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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.
| 151. |
If `y = tan^(-1) ((1 + x^(2))/(1 - x^(2))) " then " (dy)/(dx) =` ?A. `(2x)/((1 + x^(4)))`B. `(-2x)/((1 +x^(4)))`C. `(x)/((1 + x^(4)))`D. none of these |
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Answer» Correct Answer - A Putting `x^(2) = tan theta`, we get `y = tan^(-1) {tan ((pi)/(4) + theta)} = ((pi)/(4) + theta) = ((pi)/(4) + tan^(-1) x^(2))` |
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| 152. |
If `x^(p) y^(q) = (x + y)^((p + q)) " then " (dy)/(dx)=` ?A. `(x)/(y)`B. `(y)/(x)`C. `(x^(p-1))/(y^(q -1))`D. none of these |
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Answer» Correct Answer - B `p log x + q log y = (p +q) log (x + y)` |
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| 153. |
If `y = sqrt(x sin x) " then "(dy)/(dx) =` ?A. `((x cos x + sin x))/(2 sqrt(x sin x))`B. `(1)/(2) (x cos x + sin x).sqrt(x sin x)`C. `(1)/(2 sqrt(x sin x))`D. none of these |
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Answer» Correct Answer - A `y^(2) = x sin x rArr 2y.(dy)/(dx) = (x cos + sin x)` |
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| 154. |
If `y=tan^(- 1)sqrt((1-cosx)/(1+cosx)),` prove that `(dy)/(dx)=1/2.`A. `(-1)/(2)`B. `(1)/(2)`C. `(1)/((1 + x^(2)))`D. none of these |
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Answer» Correct Answer - B `y = tan^(-1) {(2 sin^(2) (.^(x)//_(2)))/(2 cos^(2) (.^(x)//_(2)))} = tan^(-1) {"tan"(x)/(2)} = (x)/(2)` |
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| 155. |
If `y = tan^(-1) ((1 - cos x)/(sin x)) " then "(dy)/(dx)=` ?A. 1B. `-1`C. `(1)/(2)`D. `(-1)/(2)` |
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Answer» Correct Answer - C `y = tan^(-1) ((1 - cos x)/(sin x)) = tan^(-1) {(2 sin^(2) (.^(x)//_(2)))/(2 sin (.^(x)//_(2)) cos (.^(x)//_(2)))} = tan^(-1) {"tan " (x)/(2)} = (x)/(2)` |
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| 156. |
If `y = x^(2) " sin " (1)/(x) " then " (dy)/(dx) =` ?A. `x "sin " (1)/(x) - " cos " (1)/(x)`B. `-"cos "(1)/(x) + 2x "sin " (1)/(x)`C. `-x " sin " (1)/(x) + " cos"(1)/(x)`D. none of these |
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Answer» Correct Answer - B `(dy)/(dx) = x^(2).("cos"(1)/(x)) ((-1)/(x^(2))) + 2x "sin " (1)/(x)` |
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| 157. |
If `y = tan^(-1){(cos x + sinx)/(cos x - sin x)} " then " (dy)/(dx) =` ?A. 1B. `-1`C. `(1)/(2)`D. `(-1)/(2)` |
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Answer» Correct Answer - A `y = tan^(-1) {(cos x + sin x)/(cos x - sinx)} = tan^(-1) {(1 + tan x)/(1 - tan x)} = tan^(-1) {tan ((pi)/(4) + x)} = ((pi)/(4) + x)` |
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| 158. |
If `y = tan^(-1){(cos x)/(1 + sinx)} " then " (dy)/(dx) =` ?A. `(1)/(2)`B. `(-1)/(2)`C. 1D. `-1` |
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Answer» Correct Answer - B `y = tan^(-1) {(sin ((pi)/(2) -x))/(1 + cos ((pi)/(2) -x))} = tan^(-1) {tan ((pi)/(4) - (x)/(2))}= ((pi)/(4) - (x)/(2))` |
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| 159. |
