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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Find the slope of the tangent to the curve `y=(x-1)/(x-2), x!=2`at `x= 10`. |
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Answer» `y = (x-1)/(x-2)` ` rArr (dy)/(dx) = ((x-2)d/(dx)(x-1)-(x-1)d/(dx)(x-2))/((x-2)^(2))` ` = ((x-2)-(x-1))/((x-2)^(2))=(-1)/((x-2)^(2))` at x = 10 Slope of tangent ` m = (-1)/((10-2)^(2)) =-1/64` |
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| 252. |
Find the slope of the tangent to the curve `y= 3x^4-4x` at `x=4`. |
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Answer» `y= 3x^(4) - 4x` ` rArr (dy)/(dx) = 12 x^(3) - 4` at x = 4 Slope of tangent ` m = 12 (4)^(3) - 4` ` = 786 - 4 = 764` |
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| 253. |
Find the equation of the normal at the point `(a m^2,a m^3)`for the curve `a y^2=x^3`. |
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Answer» Equation of curve ` ay^(2) = x^(3)` ` rArr 2ay (dy)/(dx) = 3x^(2)` ` rArr (dy)/(dx) = (3x^(2))/(2ay)` Slope of tangent at point `(am^(2), am^(3)) = (3(am^(2)^(2))/(2a(am^(3))) = (3m)/2` ` rArr" Slope of normal "=(-2)/(3m)` and equation of normal ` y-am^(3) = - 2/(3m) (x- am^(2))` ` rArr 3my - 3am^(4) = - 2x + 2am^(2)` ` rArr 2x + 3my - 3am^(4) - 2am^(2) = 0` |
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| 254. |
Find the points on the curve `x^2+y^2-2x-3=0`at which the tangents are parallel to the x-axis. |
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Answer» Equation of curve ` x^(2)+y^(2)-2x - 3 = 0`…(1) ` rArr 2x + 2y (dy)/(dx) - 2 =0` ` rArr (dy)/(dx) = (1-x)/y` Slope of tangent at point `(x, y) is m = (1-x)/y` but the tangent is parallel to x-axis. ` :. M = 0` ` rArr (1-x)/y = 0 rArr x = 1` put x = 1 in equation (1) ` 1+ y^(2) - 2 - 3 = 0` ` rArr y^(2) = 4 rArr y = pm 2` `:.` Required point = (1, 2) and (1, -2) |
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| 255. |
For the curve `y=4x^3-2x^5,`find all the points at which the tangent passes through the origin. |
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Answer» Equation of curve ` y = 4x^(3) - 2x^(5)` ….(1) ` rArr (dy)/(dx) = 12x^(2) - 10x^(4)` Slope of tangent at point `(x_(1), y_(1)) = 12x_(1)^(2) - 10x_(1)^(4)` and equation of tangent. ` y - y_(1) = (12x_(1)^(2)-10x_(1)^(4))(x-x_(1))` This tangent passes through the point (0, 0). `:. 0-y_(1)=(12x_(1)^(2)-10x_(1)^(4))(0-x_(1))` `rArr y_(1) = 12x_(1)^(3) - 10x_(1)^(5)` ....(2)` `(x_(1), y_(1))` line on curve (1) ` :. y_(1)= 2x_(1)^(5)` `rArr 12x_(1)^(3)-10x_(1)^(5)=4x_(1)^(3)-2x_(1)^(5)` `rArr 8x_(1)^(3)-8x_(1)^(5)= 0` `rArr 8x_(1)^(3) (1-x_(1)^(2))= 0` ` rArr x_(1) = 0, 1, -1` From equation (2), the corresponding values of `y_(1)` are 0, 2 and -2. `:. " Required point " -=(0, 0), (1, 2) and (-1, 2) and (-1, -2)`. |
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| 256. |
Show that the maximum and minimum values of the function `(x+1)^2/(x+3)^3` are respectively given by `2/27 and 0.` |
| Answer» Correct Answer - Min value = 0 at x = - 1 and max value = `2/27 ` at x = 3 | |
| 257. |
Find the maximum value of the function `x * e^(-x)`. |
| Answer» Correct Answer - 1/e | |
| 258. |
If the function `f(x) = x^(3)-24x^(2)+6kx-8` is maximum at x = 2 then find value of k. |
| Answer» Correct Answer - k = 14 | |
| 259. |
