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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.
| 51. |
Divide the number 84 into two parts such that the product of one part and the square of other is maximum.A. 42,42B. 40,44C. 50,34D. 28,56 |
| Answer» Correct Answer - D | |
| 52. |
Divide 16 into two parts such that the sum of their cubes is minimum. |
| Answer» Correct Answer - 8,8 | |
| 53. |
Find two positive numbers whose sum is 16 and the sum of whose squares is minimum |
| Answer» Correct Answer - 8, 8 | |
| 54. |
Find two positive numbers whose sum is 16 and thesum of whose cubes is minimum. |
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Answer» Let a number be x then other number is (16 - x). Let the sum of their cubes is S. Then ` S = x^(3) + (16 - x)^(3)` ` rArr (dS)/(dx) = 3x^(2)+3(16-x)^(2)(-1)` ` = 3x^(2) - 3(16-x)^(2)` and ` ((d^(2)S)/(dx^(2))) = 6x + 6(16-x) = 96` For minimum value ` (dS)/(dx) = 0` ` rArr 3x^(2) -3 (16-x)^(2)= 0` ` rArr x^(2) - (256+x^(2)-32x)=0` ` rArr 32x = 256 rArr x = 8` at ` x = 8, ((d^(2)S)/(dx^(2)))_(x=8) = 96 gt 0 ` ` :. ` at x = 8, S is minimum. `:. 16 - x = 16 - 8 = 8 ` Therefore numbers are 8 and 8. |
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| 55. |
If there is an error of0.1% in the measurement of the radius of a sphere, find approximately thepercentage error in the calculation of the volume of the sphere. |
| Answer» Correct Answer - 0.003 | |
| 56. |
A kite is flying at a height of 16 meters. A boy who is flying it, is carrying it horizontally at the rate of 1.2 meter/sec. If the height of the kite remains same, the string is straight and the length of string released is 20 meters, then rate at which the string being pair out, isA. 0.36 meter/secB. 0.48 meter/secC. 0.60 meter/secD. 0.72 meter/sec |
| Answer» Correct Answer - D | |
| 57. |
Find the angle of intersection of the curves `2y^(2) = x^(3) and 32x`. |
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Answer» `2y^(2)= x^(3) …(1)` `y^(2) = 32x` …(2) Solving eqs. (1) and (2) ` 64x = x^(3)` ` rArr x(x^(2)-64) = 0 rArr x=0,8,-8` `:. "At "x=0,y-0` `"At " x=8, y=pm 16` At x =- 8, y is imaginary. Therefore the points of intersection are (0, 0), (8, 16) and (8, -16). From eq.(1) `" "` From eq.(2) `3y^(2) = x^(3)" "y^(2) = 32x` ` rArr 4y (dy)/(dx) = 3x^(2) rArr 2y (dy)/(dx) = 32` `rArr (dy)/(dx) = (3x^(2))/(4y) rArr (dy)/(dx) = (16)/y` At point (8, 16) ` m_(1) = (3(8)^(2))/(4 xx 16) = 3 " "m_(2) = (16)/(16) = 1` ` and tan theta_(1) = (m_(1)-m_(2))/(1+m_(1)m_(2)) = (3-1)/(1+3) = 1/2 ` ` rArr theta_(1) = tan^(-1) 1/2*` At point (8,16) `m_(3)= (3(8)^(2))/(4 xx (-16)) = - 3 " "m_(4) = 16/((-16))=-1` `and tan theta_(2) = (m_(3)-m_(4))/(1+m_(3)m_(4)) = (-3+1)/(1+3) = (-1/2)` `rArr theta_(2) = tan^(-1) (-1/2)*` At point (0, 0), the slopes cannot be determined. |
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| 58. |
Find the equation of normals of the following curves at the given points: (i) Curve`y^(2)=4 ax" at point "(at^(2), 2at)`. (ii) Curve `y= e^(x)" at point "(0, 1)` (iii) Curve `y = x^(3)" at point "(1, 1)`. (iv) Curve `2y = 3 - x^(2)" at point "(1, 1)`. (v) Curve `16x^(2)-9y^(2) = 432` at point (6, 4). |
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Answer» Correct Answer - (i)`y+tx=2 at+at^(3)" "(ii) x+y=1` (iii)`x+3y=4" "(iv)x=y` (v)`3x+8y=50` |
