Application of Derivatives - Test Papers

 CBSE Test Paper 01

Chapter 6 Application of Derivatives

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1. The instantaneous rate of change at t = 1 for the function f (t) =te-t + 9 is 

   1. 2
   2. 9
   3. -1
   4. -0

2. The function f (x) = x2, for all real x, is 

   1. Neither decreasing nor increasing
   2. Increasing
   3. Decreasing
   4. None of these

3. The slope of the tangent to the curve x = a sint, y = a {cost+log(tant2)}$\left\{\cos t+\log (\tan \frac{t}{2})\right\}$ at the point ‘t’ is 

   1. tant2$\tan \frac{t}{2}$
   2. none of these
   3. tan t
   4. cot t

4. The function f (x) = x2 - 2x is strict decreasing in the interval 

   1. none of these
   2. R
   3. [1,∞)$[1,\mathrm{\infty })$
   4. ( −∞, 1)$(-\mathrm{\infty },1)$

5. The equation of the tangent to the curve y2 = 4ax at the point (at2, 2at) is 

   1. ty = x + at2
   2. none of these
   3. tx + y =at3
   4. ty = x - at2
6. The maximum value of (1x)x${\left(\frac{1}{x}\right)}^x$ is ________.
7. The minimum value of f if f(x) = sin x in [−π2,π2$\frac{-\pi }{2},\frac{\pi }{2}$] is ________.
8. The equation of normal to the curve y = tan x at (0, 0) is ________.

9. Find the approximate value of f(3.02) where f(x) = 3x2 + 5x + 3. 

10. If the line ax+by+c=0 is a normal to the curve xy=1,then show that either a>0,b<0 or a<0,b>0 

11. Find the interval in which the function f(x) = x2e-x is increasing. 

12. The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm. 

13. Find the approximate value of (1.999)5${\left(1.999\right)}^5$. 

14. Show that the function f(x) = 4x3 - 18x2 + 27x - 7 is always increasing on R. 

15. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius a is 2a3√$\frac{2a}{\sqrt{3}}$. 

16. A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x – coordinate. 

17. Find the equation of tangent to the curve y=x−7x2−5x+6$y=\frac{x-7}{x^2-5x+6}$ at the point, where it cuts the X-axis. 

18. Show that semi – vertical angle of right circular cone of given surface area and maximum volume is sin−1(13)${\sin }^{-1}\left(\frac{1}{3}\right)$. 