If `y = sin (x^(x)) " then " (dy)/(dx) =` ?A. `x^(x) cos (x^(x))`B. `x^(x) cos x^(x) (1 + log x)`C. `x^(x) cos x^(x) log x`D. none of these |
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Answer» Correct Answer - B Let `x^(x) = z`. Then, `y = sin z`. Then, `(dy)/(dx) = ((dy)/(dx) xx (dz)/(dx))` |
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| 160. |
If `y = cos^(2) x^(3) " then "(dy)/(dx) =` ?A. `-3 x^(2) sin (2x^(3))`B. `-3 x^(2) sin^(2) x^(3)`C. `-3x^(2) cos^(2) (2x^(3))`D. none of these |
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Answer» Correct Answer - A `y = (cos x^(3))^(2) rArr (dy)/(dx) = 2 (cos x^(3)) (-sin x^(3)) (3x^(2)) = -3x^(2) sin (2x^(3))` |
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| 161. |
If `y = sqrt((1 + sinx)/(1 - sin x)) " then " (dy)/(dx) =` ?A. `(1)/(2) sec^(2)((pi)/(4) - (x)/2)`B. `(1)/(2) "cosec"^(2) ((pi)/(4) - (x)/(2))`C. `(1)/(2) "cosec"((pi)/(4) - (x)/(2)) cot ((pi)/(4) - (x)/(2))`D. none of these |
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Answer» Correct Answer - B `y = {(1 + cos ((pi)/(2) - x))/(1 - cos ((pi)/(2) -x))}^((1)/(2)) = {(2 cos^(2) ((pi)/(4) - (x)/(2)))/(2 sin^(2) ((pi)/(4) - (x)/(2)))}^((1)/(2)) = cot ((pi)/(4) - (x)/2)` |
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| 162. |
If `y = (sin x)^(log x) " then " (dy)/(dx) =`?A. `(log x).(sinx)^((log x -1)).cos x`B. `(sin x)^(log x).{(x log x + log sin x)/(x)}`C. `(sin x)^(log x).{((x log x) cot x + log sin x)/(x)}`D. none of these |
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Answer» Correct Answer - C `log y = (log x).(log sin x)` |
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| 163. |
If `f(x)=3x^(3)-9x^(2)-27x+15`, thenA. f has maximum value 66B. f has minimum value -66C. f has maxima at x=3D. f has minima at x=-1 |
| Answer» Correct Answer - B | |
| 164. |
Examine the following functions for maxima and minima : `f(x)=3x^(3)-9x^(2)-27x+15`A. f has maximum value 66B. f has minimum value 30C. f has maxima at x=-1D. f has minima at x=-1 |
| Answer» Correct Answer - C | |
| 165. |
If `y = log (x + sqrt(x^(2) + a^(2))) " then " (dy)/(dx) =` ?A. `(1)/(2(x + sqrt(x^(2) + a^(2))))`B. `(-1)/(sqrt(x^(2) + a^(2)))`C. `(1)/(sqrt(x^(2) + a^(2)))`D. none of these |
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Answer» Correct Answer - C `(dy)/(dx) = (1)/((x + sqrt(x^(2) + a^(2)))) .{1 + (1)/(2 sqrt(x^(2) + a^(2))) xx 2x} = (1)/(sqrt(x^(2) + a^(2)))` |
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| 166. |
If `y = log ((sqrt(1 + x^(2)) + x)/(sqrt(1 + x^(2)) -x)) " then " (dy)/(dx) =` ?A. `(2)/(sqrt(1 + x^(2)))`B. `(2 sqrt(1 + x^(2)))/(x^(2))`C. `(-2)/(sqrt(1 +x^(2)))`D. none of these |
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Answer» Correct Answer - A `y = log (sqrt(1 + x^(2)) + x) - log (sqrt(1 + x^(2)) -x)`. Now, differentiate |
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| 167. |
If `y = (tan x)^(cot x) " then " (dy)/(dx)=` ?A. `cot x.(tan x)^(cot x -1). Sec^(2) x`B. `- (tan x)^(cot x). "cosec"^(2)c`C. `(tanx)^(cot x)."cosec"^(2)x (1 - log tan x)`D. none of these |
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Answer» Correct Answer - C `log y = cot x. log (tan x)` |