Prove that `tan x gt x` for all `x in [0, (pi)/(2)]` |
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Answer» Let c be an arbitrary real number such that `c in [0, (pi)/(2)]` Let f(x) `= tan x - x` for all `x in [0, c]` `:. F(x) = sec^(2) x - 1 = tan^(2) x gt 0` for all `x in [0, c]` Thus, f(x) is increasing on [0, c] Now, `x gt 0 rArr f(x) gt f(0)` `rArr f(x) gt0` `rArr tan x - x gt 0` `rArr tan x gt x` |
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| 260. |
Find the maximum value of the function `x^(1//x)`. |
| Answer» Correct Answer - `e^(1//e)` | |
| 261. |
The combined resistance R of two resistors `R_(1) and R_(2) " where " R_(1), R_(2) gt 0` is given by `(1)/(R) = (1)/(R_(1)) + (1)/(R_(2))` If `R_(1) + R_(2) = C` (constant), show that the maximum reistance R is obtained by chossing `R_(1) = R_(2)`A. `R_(1)=R_(2)`B. `R_(1)=2R_(2)`C. `R_(2)=2R_(1)`D. `R_(2)=CR_(1)` |
| Answer» Correct Answer - A | |
| 262. |
Find the maximum value of the function `(log x)/x " when "x gt 0`. |
| Answer» Correct Answer - `1/e` | |
| 263. |
If `x gt 0 and xy =1`, the minimum value of `(x + y)` isA. `-2`B. 1C. 2D. none of these |
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Answer» Correct Answer - C `xy = 1 rArr y = (1)/(x)` Let `S = x + y = x + (1)/(x)`. Then, `(ds)/(dx) = (1 - (1)/(x^(2))) = ((x^(2) -1))/(x^(2)) and (d^(2)s)/(dx^(2)) = (2)/(x^(3))` `(ds)/(dx) = 0 rArr x^(2) - 1 = 0 rArr x = +- 1` `(d^(2)s)/(dx^(2))]_((x = -1)) = -2 lt 0 and (d^(2)s)/(dx^(2))]_((x =1)) = 2 gt 0` `:.` S is minimum at x = 1 and minimum value of `S = (1 + 1) = 2` |
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| 264. |
The time `t`of a complete oscillation of a simple pendulum of length `l`is given by the equation`T=2pisqrt(1/g)`where `g`is constant. What is the percentage error in `T`when `l`is increased by 1%? |
| Answer» Correct Answer - 0.02 | |
| 265. |
There is an error of `0.2%` in measurment of the redius of a sphere. Find the percentage error in its (i) volume, (ii) surface. |
| Answer» Correct Answer - `(i) 0.6% (ii) 0.4%` | |
| 266. |
Two sides of a triangle are given. The angle between them such that the area is maximum, is given byA. `(3pi)/(2)`B. `(pi)/(2)`C. `(pi)/(3)`D. `(pi)/(3)` |
| Answer» Correct Answer - B | |
| 267. |
Find the position of the point P on seg AB of length 8 cm, so that `"AP"^(2)+"BP"^(2)` is minimum.A. AP=2BPB. AP=BPC. AP+2BP=0D. AP+BP=0 |
| Answer» Correct Answer - B | |
| 268. |
Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using derivatives.A. `pqsqrtr`B. `2pqsqrtr`C. `2rsqrtpq`D. None of these |
| Answer» Correct Answer - C | |
| 269. |
Find the maximum or minium values, if any, without using derivatives, of the functions: `sin 2x + 5` |
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Answer» Correct Answer - max. value = 4, min. value = 6 `-1 le sin 2x le 1` |
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| 270. |
Without using the derivative, find the maximum or minium values, if any of the function `f(x) = 4x^(2) - 4x + 7` for all `x in R` |
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Answer» We have `f(x) = 4x^(2) - 4x + 7 = (2x -1)^(2) + 6` Clearly, `(2x -1)^(2)` is non-negative for all `x in R` The least value of `(2x -1)^(2)` is 0 So, the least value of the function is 6 Clearly, this happens when `2x -1 = 0`, i.e., `x = (1//2)` Thus, `x = (1//2)` is a point of absolute minimum. However, f(x) does not have an absolute maximum. |
