Class 12 Maths Application of Derivatives Extra Questions

Chapter 6 Application of Derivatives

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 {tex}\left\{ {\cos t + \log (\tan \frac{t}{2})} \right\}{/tex} at the point ‘t’ is

   1. {tex}\tan \frac{t}{2}{/tex}
   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. {tex} [1,\infty ) {/tex}
   4. {tex} ({\text{ }}-\infty ,{\text{ }}1){/tex}

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 {tex}{\left( {\frac{1}{x}} \right)^x}{/tex} is ________.
7. The minimum value of f if f(x) = sin x in [{tex}\frac {-\pi}2,\frac {\pi}2{/tex}] 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 {tex}{\left( {1.999} \right)^5}{/tex}.

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 {tex}\frac{{2a}}{{\sqrt 3 }}{/tex}.

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 {tex} y = \frac { x – 7 } { x ^ { 2 } – 5 x + 6 }{/tex} 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 {tex}{\sin ^{ – 1}}\left( {\frac{1}{3}} \right){/tex}.

Chapter 6 Application of Derivatives

Solution

1. (d) 0, Explanation: {tex}f'(t) = t{e^{ – t}}( – 1) + {e^{ – t}} \Rightarrow f'(1) = – {e^{ – 1}} + {e^{ – 1}} = 0 {/tex}
2. (a) Neither decreasing nor increasing, Explanation: f(x) = x²
   {tex}\Rightarrow {/tex} 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,{tex}\infty {/tex} ) and strict decreasing on (- {tex}\infty ,0\;{/tex}].
3. (d) cot t, Explanation: Given, {tex} x=asint,y=a\left\{ { \cos t+\log (\tan \frac { t }{ 2 } ) } \right\}{/tex}
   {tex} { { dx } }{ { dt } } =a\cos t,\frac { { dy } }{ { dt } } {/tex}{tex}=a\left[ -\sin t+\frac { 1 }{ \tan \frac { t }{ 2 } } .{ sec }^{ 2 }\frac { t }{ 2 } .\frac { 1 }{ 2 } \right] {/tex}{tex}=a\left[ -\sin t+\frac { 1 }{ 2sin\frac { t }{ 2 } .cos\frac { t }{ 2 } } \right] {/tex}{tex}=a\left[ -\sin t+\frac { 1 }{ sint } \right] =a\frac { { cos }^{ 2 }t }{ sint } {/tex}
   Slope of the tangent{tex}=\frac { { dy } }{ { dx } } =\frac { { \frac { { dy } }{ { dt } } } }{ { \frac { { dx } }{ { dt } } } } =\frac { { a\frac { { cos }^{ 2 }t }{ sint } } }{ { a\cos t } } =\cot t{/tex}
4. (d) {tex} ({\text{ }}-\infty ,{\text{ }}1){/tex}, Explanation: f ‘ (x ) = 2x – 2 = 2 ( x – 1) <0 if x < 1 i.e. x {tex}x \in \left( { – \infty ,1} \right) {/tex}. Hence f is strict decreasing in{tex}\left( { – \infty ,1} \right){/tex}
5. (a) ty = x +at², Explanation: {tex}{ y }^{ 2 }=4ax{/tex}
   {tex}\\ \Rightarrow 2y\frac { dy }{ dx } =4a{/tex}
   {tex}\\ \Rightarrow \frac { { dy } }{ { dx } } =\frac { 2a }{ y }{/tex}
   {tex} \Rightarrow \frac { { dy } }{ { dx } } {/tex} at {tex}(a{ t^{ 2 } },2at){/tex} is {tex}\frac { { 2a } }{ { 2at } } =\frac { 1 }{ t } {/tex}
   {tex}\Rightarrow {/tex} Slope of tangent {tex}=m=\frac { 1 }{ t } {/tex}
   Hence, equation of tangent is {tex} y-{ y }_{ 1 }=m\left( x-{ x }_{ 1 } \right) {/tex}
   {tex}\\ \Rightarrow y-2at=\frac { 1 }{ t } (x-a{ t^{ 2 } }){/tex}