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| 168. |
If `y = x^(sqrtx) " then " (dy)/(dx) =` ?A. `sqrtx.x^((sqrtx -1))`B. `(s^(sqrtx)logx)/(2 sqrtx)`C. `x^(sqrtx){(2 + log x)/(2 sqrtx)}`D. none of these |
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Answer» Correct Answer - C `log y = sqrtx (log x)` |
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| 169. |
If `y = "log "((1 + sqrtx)/(1 - sqrtx)) " then " (dy)/(dx) =` ?A. `(1)/(sqrtx (1-x))`B. `(-1)/(x(1 - sqrtx)^(2))`C. `(-sqrtx)/(2(1 - sqrtx))`D. none of these |
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Answer» Correct Answer - A `y = log (1 + sqrtx) - log (1 - sqrtx)` `rArr (dy)/(dx) = {(1)/(2 sqrtx (1 + sqrtx)) + (1)/(2 sqrtx (1 - sqrtx))} = (1)/(2 sqrtx) .(2)/((1 -x)) = (1)/(sqrtx (1 -x))` |
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| 170. |
Find the intervals in which `f(x)=6+12 x+3x^2-2x^3`is increasing or decreasing. |
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Answer» Correct Answer - (a) f(x) is increasing on `[-1, 2]` (b) `f(x)` is decreasing on `[-oo, -1] uu [2, oo]` |
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| 171. |
If `f(x)=2x^(3)-21x^(2)+36x-20`, thenA. f has maxima at x=6B. f has minima at x=1C. f has maximum value -128D. f has minimum value -128 |
| Answer» Correct Answer - D | |
| 172. |
Find the intervals in which `f(x)=(x-1)(x-2)^2`is increasing or decreasing. |
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Answer» Correct Answer - (a) `f(x)` is increasing on `[-oo, (4)/(3)] uu [2, oo]` (b) `f(x)` is decreasing on `[(4)/(2), 2]` |
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| 173. |
Show that of all the rectangles in a given fixed circle, the square has the maximum area. ` (##NTN_MATH_XII_C06_E04_252_Q01.png" width="80%"> |
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Answer» Let r be the radius of given circle. A rectangle ABCD is inscribed in the circle whose diagonal is AC= 2r. Let ` AB = x and BC = y` `:." In "Delta ABC`, `AB^(2)+BC^(2)=AC^(2)` ` rArr x^(2) + y^(2) = (2r)^(2)` ` rArr y^(2) = 4r^(2) - x^(2)` ...(1) Now, area of rectangle `A = x * y` ` rArr A = x sqrt(4r^(2)-x^(2))` ` rArr A^(2) = x^(2) (4r^(2)-x^(2))` Let ` Z=A^(2)` ` rArr Z = 4r^(2)x^(2) - x^(4)` ` rArr (dz)/(dx) = 8r^(2)x-4x^(3)` For maxima/minima ` (dZ)/(dx) = 0` ` rArr 8r^(2)x-4x^(3) = 0 rArr x^(2)=2r^(2)` and ` ((d^(2)Z)/(dr^(2))) = 8r^(2) - 12x^(2)` at ` x^(2) =2r^(2)` ` (d^(2)Z)/(dr^(2)) = 8r^(2) - 24r^(2) =- 16r^(2) lt 0 ` ` rArr at x^(2) = 2r^(2) , Z` is maximum. ` rArr at x^(2) = 2r^(2), A^(2)` is maximum. ` rArr at x^(2) = 2r^(2), A` is maximum. Now ` x^(2) = 2r^(2)` ` rArr 2x^(2) = 4r^(2) = x^(2)+y^(2)` [From equation (1)] ` rArr x^(2) = y^(2)` ` rArr x = y ` Therefore, the rectangle with maximum area, is a square. |
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| 174. |
Show that of all therectangles of given area, the square has the smallest perimeter. |