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| 59. |
Find the angle of intersection of the curves `x y=a^2a n dx^2+y^2=2a^2` |
| Answer» Correct Answer - (i)`0^(@), (ii)90^(@), tan^(-1)(3/4)` | |
| 60. |
The two curves `x^3-3xy^2+2=0` and `3x^2y-y^3-2=0` |
| Answer» Correct Answer - `m_(1) m_(2) = -1` | |
| 61. |
A stone is thrown vertically upwards from the top of a tower `64m` high according to the law of motion given by `s=48t-16t^(2)`. The greatest height attained by the stone above ground isA. 32 metersB. 64 metersC. 36 metersD. 100 meters |
| Answer» Correct Answer - D | |
| 62. |
Water is dropped at the rate of `2m^(2)//s` into a cone of semivertical angel of `45^(@)`. The rate at which periphery of water surface changes when height of water in the cone is 2 m, isA. `0.5m//s`B. `2m//s`C. `3m//s`D. `1m//s` |
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Answer» Correct Answer - D Given that `(dV)/(dt)=2` `rArr" "(d)/(dt)((1)/(3)pir^(3))=2` `rArr" "pir^(2).(dr)/(dt)=2` `rArr" "(dr)/(dt)=(2)/(pir^(2))` `rArr" "(d)/(dt)(2pir)=(4)/(r^(2))" (i)"` When h = 2 m `rArr" "r=2m` Hence, `(d(2pir))/(dt)=(4)/(4)=1m//s` |
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| 63. |
If the length of a simple pendulum is decreased by 2%, find the percentage decrease in its period T, where `T = 2pi sqrt((l)/(g))` |
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Answer» Correct Answer - `1%` `log T = log ((2pi)/(sqrtg)) + (1)/(2) log l rArr (1)/(T).(dT)/(dl) = (1)/(2l)` `:. delta T = (T)/(2l).deltal rArr ((deltaT)/(T) xx 100) = ((1)/(2).(deltal)/(l) xx 100) = {(1)/(2) xx ((-2)/(100)) xx 100} = -1` |
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| 64. |
The radius of a sphere decreases from 10 cm to `9.9` cm. Find (i) approximate decrease in its volume. (ii) approximate decrease in its surface. |
| Answer» Correct Answer - `(i) 40 pi cm^(3) (ii) 8 pi cm^(2)` | |
| 65. |
Show that the curves `x = y^2` and `xy = k` cut at right angles; if `8k^2 = 1` |
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Answer» Let the point of intersection of the given curves be `(x_(1), y_(1))` Then, `y^(2) = x hArr 2y + (dy)/(dx) = 1 hArr ((dy)/(dx))_((x_(1)"," y_(1))) = (1)/(2y_(1)) hArr m_(1) = (1)/(2y_(1))` `xy = k hArr x (dy)/(dx) + y = 0 hArr (dy)//(dx) = (-y)/(x) hArr ((dy)/(dx))_((x_(1)"," y_(1))) = m_(2) = (-y_(1))/(x_(1))` `:. m_(1) m_(2) = -1 hArr (1)/(2y_(1)) xx (-y_(1))/(x_(1)) = -1 hArr x_(1) = (1)/(2)` `(x_(1), y_(1))` lies on `y^(2) = x hArr y_(1)^(2) = x_(1) hArr x_(1) = (1)/(2)` `(x_(1), y_(1))` lies on `xy = k hArr x_(1) y_(1) = k hArr x_(1)^(2) y_(1)^(2) = k^(2)` `hArr k^(2) = ((1)/(4) xx (1)/(2)) = (1)/(8)` `hArr 8k^(2) = 1` |
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| 66. |
The rate of increase of the radius of an air bubble is 0.5 cm/sec. Find the rate o fincrease of its volume when the radius of air bubble is 2 cm. |
| Answer» Correct Answer - `8 pi cm^(3)//sec` | |
| 67. |
The side of a square is increasing at a rate of 4cm/sec. Find the rate of increase of its area when the side of square is 10 cm. |
| Answer» Correct Answer - `80 cm^(3)//sec` | |
| 68. |
The volume of cube is increasing at a rate of `9 cm^(3)//sec`. Find the rate of increase of its surface area when the side of the cube is 10 cm. |
| Answer» Correct Answer - `3.6 cm^(2)//sec` | |
| 69. |