   {tex}\Rightarrow yt-2a{ t^{ 2 } }=x-a{ t^{ 2 } }{/tex}
   {tex}\Rightarrow yt=x+a{ t^{ 2 } }{/tex}
6. {tex}{e^{\frac{1}{e}}}{/tex}
7. -1
8. x + y = 0
9. {tex}x = 3,\Delta x = 0.02{/tex}
   {tex}f(x + \Delta x) = f(x) + f'(x)\Delta x{/tex}
   {tex}f(x + \Delta x) = (3{x^2} + 5x + 3) + (6x + 5) \times 0.02{/tex}
   Put {tex}x = 3,\Delta x = 0.02{/tex}
   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
    {tex} \Rightarrow y = \frac{1}{x}{/tex}
    {tex}\therefore\ \frac{dy}{dx}=-\frac{1}{x^2}{/tex}
    The slope of the normal = x²
    If ax+by+c=0 is normal to the curve xy=1,then
    {tex}x^2=-\frac{a}{b} \ [\because slope\ of\ normal\ =-\frac{coeff.\ of\ x}{coeff. of\ y}]{/tex}
    {tex}\therefore \; – \frac{a}{b} > 0{/tex}
    {tex}\Rightarrow\ a>0,b<0 \ or\ a<0,b>0{/tex}
11. f(x) = x²e^-x
    Differentiating w.r.t x, we get,
    f'(x) = {tex}-x^2e^{-x}+2xe^{-x}=xe^{-x}(2-x){/tex}
    For increasing function, f'(x){tex}\geq0{/tex}
    {tex}xe^{-x}(2-x)\geq0{/tex}
    {tex}x(2-x)\geq0{/tex} [{tex}\because\ e^{-x}{/tex} is always positive] {tex}x(x-2)\leq0{/tex} [ since – ( x – 2) will change the inequality )
    Here x < 0 & (x – 2) > 0 {tex}\Rightarrow{/tex} x < 0 & x > 2 {tex}\Rightarrow{/tex} 0 < x < 2
    But when x > 0 & (x – 2) < 0 {tex}\Rightarrow{/tex} x > 0 & x < 2
    {tex}0\le x \leq\ 2{/tex}
12. Let r be the radius of sphere and V be its volume.
    Then V = {tex}\frac { 4 } { 3 } \pi r ^ { 3 }{/tex}……..(i)
    Given, {tex}\frac { d V } { d t }{/tex} = 3 cm³/s
    Differentiating (i) both sides w.r.t x,we get,
    {tex}\frac { d V } { d t } = \frac { 4 } { 3 } \pi \left( 3 r ^ { 2 } \right) \frac { d r } { d t }{/tex}
    {tex}\Rightarrow \quad 3 = \frac { 4 } { 3 } ( 3 \pi r ^2 ) \frac { d r } { d t }{/tex}
    {tex}\Rightarrow \quad \frac { d r } { d t } = \frac { 3 } { 4 \pi r ^ { 2 } }{/tex}…….(ii)
    Now, let S be the surface area of sphere, then S = {tex}4 \pi r ^ { 2 }{/tex}
    {tex}\Rightarrow \quad \frac { d S } { d t } = 4 \pi ( 2 r ) \frac { d r } { d t }{/tex}
    {tex}\Rightarrow \quad \frac { d S } { d t } = 8 \pi r \left( \frac { 3 } { 4 \pi r ^ { 2 } } \right){/tex}[using Eq.(ii)] {tex}\Rightarrow \quad \left( \frac { d S } { d t } \right) = \frac { 6 } { r }{/tex}
    when r = 2, then {tex}\frac { d S } { d t } = \frac { 6 } { 2 }{/tex} = 3 cm²/s
    Therefore,the rate of inrcrease of the surface area of sphere is 3 cm{tex}^2{/tex}/s when it’s radius is 2 cm
13. Let x = 2
    and {tex}\Delta x = – 0.001\,\left[ {\because 2 – 0.001 = 1.999} \right]{/tex}
    let y = x⁵
    On differentiating both sides w.r.t. x, we get
    {tex}\frac{{dy}}{{dx}} = 5{x^4}{/tex}
    Now, {tex}\Delta y = \frac{{dy}}{{dx}}.\Delta x = 5{x^4} \times \Delta x{/tex}
    {tex} = 5 \times {2^4} \times \left[ { – 0.001} \right]{/tex}
    {tex} = – 80 \times 0.001 = – 0.080{/tex}
    {tex}\therefore {\left( {1.999} \right)^5} = y + \Delta y{/tex}
    {tex}= {2^5} + \left( { – 0.080} \right){/tex}
