(1,3)$ is 8 .

$\begin{aligned} & \text { Given, } f(x)=|x|+\left|x^2-1\right| \\ & \Rightarrow f(x)=|x|+|x-1||x+1| \end{aligned}$

$\begin{aligned} f(x) & =x^2+x-1, x \geq 1 \\ & =1-x^2+x, 0 \leq x < 1 \\ & =1-x^2-x,-1 < x < 0 \\ & =x^2-x-1, x \leq-1 \\ f^{\prime}(x) & =2 x+1 \quad(+v e) \\ & =1-2 x \quad x > \frac{1}{2}(+v e) \\ & =-2 x-1 \quad x > -\frac{1}{2}(-v e) ; x < \frac{-1}{2}(+v e) \\ & =2 x-1 \quad(-v e) \end{aligned}$

From the figure, total number of points at which $f$ attains either a local maximum or a local minimum is 5.

Since, $p(x)$ has a local maximum at $x=1$ and $a$ local minimum at $x=3$.

$\begin{aligned} \text { Let } & p^{\prime}(x)=k(x-1)(x-3) \\ \Rightarrow & p^{\prime}(x)=k\left(x^2-4 x+3\right) \\ \Rightarrow & p(x)=k\left(\frac{x^3}{3}-\frac{4 x^2}{2}+3 x\right)+c \\ \Rightarrow & p(x)=k\left(\frac{x^3}{3}-2 x^2+3 x\right)+c \end{aligned}$

Now, $p(1)=6$

$\Rightarrow k\left(\frac{1^3}{3}-2(1)^2+3(1)\right)+c=6$

$\begin{array}{rrr} \Rightarrow & k\left(\frac{1}{3}-2+3\right)+c & =6 \\ \Rightarrow & \frac{4}{3} k+c & =6 \end{array}$

Also, $p(3)=2$

$\begin{aligned} \Rightarrow \quad & k\left(\frac{3^3}{3}-2(3)^2+3(3)\right)+c =2 \\ \Rightarrow \quad & k(0)+c =2 \\ \Rightarrow \quad & c =2 \end{aligned}$

$\begin{aligned} \text { Now, } \quad & \frac{4}{3} k+2 =6 \\ \Rightarrow \quad & \frac{4}{3} k =4 \end{aligned}$

$ \begin{array}{rlrl} \Rightarrow \quad & k =3 \\ \therefore \quad & p^{\prime}(0) =3\left(0^2-4(0)+3\right) \\ \qquad =9 \end{array}$

The equation of the tangent at (x, y) to the given curve y = f(x) is

$Y - y = {{dy} \over {dx}}(X - x)$

Y-intercept $ = y - x{{dy} \over {dx}}$

According to the question

${x^3} = y - x{{dy} \over {dx}}$

$ \Rightarrow {{dy} \over {dx}} - {y \over x} = - {x^2}$

which is linear in x.

$IF = {e^{\int {{{ - 1} \over x}dx} }} = {1 \over x}$

$\therefore$ Required solution is

$y\,.\,{1 \over x} = \int { - {x^2}\,.\,{1 \over x}dx} $

$ \Rightarrow {y \over x} = {{ - {x^2}} \over 2} + c$

$ \Rightarrow y = {{ - {x^3}} \over 2} + cx$

At x = 1, y = 1,

$1 = {{ - 1} \over 2} + c$

$ \Rightarrow c = {3 \over 2}$

Now, $f( - 3) = {{27} \over 2} + {3 \over 2}( - 3)$

$ = {{27 - 9} \over 2} = 9$

Let $g(x) = {e^{f(x)}},\,\forall x \in R$

$ \Rightarrow g'(x) = {e^{f(x)}}\,.\,f'(x)$

$\Rightarrow$ f'(x) changes its sign from positive to negative in the neighbourhood of x = 2009

$\Rightarrow$ f(x) has local maxima at x = 2009

So, the number of local maximum is one.

We have,

$f'(x) = 6(x - 2)(x - 3)$

Hence, $f(x)$ is increasing in $(3,\infty )$.

Also, $A = \{ 4 \le x \le 5\} $

Therefore, ${f_{\max }} = f(5) = 7$

Let us consider

$P(x) = a{x^4} + b{x^3} + c{x^2} + dx + e$

$P'(1) = P'(2) = 0$

$\mathop {\lim }\limits_{x \to 0} \left( {{{{x^2} + P(x)} \over {{x^2}}}} \right) = 2$

$ \Rightarrow P(0) = 0 \Rightarrow e = 0$

$\mathop {\lim }\limits_{x \to 0} \left( {{{2x + P'(x)} \over {2x}}} \right) = 2$

$ \Rightarrow P'(0) = 0 \Rightarrow d = 0$

$\mathop {\lim }\limits_{x \to 0} \left( {{{2 + P''(x)} \over 2}} \right) = 2$

$ \Rightarrow c = 1$

On solving, we get $a = 1/4,b = - 1$. Thus,

$P(x) = {{{x^4}} \over 4} - {x^3} + {x^2} \Rightarrow P(2) = 0$

$ \begin{aligned} & \quad g(x)=\left(f^{\prime}(x)\right)^2+f^{\prime \prime}(x) f(x) \\ & \Rightarrow \quad \frac{d}{d x}\left[f^{\prime}(x) \cdot f(x)\right]=\left[f^{\prime}(x)\right]^2+f^{\prime \prime}(x) f(x) \\ & \Rightarrow \quad g(x)=\frac{d}{d x}\left[f^{\prime}(x) \cdot f(x)\right] \\ & \text { Let, } \quad h(x)=f^{\prime}(x) f(x) \end{aligned} $

Now, estimate graph of $f(x)$

From given data.

$ \begin{aligned} & f(a)=0 \\ & f(b)=2 \\ & f(c)=-1 \\ & f(d)=2 \\ & f(e)=0 \end{aligned} $

Also a < b < c < d < e

$\Rightarrow f(x)$ has minimum four zeros (as the graph intersecting $x$-axis 4 times)

There must be there critical points

Where $f^{\prime}(x)=0$ (from graph)

Therefore, $h(x)$ must have 7 minimum solutions.

Now, $g(x)=h^{\prime}(x)$ must have 6 minimum zeroes.

$\begin{aligned} & g(x)=\left(f^{\prime}(x)\right)^{2}+f^{\prime \prime}(x) f(x) \\ & \Rightarrow \frac{d}{d x}\left[f^{\prime}(x) \cdot f(x)\right]=\left[f^{\prime}(x)\right]^{2}+f^{\prime \prime}(x) f(x) \\ & \Rightarrow \quad g(x)=\frac{d}{d x}\left[f^{\prime}(x) \cdot f(x)\right] \\ & \text { Let, } \quad h(x)=f^{\prime}(x) f(x) \\ & \text { Now, estimate graph of } f(x) \\ & \text { From given data. } \\ & f(a)=0 \\ & f(b)=2 \\ & f(c)=-1 \\ & f(d)=2 \\ & f(e)=0 \end{aligned}$

Also $a < b < c < d < e$

$\Rightarrow f(x)$ has minimum four zeros (as the graph intersecting $x$-axis 4 times)

There must be there critical points

Where $f'(x)=0$ (from graph)

Therefore, $h(x)$ must have 7 minimum solutions.

Now, $g(x)=h'(x)$ must have 6 minimum zeroes.