The side of a cube is increasing at a rate of 4 cm/sec. Find the rate of change of the volume of the cube when its side is 5 cm. |
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Answer» Let x =side of the cube and V = volume of the cube. `:. V = x^(3)` Give that x = 5 cm and `(dx)/(dt)` = 4cm/sec. Now `(dV)/(dt)=3x^(2).(dx)/(dt)` At x = 5 cm, `(dV)/(dt)=3(5)^(2).4 cm^(3)`/sec =300 `cm^(3)`/sec Therefore the volume will be increased at 300 `cm^(3)`/sec. |
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| 70. |
Find the solpe of tangent at point (1, 1) of the curve `x^(2)= y`. |
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Answer» `x^(2)= y` `rArr (dy)/(dx) = 2x` At point (1, 1), the slope of tangent ` m= ((dy)/(dx))_"(1,1)" = 2 xx 1 =2`. |
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| 71. |
In `Delta ABC`, the side c and angle C are constant. If there are small changes in the remaining sides and angles, then show that `(da)/(cos A) +(db)/(cos B)= 0`. |
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Answer» We know that `a/(sin A) = b/(sin B) = c/(sin C)` But c and C are constant. Therefore `a/(sin A)= b/(sin B) = k` ` rArr a=k sin A and b=k sin B` ` rArr (da)/(dA) = k cos A and (db)/(dB) k cos B` `rArr (da)/(cos A) = k * dA and (db)/(cos B) = k * dB` `rArr (da)/(cos A) + (db)/(cos B) = k(dA+ dB)` ` = k d (A+B) - k d (pi - C)` ` = k * (0)` `=0` ` :. (da)/(cos A) + (db)/(cos B) = 0`. |
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| 72. |
If in a triangle `A B C ,`the side `c`and the angle `C`remain constant, while the remaining elements are changed slightly,show that`(d a)/(cosA)+(d b)/(cosB)=0.` |
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Answer» As given , we have `(c)/(sinC) = k` (constant) `:. (a)/(sin A) = (b)/(sin B) = k rArr a = k sin A and b = k sin B` `:. da = k cos A .dA and db = k cos B. dB` or `(da)/(cos A) + (db)/(cos B) = k (dA + dB) = kd (A + B) = kd (pi - C) = 0` Hence, `(da)/(cos A) + (db)/(cos B) = 0` |
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| 73. |
The perimeter of a triangle is 10 cm. If one of the side is 4 cm, then for its maximum area, remaining two sides areA. 3 cm, 3 cmB. 2 cm, 4 cmC. 5 cm, 1 cmD. 3.5 cm, 2.5 cm |
| Answer» Correct Answer - A | |
| 74. |
If the displacement of a particle is `x=t^(3)-4t^(2)-5t`, then the velocity of particle at t=2 isA. `-9" units/sec"`B. 9 units/secC. `-18" units/sec"`D. 18 units/sec |
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Answer» Correct Answer - Ax=t^3-4t^2-5t dx/dt=3t^2-8t-5 t=2 dy/dx=3(2)^2-8(2)-5=-9units/sec |
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| 75. |
If the displacement of a particle is `x=t^(3)-4t^(2)-5t`, then the acceleration of particle at t=2 isA. `2" units/sec"^(2)`B. `4" units/sec"^(2)`C. `-2" units/sec"^(2)`D. `-4" units/sec"^(2)` |
| Answer» Correct Answer - B | |
| 76. |
The radius of a circle is increasing uniformly at the rate of 0.3 centimetre per second. At what rate is the area increasing when the radius is 10 cm ? (Take `pi = 3.14`) |
| Answer» Correct Answer - `(dA)/(dt) = 18.84 cm^(2)//s` | |
| 77. |
The surface of a spharical balloon is increase of its volume when its radius is 6 cm. |
| Answer» Correct Answer - `6 cm^(3)//sec` | |
| 78. |