    = 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
    {tex}\Rightarrow{/tex}f'(x) = 3(4x² -12 + 9)
    {tex}\Rightarrow{/tex} f'(x) = 3(x – 3)²
    {tex}\Rightarrow{/tex} f'(x) {tex}\geq{/tex} 0
    Since, a perfect square number cannot be negative] {tex}\therefore{/tex} Given function f(x) is an increasing function on R.
15. 
    {tex}v = \pi {r^2}.2x\,\,\left[ \begin{gathered} \because OL = x \hfill \\ LM = 2x \hfill \\ \end{gathered} \right]{/tex}
    {tex} = \pi .\left( {{a^2} – {x^2}} \right).2x{/tex}
    {tex}V = 2\pi \left( {{a^2}x – {x^3}} \right){/tex}
    {tex}\frac{{dv}}{{dx}} = 2\pi \left( {{a^2} – 3{x^2}} \right){/tex}
    {tex}\frac{{{d^2}v}}{{d{x^2}}} = 2\pi \left[ {0 – 6x} \right]{/tex}
    {tex} = – 12\pi x{/tex}
    For maximum/minimum
    {tex}\frac{{dv}}{{dx}} = 0{/tex}
    {tex}2\pi \left[ {{a^2} – 3{x^2}} \right] = 0{/tex}
    {tex}{a^2} = 3{x^2} \Rightarrow \sqrt {\frac{{{a^2}}}{3}} = x{/tex}
    {tex} \Rightarrow x = \frac{a}{{\sqrt 3 }}{/tex}
    {tex}{\left. {\frac{{{d^2}v}}{{d{x^2}}}} \right]_{x = \frac{a}{{\sqrt 3 }}}} = – 12\pi .\frac{a}{{\sqrt 3 }}{/tex}
    = – tive maximum
    Volume is maximum at {tex}x = \frac{a}{{\sqrt 3 }}{/tex}
    Height of cylinder of maximum volume is
    = 2x
    {tex} = 2 \times \frac{a}{{\sqrt 3 }}{/tex}
    {tex} = \frac{{2a}}{{\sqrt 3 }}{/tex}
16. Given curve is 6y = x³ + 2 …(i)
    so, {tex}6\frac{{dy}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}{/tex}
    {tex}\Rightarrow 6 \times 8\frac{{dx}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}\,\,\left[ {\because \frac{{dy}}{{dt}} = 8\frac{{dx}}{{dt}}} \right]{/tex}
    {tex}\Rightarrow 16 = {x^2}{/tex}
    {tex}\Rightarrow x = \pm 4{/tex}
    Put the value of x in equation (1)
    When x = 4
    6y = ( 4 )³ + 2
    {tex}\Rightarrow{/tex} 6y = 64 + 2
    {tex}\Rightarrow{/tex} 6y = 66
    {tex}\therefore{/tex} {tex}y = \frac{{66}}{6} = 11{/tex}
    So, point is (4, 11)
    Now, When x = – 4
    6y = ( – 4}³ + 2
    = – 64 + 2
    {tex}\therefore{/tex} {tex}y = \frac{{ – 62}}{6}= \frac{-31}{3}{/tex}
    So the point is {tex}\left( { – 4,\frac{{ – 31}}{3}} \right){/tex}
17. Given equation of curve is
    {tex} y = \frac { x – 7 } { x ^ { 2 } – 5 x + 6 }{/tex}…….(i)
    On differentiating both sides w.r.t. x, we get
    {tex} \frac { d y } { d x } = \frac { \left( x ^ { 2 } – 5 x + 6 \right) \cdot 1 – ( x – 7 ) ( 2 x – 5 ) } { \left( x ^ { 2 } – 5 x + 6 \right) ^ { 2 } }{/tex}{tex}\left[ \because \frac { d } { d x } \left( \frac { u } { v } \right) = \frac { v \frac { d u } { d x } – u \frac { d v } { d x } } { v ^ { 2 } } \right]{/tex}
    {tex}\Rightarrow \frac { d y } { d x } = \frac { \left[ \left( x ^ { 2 } – 5 x + 6 \right) – y \left( x ^ { 2 } – 5 x + 6 \right) (2x +5)\right] } { \left( x ^ { 2 } – 5 x + 6 \right) ^ { 2 } }{/tex}
    {tex}\Rightarrow \frac { d y } { d x } = \frac { 1 – ( 2 x – 5 ) y } { x ^ { 2 } – 5 x + 6 }{/tex}[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