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| 271. |
Find the points of local maxima or local minima and the corresponding local maximum and minimum values of each of the following functions: `f(x) = x^(2)` |
| Answer» Correct Answer - local min, value is 0 at x = 0 | |
| 272. |
Find the points of local maxima or local minima and the corresponding local maximum and minimum values of each of the following functions: `f(x) = -x^(3) + 12x^(2) - 5` |
| Answer» Correct Answer - local max. value is 251 at x = 8 and local min. value is `-5 " at " x = 0` | |
| 273. |
If an error of `k %`is made in measuring the radius of a sphere, then percentage error in itsvolume.k% (b) 3k% (c) 3k% (d) `k/3%`A. `2%`B. `4%`C. `6%`D. `8%` |
| Answer» Correct Answer - C | |
| 274. |
In the interval (-1, 1), the function ` f(x) = x^(2) - x + 4` is :A. increasingB. decreasingC. neither increasing nor decreasingD. None of the above |
| Answer» Correct Answer - C | |
| 275. |
The radius of an air bubble is increasing at the rate of `1/2c m//s`. At what rate is the volume of the bubble increasing when the radius is 1 cm? |
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Answer» Let r be the radius and V be the volume of the air bubble. Given ` (dr)/(dt) = 1/2 ` cm/sec Now, ` V= 4/3 pi r^(3)" "rArr (dV)/(dt) = 4 pi r^(2) (dr)/(dt)` at r = 1 cm ` (dV)/(dt) = 4 pi (1)^(2)* 1/2 = 2pi cm^(3)//sec` Therefore, the volume of bubble is increasing at the rate of ` 2pi cm^(3) //sec` |
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| 276. |
The total cost C (x) in Rupees associated with the production of x units of an item is given by `C(x)=0. 007 x^3-0. 003 x^2+15 x+4000`. Find the marginal cost when 17 units are produced |
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Answer» `C(x) = 0.007x^(3) - 0.003x^(2)+ 15x+ 4000` Marginal cost ` = (dC)/(dx) = 0.021x^(2) - 0.006x+15` at x = 17, the marginal cost ` = 0.021 (17)^(2) - 0.006(17) + 15` ` = 6.069- 0.102 +15` ` = Rs. 20.967` |
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| 277. |
A jet of an enemy isflying along the curve `y=x^2+2`. A soldier is placedat the point `(3, 2)`. What is the shortestdistance between the soldier and the jet? |
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Answer» Correct Answer - `(1, 3)` Find a point `(x, x^(2) + 2)` nearest to (3, 2) |
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| 278. |
A balloon, which always remains spherical, has a variable diameter `3/2(2x+1)`.Find the rate of change of its volume with respect to x.A. `3pi(2x+1)^(2)`B. `8/9 pi (2x+1)^(2)`C. `8/27 pi(2x+1)^(2)`D. `8/3 pi (2x+1)^(2)` |
| Answer» Correct Answer - C | |
| 279. |
The perimeter of a rectangle is 40 cm. Find the dimensions of the rectangle if its area is maximum. |
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Answer» Let l and b are the sides of rectangle. ` :. 2(l+b) = 40 ` `rArr l+b = 20` …(1) If the area of rectangle is A, then `A= l*b` `l(20-l)` [From eq. (1)] `=20 l - l^(2)` `rArr (dA)/(dl) = 20 - 2l` For maxima/minima `(dA)/(dl) = 0` ` rArr 20 - 2l=0` ` rArr l = 10 cm` And `(d^(2)A)/(dl^(2)) = - 2 lt 0` `:." At " l = 10 cm`, is maximum. Therefore, for maximum area, the sides of rectangle are 10 cm and 10 cm. |
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| 280. |