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Answer» Let A be the fixed area. Consider a rectangle with sides x and y, and area A Let P be the perimeter of the this rectangle Then, `A = xy rArr y = (A)/(x)`..(i) And, `P = 2x + 2y = 2x + (2A)/(x)`[using (i)] Now, `P = (2x + (2A)/(x)) rArr (dP)/(dx) = (2 - (2A)/(x^(2))) and (d^(2)P)/(dx^(2)) = (4A)/(x^(3))` Now, `(dP)/(dx) = 0 rArr 2 - (2A)/(x^(2)) = 0 rArr x = sqrtA` and, `[(d^(2)P)/(dx^(2))]_(x = sqrtA) = (4A)/(A^(3//2)) = (4)/(sqrtA) gt 0` So, `x = sqrtA` is a point of minimum. Moreover, `x = sqrtA rArr y = (A)/(x) = (A)/(sqrtA) = sqrtA = x` So, when the perimeter is smallest, the rectangle is a square. |
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| 175. |
Find the intervals in which the function `f(x)=x^4-(x^3)/3`is increasing or decreasing. |
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Answer» Correct Answer - (a) `f(x)` is increasing on `[(1)/(4),oo]` (b) f(x) is decreasing on `[-oo, (1)/(4)]` |
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| 176. |
Determine the intervals in which the function `f(x)=x^4-8x^3+22 x^2-24 x+21`is decreasing or increasing. |
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Answer» Correct Answer - (a) `f(x)` is increasing on `[1, 2] uu [3, oo]` (b) `f(x)` is decreasing on `[-oo, 1] uu [2, 3]` |
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| 177. |
Find the intervals on which the function `f(x) = 6 - 9x - x^(2)` is (a) strictly increasing (b) strictly decreasing |
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Answer» Correct Answer - (a) f(x) is strictly increasing on `[-oo, -(9)/(2)]` (b) f(x) is strictly decreasing on `[-(9)/(2), oo]` |
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| 178. |
Show that the function f(x) = 2x+1 is strictly increasing on R. |
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Answer» f(x) 2x+1 Let `x_(1), x_(2) in R and x_(1)ltx_(2)` `"Now "x_(1)ltx_(2)` `rArr 2x_(1)lt 2x_(2)` `rArr 2x_(1)+1 lt 2x_(2) +1 ` `rArr f(x_(1))lt f (x_(2))` `:. " "x_(1)lt x_(2)` `rArr f(x_(1))lt f(x_(2))forall x_(1), x_(2) in R` `rArr ` f(x) is strictly increasing on R. |
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| 179. |
Using differentials,find the approximate value of `sqrt(0. 48)` |
| Answer» Correct Answer - `0.693` | |
| 180. |
Using differentials,find the approximate value of `(82)^(1/4)`upto 3 places ofdecimal .A. `3.008`B. `3.009`C. `3.010`D. None of these |
| Answer» Correct Answer - B | |
| 181. |
Use differentials to approximate `sqrt(25. 2)` |
| Answer» Correct Answer - `5.02` | |
| 182. |
Using differentials, find the approximate values of the following: `(i) root(4)15 (ii)(82)^(1//4)` |
| Answer» Correct Answer - `(i)63/32 (ii)(325)/(108)` | |
| 183. |
Prove that f (x) = `e^(3x)` is strictly increasing function on R. |
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Answer» `f(x) = e^(3x)` `"Let "x_91), x_(2) in R and x_(1) lt x_(2)` `"Now "x_(1)lt x_(2)` `rArr " "3x_(1) lt 3x_(2)` `rArr" "e^(3x_(1))lt e^(3x_(2))` `rArr" "f(x_(1))lt f(x_(2))` `:. " "x_(1) lt x_(2)` `rArr" "f(x_(1)) lt f(x_(2)) forall x_(1), x_(2) in R` `rArr` f(x) is strictly increasing function on R. |
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| 184. |
Show that the function f(x) = 3 - 2x is strictly decreasing function on R. |