    {tex}\frac { x – 7 } { x ^ { 2 } – 5 x + 6 } = 0{/tex}
    {tex}\Rightarrow{/tex} x= 7
    So, curve passes through the point (7, 0).
    Now, slope of tangent at (7,0) is
    {tex}m = \left( \frac { d y } { d x } \right) _ { ( 2,0 ) } = \frac { 1 – 0 } { 49 – 35 + 6 } = \frac { 1 } { 20 }{/tex}
    Hence, the required equation of tangent passing through the point (7, 0) having slope 1/20 is
    y – 0 = {tex}\frac{1}{20}{/tex}(x – 7)
    {tex} \Rightarrow{/tex} 20y = x – 7
    {tex} \therefore{/tex} x – 20y = 7
18. 
    {tex}s = \pi {r^2} + \pi rl{/tex} (given)
    {tex}\Rightarrow{/tex}{tex}l = \frac{{s – \pi {r^2}}}{{\pi r}}{/tex}
    Let v be the volume
    {tex}v = \frac{1}{3}\pi {r^2}h{/tex}
    {tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}{h^2}\,\,\left[ {{h^2} = {l^2} – {r^2}} \right]{/tex}
    {tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}\,\left( {{l^2} – {r^2}} \right){/tex}
    {tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}\left[ {\,{{\left( {\frac{{s – \pi {r^2}}}{{\pi r}}} \right)}^2} – {r^2}} \right]{/tex}
    {tex}= \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]{/tex}
    {tex}= \frac{1}{9}{r^2}\left[ {{{\left( {s – \pi {r^2}} \right)}^2} – {\pi ^2}{r^4}} \right]{/tex}
    {tex}= \frac{1}{9}{r^2}\left[ {{s^2} + {\pi ^2}{r^4} – 2s\pi {r^2} – {\pi ^2}{r^4}} \right]{/tex}
    {tex}= \frac{1}{9}{r^2}\left[ {{s^2} – 2s\pi {r^2}} \right]{/tex}
    {tex}z = \frac{1}{9}\left[ {{s^2}{r^2} – 2s\pi {r^4}} \right]{/tex}
    {tex}\left[ {\because {v^2} = z} \right]{/tex}
    Now {tex}\frac{{dz}}{{dr}} = \frac{1}{9}\,\left[ {2r{s^2} – 8s\pi {r^3}} \right]{/tex}
    {tex}0 = \frac{1}{9}\,\left[ {2r{s^2} – 8s\pi {r^3}} \right]{/tex}
    {tex}8s\pi {r^2} = 2r{s^2}{/tex}
    {tex}\implies{/tex} {tex}4\pi {r^2} = s{/tex}
    Now {tex}\frac{{{d^2}z}}{{d{x^2}}} = \frac{1}{9}\,\left[ {2{s^2} – 24s\pi {r^2}} \right]{/tex}
    {tex}{\left. {\frac{{{d^2}z}}{{d{x^2}}}} \right]_{{r^2} = \frac{s}{{4\pi }}}} = \frac{1}{9}\,\left[ {{{25}^2} – 24\pi .\frac{5}{{4\pi }}} \right]{/tex}
    = + ve
    Hence minimum
    Now {tex}s = 4\pi {r^2}{/tex}
    We have {tex}s = \pi rl + \pi {r^2}{/tex}
    {tex}4\pi {r^2} = \pi rl + \pi {r^2}{/tex}
    {tex}\Rightarrow{/tex} {tex}3\pi {r^2} = \pi rl{/tex}
    {tex}\Rightarrow{/tex} 3 r = l
    {tex}\Rightarrow{/tex} {tex}\frac{r}{l} = \frac{1}{3}{/tex}
    {tex}\Rightarrow{/tex} {tex}\sin \alpha = \frac{1}{3}{/tex}
    {tex}\therefore{/tex} {tex}\alpha = {\sin ^{ – 1}}\left( {\frac{1}{3}} \right){/tex}

Chapter Wise Important Questions Class 12 Maths Part I and Part II

1. Relations and Functions
2. Inverse Trigonometric Functions
3. Matrices
4. Determinants
5. Continuity and Differentiability
6. Application of Derivatives
7. Integrals
8. Application of Integrals
9. Differential Equations
10. Vector Algebra
11. Three Dimensional Geometry
12. Linear Programming
13. Probability