Find the dimension of the rectangle of area `96 cm^(2)` whose perimeter is the least, Also, find the perimeter of the rectangle |
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Answer» Correct Answer - length = `4 sqrt6 cm`, breadth `= 4 sqrt6 cm`, perimeter `= 16 sqrt6 cm` Let P be the fixed perimeter and x, y be the sides Then, `96 = xy rArr y = (96)/(x)` `P = 2 (x + y) rArr P = 2 (x + (96)/(x))` `P = 2(x + y) rArr y = (1)/(2) (P - 2x) " " :. A = xy rArr A = (1)/(2) x (P - 2x)` |
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| 281. |
6) The volume of a spherical balloon is increasing at the rate of 20cm / sec. Find the rate of change of its surface area at the instant when its radius is 8 cm. |
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Answer» At any instant t, let r be the radius, V the volume and S the surface area of the balloon. Then, `(dV)/(dt) = 20 cm^(3)//sec` (given ) ...(i) Now, `V = (4)/(3) pi r^(3) rArr (dV)/(dt) = (dV)/(dr).(dr)/(dt)` `rArr 20 = (d)/(dr) ((4)/(3) pi r^(3)).(dr)/(dt)` `rArr 20 = (4)/(3) pi xx 3r^(2) xx (dr)/(dt) = 4 pi r^(2). (dr)/(dt)` `rArr (dr)/(dt) = (5)/(pir^(2))`...(ii) `:. S = 5pir^(2) rArr (dS)/(dt) = (dS)/(dr) .(dr)/(dt)` `= (d)/(dr) (4pir^(2)).(5)/(pir^(2))` `= (8 pi rxx (5)/(pir^(2))) = (40)/(r)` `rArr [(dS)/(dt)]_(r =8 cm) = ((40)/(8)) cm^(2)//sec = 5 cm^(2)//sec` Hence, the rate of change of surface area at the instant when `r = 8 cm" is " 5 cm^(2)//sec` |
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| 282. |
A balloon, which always remains spherical, has avariable radius. Find the rate at which its volume is increasing with theradius when the later is 10 cm |
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Answer» Let r be the radius and V be the volume of the balloon. `:. V= 4/3 pi r^(3)` `rArr (dV)/(dr) = 4 pi r^(2)` at r 10 cm ` (dV)/(dr) = 4 pi (10)^(2) = 400 pi` ` :. ` Volume of balloon is changing at the rate of ` 400 pi cm^(3)`/sec. |
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| 283. |
The radius of a balloon is increasing at the rateof 10 cm/sec. At what rate is the surface area of the balloon increasing whenthe radius is 15 cm? |
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Answer» Correct Answer - `1200pi cm^(2)//s` `S = 4pi r^(2) rArr (dS)/(dt) = 8pir.(dr)/(dt) = (8pi xx 15 xx 10) cm^(2)//s` |
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| 284. |
A rectangle has an area of 50 `"cm"^(2)`. Find its dimensions for least perimeter.A. 5 cm, 5 cmB. `sqrt(2) cm, sqrt(2) cm`C. `5 sqrt(2) cm, 5sqrt(2) cm`D. `(5)/(sqrt(2)) cm, (5)/(sqrt(2)) cm` |
| Answer» Correct Answer - C | |
| 285. |
The volume of a spherical balloon is increasing at a rate of ` 25 cm^(3)//sec`. Find the rate of increase of its curved surface when the radius of balloon is 5 cm. |
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Answer» Correct Answer - `(dS)/(dt) = 10 cm^(2)//s` `V = (4)/(3) pir^(3) rArr (dV)/(dt) = (4pi r^(2)).(dr)/(dt) .(dr)/(dt) rArr (dr)/(dt) = (25)/(4pir^(2))` `S = 4 pi r^(2) rArr (dS)/(dt) = 8pi r.(dr)/(dt) = (8pi r.(25)/(4pir^(2))) = (50)/(r)` `rArr [(dS)/(dt)]_(r =5) = ((50)/(5)) cm^(2)//s` |
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| 286. |
A balloon which always remains spherical, isbeing inflated by pumping in 900 cubic centimetres of gas per second. Findthe rate at which the radius of the balloon is increasing when the radius is15 cm. |