The length of a rectangle is decreasing at a rate of 3 cm/sec and breadth is increasing at a rate of 4 cm/sec. Find the rate of change of its (a) perimeter (b) area, when the length and breadth of rectangle are 7 cm and 8 cm respectively. |
| Answer» Correct Answer - `(a) 2 cm//sec (b) 4 cm^(2)//sec`,Increase | |
| 79. |
The side of a square sheet of metal is increasing at 3 centimetres per minute. At what rate is the area increasing when the side is 10 cm long ? |
| Answer» Correct Answer - `(dA)/(dt) = 60 cm^(2)//min` | |
| 80. |
The radius of a cylinder is increasing at the rate of `3ms^(-1)` and its altitude is decreasing at the rate of `4ms^(-1)`. The rate of change of volume when radius is `4m` and altitude is 6m isA. `144 m^(3)/sec`B. `64 pi m^(3)//sec`C. `80 pi m^(3)//sec`D. `-80 pi m^(3)//sec` |
| Answer» Correct Answer - C | |
| 81. |
If a particle moving in a straight line and its distance x cms from a fixed point O on the line is given by `x=sqrt(1+t^(2))` cms, then acceleration of the particle at t sec. isA. `(1)/(x^(2))" cm/sec"^(2)`B. `(-1)/(x^(2))" cm/sec"^(2)`C. `(1)/(x^(3))" cm/sec"^(2)`D. `(-1)/(x^(3))" cm/sec"^(2)` |
| Answer» Correct Answer - C | |
| 82. |
Verify LMVT for the following function `f(x)=x^2-3x-1, x in[-11/7, 13/7]`A. `(1)/(7)`B. `(-1)/(7)`C. `(2)/(7)`D. `(-2)/(7)` |
| Answer» Correct Answer - A | |
| 83. |
Verify `LMVT` for the following functions: `f(x)=x(2-x), x in [0, 1]`A. `(1)/(4)`B. `(2)/(3)`C. `(1)/(2)`D. `(1)/(3)` |
| Answer» Correct Answer - C | |
| 84. |
Find c if LMVT is applicable for `f(x)=x(x-1)(x-2), x in [0, 1 /2]`A. `1+(sqrt(21))/(6)`B. `1-(sqrt(21))/(6)`C. `1pm (sqrt(21))/(6)`D. `1-(sqrt(7))/(2)` |
| Answer» Correct Answer - B | |
| 85. |
If LMVT is applicable for `f(x)=x(x+4)^(2), x in [0,4],` then c=A. `(-4-2sqrt(3))/(3)`B. `(-4+2sqrt(3))/(3)`C. `(-8+4sqrt(13))/(3)`D. `(-8-4sqrt(13))/(3)` |
| Answer» Correct Answer - C | |
| 86. |
The slope of the normal to the curve `y=2x^2+3`sin x at `x = 0`is(A) 3 (B) `1/3` (C)`-3` (D) `-1/3`A. 3B. `1/3`C. `-3`D. `-1/3` |
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Answer» Correct Answer - D Equation of curve `y = 2x^(2) + 3 sin x` ` rArr (dy)/(dx) = 4x + 3 cos x` slope of tangent at `x=0 is 0+3=3` ` rArr "at "x=0," slope of normal "=-1/3` |
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| 87. |
The line `y = x + 1`is a tangent to the curve `y^2=4x`at the point(A) `(1, 2)` (B)`(2, 1)` (C) `(1, 2)` (D) `( 1, 2)`A. `(1, 2)`B. `(2, 1)`C. `(1, -2)`D. `(-1, 2)` |
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Answer» Correct Answer - A Equation of curve `y^(2) = 4x` ….(1) ` rArr 2y(dy)/(dx) = 4 rArr (dy)/(dx) = 2/y` Slope of tangent at point `(x, y) = 2/y` Equation of given tangent ` y = x + 1` Its slope = 1 ` :. 2/y = 1 rArr y= 2` put y = 2 in equation (1) `2^(2) = 4x rArr x = 1` ` :.` Point of contact = (1, 2) |
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| 88. |
Find the approximate change in the surface area of acube of side x metres caused by decreasing the side by 1%. |
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Answer» Total surface area of a cube of edge x is `S = 6x^(2)` `rArr (dS)/(dx) = 12x` Change in the surface area ` Delta S (dS)/(dx) * Delta x = 12x * Delta x` ` = 12x (-1/100 xx x)` ` = 0.12x^(2) m^(2)` . |
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| 89. |
Find the approximate change in the volume V of acube of side x metres caused by increasing the side by 1%. |