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Answer» f(x) = 3 -2x Let `x_(1), x_(2) in R and x_(1) lt x_(2)` Now `x_(1) lt x_(2)` ` rArr -2x_(1) gt - 2x_(2)` `rArr 3-2x_(1) gt3-2x_(2)` `rArr f(x_(1)) gt f(x_(2))` `:. X_(1) le x_(2)` `rArr f(x_(1)) gt f(x_(2)) forall x_(1), x_(2) in R` ` rArr` f(x) is strictly decreasing function on R. |
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| 185. |
Using differentials, find the approximate value of ` sqrt(0.037)`. |
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Answer» Let y = `sqrtx` Let x =0.04 and ` x+ Deltax = 0.037` ltbr. `rArr 0.04 + Deltax = 0.037` ` rArr Deltax =- 0.003` At x = 0.04, ` y= sqrt(0.04) = 0.2` Now `y=sqrtx` `rArr (dy)/(dx) = 1/(2sqrtx)` At x= 0.04, ` (dy)/(dx) = 1/(2sqrt(0.04))= 1/(0.4) = 2.5` `:. Deltay = (dy)/(dx) . Deltax = 2.5 xx (-0.003) =- 0.0075` and ` y+ Deltay = 0.2 - 0.0075` ` rArr sqrt(0).037 = 0.1925`. |
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| 186. |
Using differentials, find the approximate value of `root(3)127`. |
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Answer» Let `y = root(3)x = x^(1//3)` Let x = 125 and x + ` Deltax = 127` ` rArr Deltax = 2` ` At x = 125` `y = root(3)125 =5` Now `y= x^(1//3)` `rArr (dy)/(dx) = 1/3 x^(-2//3) = 1/ (3x^(2//3))` At x = 125, ` (dy)/(dx) = 1/(3(125)^(2//3)) = 1/75` `:. Delta y = (dy)/(dx) . Deltax = 1/75 xx 2 = 2/75 = 0.026` ` and y+Deltay = 5 + 0.026` ` rArr (127)^(1//3)= 5.026`. |
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| 187. |
Use differentials andfind approximate value of `(29)^(1//3)` |
| Answer» Correct Answer - `3.074` | |
| 188. |
The sides of an equilateral triangle are increasing atthe rate of 2 cm/sec. Find the rate at which the area increases, when theside is 10 cm. |
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Answer» Correct Answer - `10 sqrt3 cm^(2)//s` `A = (sqrt3)/(4) a^(2) rArr (dA)/(da) = (sqrt3)/(2) a rArr (dA)/(dt).(dt)/(da) = (sqrt3)/(2) a` `rArr (dA)/(dt) xx (1)/(2) = (sqrt3)/(2) a rArr (dA)/(dt) = sqrt3 a rArr [(dA)/(dt)]_(a = 10) = sqrt3 xx 10 cm^(2)` |
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| 189. |
If `f(x)=a+bx^(2)+cx^(4)+dx^(6),` where a,b,c,d all are positive constants, thenA. f has only one point of mimimumB. f has only one point of maximumC. f has only two point of mimimumD. f has no point of mimimum |
| Answer» Correct Answer - A | |
| 190. |
If the rate of increase of area of a circle is not constant but the rate of increase of perimeter is constant, then the rate of increase of area variesA. as the square of the perimeterB. as the radiusC. inversely as the perimeterD. inversely as the radius |
| Answer» Correct Answer - B | |
| 191. |
A stone is dropped into a pond. Waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of 2 inches/sec. The rate at which the area is increasing after 5 seconds, isA. `20 pi" inches"^(2)//sec`B. `40pi" inches"^(2)//sec`C. `5pi" inches"^(2)//sec`D. `10pi" inches"^(2)//sec` |
| Answer» Correct Answer - B | |
| 192. |
A stone is dropped into a pond. Waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of 2 inches/sec. If the radius is 5 inches, then the rate at which the area increasing, isA. `20 pi" inches"^(2)//sec`B. `40pi" inches"^(2)//sec`C. `5pi" inches"^(2)//sec`D. `10pi" inches"^(2)//sec` |