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Answer» Correct Answer - `(dr)/(dt) = 0.32 cm//s` `V =(4)/(3) pi r^(3) rArr (dV)/(dt) = 4pi r^(2).(dr)/(dt) rArr (dr)/(dt) = (900)/(4pir^(2)) = (225)/(pir^(2))` `rArr [(dr)/(dt)]_(r = 15) = (225)/(pi xx (15)^(2)) = (1)/(pi) cm//s` |
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| 287. |
(i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. Find the dimensions of this box if we want to use least material for construction: (ii) A window is in the form of a rectangle surmounted by a semi-circle. Its perimeter is 40m. Find the dimensions of this window from which maximum light can admit. |
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Answer» Correct Answer - (i) 8 cm, 8 cm, 4 cm (ii) Length = `80/(pi+4)" metre, Breadth "40/(pi+4)" metre,Radius "40/(pi+4)` metre |
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| 288. |
The volume of a closed square based rectangular box is 1000 cubic metre. The cost of constructing the base is 15 paise per square metre and the cost of constructing the top is 25 paise per square metre. The cost of constructing its sides is 20 paise per square metre and the cost of constructing the box is Rs. 3.Find the dimensions of box for minimum cost of construction. |
| Answer» Correct Answer - 10 metre, 10 metre, 10 metre | |
| 289. |
A closed cylinder hasvolume 2156cm3. What will be the radius of its base so that itstotal surface area is minimum? |
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Answer» Let r be the radius and h the height of the cylinder. Then, `pi r^(2) h = 2156 rArr h = ((2156)/(pir^(2)))`. Let S be its total surface area. Then, `S = (2pi r^(2) + 2 pi r h) = (2pi r^(2) + 2pi r xx (2156)/(pi r^(2))) = (2pi r^(2) + (4312)/(r))` `:. (dS)/(dr) = (4pi r - (4312)/(r^(2))) and (d^(2)S)/(dr^(2)) = (4pi + (8624)/(r^(3)))` Now, `(dS)/(dr) = 0 rArr (4pi r - (4312)/(r^(2))) = 0` `rArr 4 pi r^(3) = 4312 rArr r = ((4312)/(4pi))^(1//3) = ((4312 xx 7)/(4 xx 22))^(1//3) = 7` And, `[(d^(2)S)/(dr^(2))]_(r =7) = (4pi + (8624)/(343)) gt 0 " " :. S` is maximum when r = 7 |
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| 290. |
If h is the height and r is the base radius of an open right circular cylinder of given volume has least surface area, thenA. h=3rB. h=4rC. h=rD. h=2r |
| Answer» Correct Answer - C | |
| 291. |
The semi-vertical angle of a cone remains constant. If its height increases by 2%, then find the approximate increase in its volume. |
| Answer» Correct Answer - `6%` | |
| 292. |
A window isin the form of a rectangle surmounted by a semi-circle. If the totalperimeter of the window is `30" m",`find the dimensions of the window so thatmaximum light is admitted.A. `(20)/(4+pi) m, (20)/(4+pi)m`B. `(30)/(4+pi) m, (30)/(4+pi)m`C. `(20)/(4+pi) m, (40)/(4+pi) m`D. `(30)/(4+pi) m, (60)/(4+ pi) m` |
| Answer» Correct Answer - B | |
| 293. |
Show that the surfacearea of a closed cuboid with square base and given volume is minimum, when itis a cube. |
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Answer» Let V be the fixed volume of a closed cuboid with length a, breadth a and height h. Let S be its surface area. Then, `V = (a xx a xx h) or h = (V)/(a^(2))`..(i) Now, `S = 2 (a^(2) + ah + ah) = 2 (a^(2) + 2ah) = 2 (a^(2) + (2V)/(a))`[using (i)] i.e., `S = 2 (a^(2) + (2V)/(a))." ":. (dS)/(da) = 2 (2a - (2V)/(a^(2))) and (d^(2)S)/(da^(2)) = (4 + (8V)/(a^(3)))` Now, `(dS)/(da) = 0 rArr V = a^(3) rArr a xx a xx h = a^(3) rArr h = a` Now, when `h = a`, we have `V = a^(3)` `:. [(d^(2)S)/(da^(2))]_(h = a) = (4 + (8a^(3))/(a^(3))) = 12 gt 0` So, S is minimum when length = a, breadth = a and height = a, i.e., when it is a cube |