CBSE Test Paper 01
Chapter 6 Application of Derivatives

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Solution

1. (d) 0, Explanation: f′(t)=te−t(−1)+e−t⇒f′(1)=−e−1+e−1=0$f^{\prime }(t)=t{e}^{-t}(-1)+e^{-t}\Rightarrow f^{\prime }(1)=-e^{-1}+e^{-1}=0$
2. (a) Neither decreasing nor increasing, Explanation: f(x) = x2
   ⇒$\Rightarrow$ f'(x) = 2x for all x in R.
   Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x< 0 for x < 0, therefore on R, f is neither increasing nor decreasing. Infact , f is strict increasing on [ 0,∞$\mathrm{\infty }$ ) and strict decreasing on (- ∞,0$\mathrm{\infty },0$].
3. (d) cot t, Explanation: Given, x=asint,y=a{cost+log(tant2)}$x=asint,y=a\left\{\cos t+\log (\tan \frac{t}{2})\right\}$
   dxdt=acost,dydt$dxdt=a\cos t,\frac{dy}{dt}$=a[−sint+1tant2.sec2t2.12]$=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}}.{sec}^2\frac{t}{2}.\frac{1}{2}\right]$=a[−sint+12sint2.cost2]$=a\left[-\sin t+\frac{1}{2sin\frac{t}{2}.cos\frac{t}{2}}\right]$=a[−sint+1sint]=acos2tsint$=a\left[-\sin t+\frac{1}{sint}\right]=a\frac{{cos}^{2}t}{sint}$
   Slope of the tangent=dydx=dydtdxdt=acos2tsintacost=cott$=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a\frac{{cos}^{2}t}{sint}}{a\cos t}=\cot t$
4. (d) ( −∞, 1)$(-\mathrm{\infty },1)$, Explanation: f ‘ (x ) = 2x – 2 = 2 ( x - 1) <0 if x < 1 i.e. x x∈(−∞,1)$x\in \left(-\mathrm{\infty },1\right)$. Hence f is strict decreasing in(−∞,1)$\left(-\mathrm{\infty },1\right)$
5. (a) ty = x +at2, Explanation: y2=4ax$y^2=4ax$
   ⇒2ydydx=4a$\Rightarrow 2y\frac{dy}{dx}=4a$
   ⇒dydx=2ay$\Rightarrow \frac{dy}{dx}=\frac{2a}{y}$
   ⇒dydx$\Rightarrow \frac{dy}{dx}$ at (at2,2at)$(a{t}^2,2at)$ is 2a2at=1t$\frac{2a}{2at}=\frac{1}{t}$
   ⇒$\Rightarrow$ Slope of tangent =m=1t$=m=\frac{1}{t}$
   Hence, equation of tangent is y−y1=m(x−x1)$y-y_1=m\left(x-x_1\right)$
   ⇒y−2at=1t(x−at2)$\Rightarrow y-2at=\frac{1}{t}(x-a{t}^2)$
   ⇒yt−2at2=x−at2$\Rightarrow yt-2a{t}^2=x-a{t}^2$
   ⇒yt=x+at2$\Rightarrow yt=x+a{t}^2$
6. e1e$e^{\frac{1}{e}}$
7. -1
8. x + y = 0
9. x=3,Δx=0.02$x=3,\mathrm{\Delta }x=0.02$
   f(x+Δx)=f(x)+f′(x)Δx$f(x+\mathrm{\Delta }x)=f(x)+f^{\prime }(x)\mathrm{\Delta }x$
   f(x+Δx)=(3x2+5x+3)+(6x+5)×0.02$f(x+\mathrm{\Delta }x)=(3x^2+5x+3)+(6x+5)\times 0.02$
   Put x=3,Δx=0.02$x=3,\mathrm{\Delta }x=0.02$
   f(3.02)={3(9)+5(3)+3}+{6(3)+5}×0.02 =45+0.46
   f(3.02) = 45.46
10. we have, xy =1
    ⇒y=1x$\Rightarrow y=\frac{1}{x}$
    ∴ dydx=−1x2$\therefore \frac{dy}{dx}=-\frac{1}{x^2}$
    The slope of the normal = x2
    If ax+by+c=0 is normal to the curve xy=1,then
    x2=−ab [∵slope of normal =−coeff. of xcoeff.of y]$x^2=-\frac{a}{b}[\because slopeofnormal=-\frac{coeff.ofx}{coeff.ofy}]$
    ∴−ab>0$\therefore -\frac{a}{b}>0$
    ⇒ a>0,b<0 or a<0,b>0$\Rightarrow a>0,b<0ora<0,b>0$
11. f(x) = x2e-x
    Differentiating w.r.t x, we get,
    f'(x) = −x2e−x+2xe−x=xe−x(2−x)$-x^2{e}^{-x}+2x{e}^{-x}=x{e}^{-x}(2-x)$
    For increasing function, f'(x)≥0$\ge 0$
    xe−x(2−x)≥0$x{e}^{-x}(2-x)\ge 0$
    x(2−x)≥0$x(2-x)\ge 0$ [∵ e−x$\because e^{-x}$ is always positive]
    x(x−2)≤0$x(x-2)\le 0$ [ since - ( x - 2) will change the inequality )
    Here x < 0 & (x - 2) > 0 ⇒$\Rightarrow$ x < 0 & x > 2 ⇒$\Rightarrow$ 0 < x < 2