| Answer» Correct Answer - A | |
| 193. |
If displacement of particle is `s=(t^(3))/(3)-(t^(2))/(2)-(t)/(2)+6`, then acceleration of the particle when its velocity is `(3)/(2)`, isA. `-1" units/sec"^(2)`B. `0" units/sec"^(2)`C. `4" units/sec"^(2)`D. `3" units/sec"^(2)` |
| Answer» Correct Answer - D | |
| 194. |
If displacement of particle is `s=(t^(3))/(3)-(t^(2))/(2)-(t)/(2)+6`, then displacement of the particle when velocity is `(3)/(2)` , isA. `(17)/(3)` unitsB. `(-17)/(3)` unitsC. `(11)/(3)` unitsD. `(-11)/(3)` units |
| Answer» Correct Answer - A | |
| 195. |
If displacement of particle is `s=(t^(3))/(3)-(t^(2))/(2)-(t)/(2)+6`, then velocity of the particle at t=4 sec. isA. 11.5 units/secB. 14.5 units/secC. 19.5 units/secD. 16.5 units/sec |
| Answer» Correct Answer - A | |
| 196. |
A particle moves under the law `s=t^(3)-4t^(2)-5t.` If its acceleration is `4" units/sec^(2),` then its velocity isA. 9 units/secB. `-9` units/secC. 18 units/secD. `-18` units/sec |
| Answer» Correct Answer - B | |
| 197. |
A particle moves under the law `s=t^(3)-4t^(2)-5t.` If its acceleration is `4" units/sec"^(2)`, then its displacement isA. 9 unitsB. `-9` unitsC. 18 unitsD. `-18` units |
| Answer» Correct Answer - D | |
| 198. |
Find the equation of the tangent line to the curve `y=x^2-2x+7`which is(a) parallel to the line `2x y + 9 = 0`(b) perpendicular to the line `5y 15 x = 13`. |
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Answer» Equation of curve ` y=x^(2) - 2x+7` …(1) ` rArr (dy)/(dx) = 2x -2` (a) Line 2x-y +9= 0 `rArr y = 2x +9` Its slope m=2 Slope of parallel line = 2 `:. 2x - 2 = 2` `rArr 2x - 2 = 2` ` rArr 2x = 4 ` ` rArr x = 2 ` put x = 2 in equation (1) ` y= 2^(2)-2 xx 2+7=7` and equation of tangent at point (2, 7) ` y-7=2(x-2)` ` rArr y- 7 = 2x - 4` `rArr 2x - y + 3 = 0` (b) Equation of line `5y - 15x = 13` ` rArr y = 3x + (13)/5` Its slope = 3 Slope of perpendicular line ` = - 1/3` `2x - 2 =- 1/3` ` rArr 2x = - 1/3 + 2 = 5/3 ` ` rArr x = 5/6` put ` x= 5/6` in equation (1) ` y=(5/6)^(2)-2xx5/6 + 7 = (25-60+252)/(36) = (217)/36` and equation of tangent at point `(5/6,(217)/36)` ` y- (217)/36 = - 1/3 (x-5/6)` (36y- 217)/36 =-((6x-5)/18)` ` rArr 12x+ 36y-227 = 0` |
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| 199. |
Find the slope of the tangent to curve `y=x^3-x+1` at the point whose x-coordinate is 2. |
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Answer» `y= x^(3)-x+1` ` rArr (dy)/(dx) = 3x^(2) - 1` at x = 2 Slope of tangent `m = 3 (2)^(2)-1 = 11` |
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| 200. |
Find the points on the curve `y=x^3`at whichthe slope of the tangent is equal to the `y `coordinate of the point. |
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Answer» Equation of curve `y = x^(3)` …(1) ` rArr (dy)/(dx) = 3x^(2)` Slope of tangent at point `(x, y) = 3x^(2)` Given ` 3x^(2) = y` ` rArr 3x^(2) = x^(3)` From equation (1) ` rArr x^(2) (3-x)=0` ` rArr x = 0 or x= 3` put x = 0 in equation (1) we get y = 0 put x = 3 in equation (1) we get ` y=3^(3) = 27` `:. ` Required point = (0, 0) or (3, 27) . |
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