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| 294. |
Find the maximum volume of right circular cylinder, if the sum of its radius and height is 6 units.A. `4pi` cu cmB. `8pi` cu cmC. `16 pi` cu cmD. `32 pi` cu cm |
| Answer» Correct Answer - D | |
| 295. |
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is `sin^(-1)(1/3)`. |
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Answer» Let r be the radius, l the slant height and h the height of the cone. Let S denote the surface area and V the volume of the cone. Then, `S = (pi r^(2) + pi rl) =` constant `:. l = ((S)/(pi r) - r)`...(i) Now, `V = (1)/(3) pi r^(2) h = (1)/(3) pi r^(2) sqrt(l^(2) - r^(2))` `:. V^(2) = (1)/(9) pi^(2) r^(4) (l^(2) - r^(2)) = (1)/(9) pi^(2) r^(4).{((S)/(pi r) - r)^(2) - r^(2)}` [using (i)] `=(1)/(9) S (Sr^(2) - 2pi r^(4))` Thus, `V^(2) = ((1)/(9) S^(2) r^(2) - (2pi)/(9) Sr^(4))` `:. 2V .(dV)/(dr) = ((2)/(9) S^(2) r - (8 p S)/(9) r^(3)) = (2rS)/(9) (S - 4pi r^(2))`...(ii) `:. (dV)/(dr) = 0 rArr r = 0 or (S - 4pi r^(2)) = 0 rArr r^(2) = (S)/(4pi)` (neglecting `r = 0`) On differentiating (ii), we get `2((dV)/(dr))^(2) + 2V.(d^(2)V)/(dr^(2)) = (1)/(9) S (2S - 24 pi r^(2))` Putting `(dV)/(dr) = 0 and r^(2) = (S)/(4pi)`, we get `2V.(d^(2)V)/(dr^(2)) = (1)/(9) S (2S - 6S) = - (4)/(9) S^(2) lt 0` `:.` when the volume is maximum, we have `r^(2) = (S)/(4pi) = ((pi r^(2) + pi rl))/(4pi) rArr l = 3r` Now if `alpha` is the semivertical angle of the cone then `(r)/(l) = sin alpha rArr (r)/(3r) = sin alpha rArr sin alpha = (1)/(3) rArr alpha = sin^(-1) ((1)/(3))` Hence, the semivertical angle of a right cone of a given surface and maximum volume is `sin^(-1) (1//3)` |
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| 296. |
Sand is pouring at the rate of `12 cm^(3)"/sec"`. The falling sand forms a cone on the ground in such a way that the height of the cone is always `((1)/(6))^(th)` of the radius of the base. If the height of sand is 4 cm, then the rate at which height of sand increasing, isA. `(1)/(12 pi)` cm/secB. `(1)/(6pi)` cm/secC. `(1)/(48 pi)` cm/secD. `(1)/(24 pi)` cm/sec |
| Answer» Correct Answer - C | |
| 297. |
An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of `pi a^3 cu.cm` of water.Find the dimensions so that the quantity of metal sheet required is a minimum.A. Radius=2a cm, Height =2a cmB. Radius =2a cm, Height=a cmC. Radius =a cm, Height =a cmD. Radius = acm, Height = 2a cm |
| Answer» Correct Answer - C | |
| 298. |
A right circular cone have slant height 3 cm. Then its volume is maximum at heightA. 3 cmB. `sqrt(3)` cmC. `(1)/(3)` cmD. `(1)/(sqrt(3))` cm |
| Answer» Correct Answer - B | |
| 299. |
From mean value theoren : `f(b)-f(a)=(b-a)f^(prime)(x_1); a lt x_1 lt b` if `f(x)=1/x` , then `x_1` is equal toA. `(a+b)/(2)`B. `sqrt(ab)`C. `(2ab)/(a+b)`D. `(b-a)/(b+a)` |
| Answer» Correct Answer - B | |
| 300. |
For the function `f(x) = x + 1/x, x in [1,3]` , the value of c for mean value therorem isA. `(2)/(sqrt(3))`B. `(4)/(sqrt(3))`C. `-sqrt(3)`D. `sqrt(3)` |
| Answer» Correct Answer - D | |