    But when x > 0 & (x - 2) < 0 ⇒$\Rightarrow$ x > 0 & x < 2
    0≤x≤ 2$0\le x\le 2$
12. Let r be the radius of sphere and V be its volume.
    Then V = 43πr3$\frac{4}{3}\pi r^3$........(i)
    Given, dVdt$\frac{dV}{dt}$ = 3 cm3/s
    Differentiating (i) both sides w.r.t x,we get,
    dVdt=43π(3r2)drdt$\frac{dV}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}$
    ⇒3=43(3πr2)drdt$\Rightarrow 3=\frac{4}{3}(3\pi r^2)\frac{dr}{dt}$
    ⇒drdt=34πr2$\Rightarrow \frac{dr}{dt}=\frac{3}{4\pi r^2}$.......(ii)
    Now, let S be the surface area of sphere, then S = 4πr2$4\pi r^2$
    ⇒dSdt=4π(2r)drdt$\Rightarrow \frac{dS}{dt}=4\pi (2r)\frac{dr}{dt}$
    ⇒dSdt=8πr(34πr2)$\Rightarrow \frac{dS}{dt}=8\pi r\left(\frac{3}{4\pi r^2}\right)$[using Eq.(ii)]
    ⇒(dSdt)=6r$\Rightarrow \left(\frac{dS}{dt}\right)=\frac{6}{r}$
    when r = 2, then dSdt=62$\frac{dS}{dt}=\frac{6}{2}$ = 3 cm2/s
    Therefore,the rate of inrcrease of the surface area of sphere is 3 cm2${}^2$/s when it's radius is 2 cm
13. Let x = 2
    and Δx=−0.001[∵2−0.001=1.999]$\mathrm{\Delta }x=-0.001\left[\because 2-0.001=1.999\right]$
    let y = x5
    On differentiating both sides w.r.t. x, we get
    dydx=5x4$\frac{dy}{dx}=5x^4$
    Now, Δy=dydx.Δx=5x4×Δx$\mathrm{\Delta }y=\frac{dy}{dx}.\mathrm{\Delta }x=5x^4\times \mathrm{\Delta }x$
    =5×24×[−0.001]$=5\times 2^4\times \left[-0.001\right]$
    =−80×0.001=−0.080$=-80\times 0.001=-0.080$
    ∴(1.999)5=y+Δy$\therefore {\left(1.999\right)}^5=y+\mathrm{\Delta }y$
    =25+(−0.080)$=2^5+\left(-0.080\right)$
    = 32 - 0.080 = 31.920
14. Here, f(x) =4x3 - 18x2 + 27x - 7
    On differentiating both sides w.r.t. x, we get
    f'(x) = 12x2 - 36x + 27
    ⇒$\Rightarrow$f'(x) = 3(4x2 -12 + 9)
    ⇒$\Rightarrow$ f'(x) = 3(x - 3)2
    ⇒$\Rightarrow$ f'(x) ≥$\ge$ 0
    Since, a perfect square number cannot be negative]
    ∴$\therefore$ Given function f(x) is an increasing function on R.
15. 
    v=πr2.2x[∵OL=xLM=2x]$v=\pi r^2.2x\left[\begin{array}{c}\because OL=x\\ LM=2x\end{array}\right]$
    =π.(a2−x2).2x$=\pi .\left(a^2-x^2\right).2x$
    V=2π(a2x−x3)$V=2\pi \left(a^{2}x-x^3\right)$
    dvdx=2π(a2−3x2)$\frac{dv}{dx}=2\pi \left(a^2-3x^2\right)$
    d2vdx2=2π[0−6x]$\frac{d^{2}v}{d{x}^2}=2\pi \left[0-6x\right]$
    =−12πx$=-12\pi x$
    For maximum/minimum
    dvdx=0$\frac{dv}{dx}=0$
    2π[a2−3x2]=0$2\pi \left[a^2-3x^2\right]=0$
    a2=3x2⇒a23−−√=x$a^2=3x^2\Rightarrow \sqrt{\frac{a^2}{3}}=x$
    ⇒x=a3√$\Rightarrow x=\frac{a}{\sqrt{3}}$
    d2vdx2]x=a3√=−12π.a3√${\frac{d^{2}v}{d{x}^2}]}_{x=\frac{a}{\sqrt{3}}}=-12\pi .\frac{a}{\sqrt{3}}$
    = - tive maximum
    Volume is maximum at x=a3√$x=\frac{a}{\sqrt{3}}$
    Height of cylinder of maximum volume is
    = 2x
    =2×a3√$=2\times \frac{a}{\sqrt{3}}$
    =2a3√$=\frac{2a}{\sqrt{3}}$
16. Given curve is 6y = x3 + 2 ...(i)
    so, 6dydt=3x2dxdt$6\frac{dy}{dt}=3x^2\frac{dx}{dt}$
    ⇒6×8dxdt=3x2dxdt[∵dydt=8dxdt]$\Rightarrow 6\times 8\frac{dx}{dt}=3x^2\frac{dx}{dt}\left[\because \frac{dy}{dt}=8\frac{dx}{dt}\right]$
    ⇒16=x2$\Rightarrow 16=x^2$
    ⇒x=±4$\Rightarrow x=\pm 4$
    Put the value of x in equation (1)
    When x = 4
    6y = ( 4 )3 + 2
    ⇒$\Rightarrow$ 6y = 64 + 2
    ⇒$\Rightarrow$ 6y = 66
    ∴$\therefore$ y=666=11$y=\frac{66}{6}=11$
    So, point is (4, 11)
    Now, When x = - 4
    6y = ( - 4}3 + 2
    = - 64 + 2
    ∴$\therefore$ y=−626=−313$y=\frac{-62}{6}=\frac{-31}{3}$
    So the point is (−4,−313)$\left(-4,\frac{-31}{3}\right)$
17. Given equation of curve is
    y=x−7x2−5x+6$y=\frac{x-7}{x^2-5x+6}$.......(i)
    On differentiating both sides w.r.t. x, we get
    dydx=(x2−5x+6)⋅1−(x−7)(2x−5)(x2−5x+6)2$\frac{dy}{dx}=\frac{\left(x^2-5x+6\right)\cdot 1-(x-7)(2x-5)}{{\left(x^2-5x+6\right)}^2}$[∵ddx(uv)=vdudx−udvdxv2]$\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\right]$
    ⇒dydx=[(x2−5x+6)−y(x2−5x+6)(2x+5)](x2−5x+6)2$\Rightarrow \frac{dy}{dx}=\frac{\left[\left(x^2-5x+6\right)-y\left(x^2-5x+6\right)(2x+5)\right]}{{\left(x^2-5x+6\right)}^2}$
    ⇒dydx=1−(2x−5)yx2−5x+6$\Rightarrow \frac{dy}{dx}=\frac{1-(2x-5)y}{x^2-5x+6}$[dividing numerator and denominator by x2 - 5x + 6]
    Also, given that curve cuts X-axis, so its y-coordinate is zero.
    Put y = 0 in Eq. (i), we get
    x−7x2−5x+6=0$\frac{x-7}{x^2-5x+6}=0$
    ⇒$\Rightarrow$ x= 7
    So, curve passes through the point (7, 0).
    Now, slope of tangent at (7,0) is
    m=(dydx)(2,0)=1−049−35+6=120$m={\left(\frac{dy}{dx}\right)}_{(2,0)}=\frac{1-0}{49-35+6}=\frac{1}{20}$
    Hence, the required equation of tangent passing through the point (7, 0) having slope 1/20 is
    y - 0 = 120$\frac{1}{20}$(x - 7)
    ⇒$\Rightarrow$ 20y = x - 7
    ∴$\therefore$ x - 20y = 7
18. 
    s=πr2+πrl$s=\pi r^2+\pi rl$ (given)
    ⇒$\Rightarrow$l=s−πr2πr$l=\frac{s-\pi r^2}{\pi r}$
    Let v be the volume
    v=13πr2h$v=\frac{1}{3}\pi r^{2}h$
    ⇒$\Rightarrow$v2=19π2r4h2[h2=l2−r2]$v^2=\frac{1}{9}{\pi }^2{r}^4{h}^2\left[h^2=l^2-r^2\right]$
    ⇒$\Rightarrow$v2=19π2r4(l2−r2)$v^2=\frac{1}{9}{\pi }^2{r}^4\left(l^2-r^2\right)$
    ⇒$\Rightarrow$v2=19π2r4[(s−πr2πr)2−r2]$v^2=\frac{1}{9}{\pi }^2{r}^4\left[{\left(\frac{s-\pi r^2}{\pi r}\right)}^2-r^2\right]$
    =19π2r4[(s−πr2)2π2r2−r21]$=\frac{1}{9}{\pi }^2{r}^4\left[\frac{{\left(s-\pi r^2\right)}^2}{{\pi }^2{r}^2}-\frac{r^2}{1}\right]$
    =19r2[(s−πr2)2−π2r4]$=\frac{1}{9}{r}^2\left[{\left(s-\pi r^2\right)}^2-{\pi }^2{r}^4\right]$
    =19r2[s2+π2r4−2sπr2−π2r4]$=\frac{1}{9}{r}^2\left[s^2+{\pi }^2{r}^4-2s\pi r^2-{\pi }^2{r}^4\right]$
    =19r2[s2−2sπr2]$=\frac{1}{9}{r}^2\left[s^2-2s\pi r^2\right]$
    z=19[s2r2−2sπr4]$z=\frac{1}{9}\left[s^2{r}^2-2s\pi r^4\right]$
    [∵v2=z]$\left[\because v^2=z\right]$
    Now dzdr=19[2rs2−8sπr3]$\frac{dz}{dr}=\frac{1}{9}\left[2r{s}^2-8s\pi r^3\right]$
    0=19[2rs2−8sπr3]$0=\frac{1}{9}\left[2r{s}^2-8s\pi r^3\right]$
    8sπr2=2rs2$8s\pi r^2=2r{s}^2$
    ⟹$⟹$ 4πr2=s$4\pi r^2=s$
    Now d2zdx2=19[2s2−24sπr2]$\frac{d^{2}z}{d{x}^2}=\frac{1}{9}\left[2s^2-24s\pi r^2\right]$
    d2zdx2]r2=s4π=19[252−24π.54π]${\frac{d^{2}z}{d{x}^2}]}_{r^2=\frac{s}{4\pi }}=\frac{1}{9}\left[{25}^2-24\pi .\frac{5}{4\pi }\right]$
    = + ve
    Hence minimum
    Now s=4πr2$s=4\pi r^2$
    We have s=πrl+πr2$s=\pi rl+\pi r^2$
    4πr2=πrl+πr2$4\pi r^2=\pi rl+\pi r^2$
    ⇒$\Rightarrow$ 3πr2=πrl$3\pi r^2=\pi rl$
    ⇒$\Rightarrow$ 3 r = l
    ⇒$\Rightarrow$ rl=13$\frac{r}{l}=\frac{1}{3}$
    ⇒$\Rightarrow$ sinα=13$\sin \alpha =\frac{1}{3}$
    ∴$\therefore$ α=sin−1(13)