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Answer» Given that `(dx)/x xx 100 = 1%` Let volume of cube be V. `:. V=x^(3)` `rArr (dV0/(dx) = 3x^(2)` ` rArr (dV)/V = (3*x^(2)dx)/V = (3x^(2)dx)/x^(3) = 3 * (dx)/x` ` rArr (dV)/V xx 100 = 3 * (dx)/x xx 100` ` rArr" Percentage increase in volume " 3 xx 1% = 3%` . |
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| 90. |
Find the equation of tangent to the curve `x=sin3t , y=cos2t `at `t=pi/4` |
| Answer» Correct Answer - `4x - 3 sqrt2y - 2 sqrt2 = 0` | |
| 91. |
Find the maximum or minium values, if any, without using derivatives, of the functions: `-(x -3)^(2) + 9` |
| Answer» Correct Answer - max. value =9 | |
| 92. |
Find the maximum or minium values, if any, without using derivatives, of the functions: `-|x +4| + 6` |
| Answer» Correct Answer - max. value = 6 | |
| 93. |
Find the maximum or minium values, if any, without using derivatives, of the functions: `(5x -1)^(2) + 4` |
| Answer» Correct Answer - min. value = 4 | |
| 94. |
Find the maximum or minium values, if any, without using derivatives, of the functions: `|sin 4x + 3|` |
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Answer» Correct Answer - max. value = 4, min. value = 2 `-1 le sin 4x le 1 rArr (-1 + 3) le (sin 4x + 3) le (1 + 3)` |
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| 95. |
Amongst all pairs of positive number with sum 24, find those whose product is maximum |
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Answer» Let the number be x and `(24 -x)` Let `P = x (24 -x) = (24 x - x^(2))` then, `(dP)/(dx) = (24 - 2x) and (d^(2)P)/(dx^(2)) = -2` Now, `(dP)/(dx) = 0 rArr (24 - 2x) = 0 rArr x = 12` Thus, `{(d^(2)P)/(dx^(2))}_(x = 12) = -2 lt 0` `:. x = 12`, is point of maximum Hence, the required number are 12 and 12 |
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| 96. |
Show that `s in^p theta cos^q theta`attains a maximum, when`theta=tan^(-1)sqrt(p/q)`. |
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Answer» Let `y = sin^(p) theta cos^(q) theta` Then, `(dy)/(dtheta) = p sin^(p -1)theta cos^(q +1) cos^(q -1)theta sin^(p +1) theta` `= (sin^(p-1)theta cos^(q-1)theta) (p cos^(2) theta - q sin ^(2) theta)` Now, for maxima or minima, we have `(dy)/(d theta) = 0` But `(dy)/(d theta) = 0 rArr sin^( p-1) theta = 0 or cos^(q -1) theta = 0 or p cos^(2) theta - q sin^(2) theta = 0` `rArr theta = 0 or theta = (pi)/(2) or theta = tan^(-1) sqrt((p)/(q))` Moreover, we may write `(dy)/(d theta) = (y (p cos^(2) theta - q sin^(2) theta))/(sin theta cos theta) = y (p cot theta - q tan theta)` `:. (d^(2)y)/(d theta^(2)) = y (-p " cosec"^(2) theta - q sec^(2) theta) + (p cot theta - q tan theta) .(dy)/(d theta)` Thus, at `theta = tan^(-1) sqrt((p)/(q))`, we have `(dy)/(d theta) = 0` and therefore, `[(d^(2) y)/(d theta^(2)) " at " theta = tan^(-1) sqrt((p)/(q))] = sin^(p) theta cos^(q) theta (-p " cosec"^(2) theta - q sec^(2) theta) lt 0` Hence, y is maximum when `theta = tan^(-1) sqrt((p)/(q))` |
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| 97. |
The area S of a triangle is calculated by measuring the sides b and c, and `angleA`. If there be an error `deltaA` in the measurement of `angleA`, show that the relative error in area is given by `(deltaS)/(S) = cot A.deltaA` |
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Answer» `S = (1)/(2) bc sin A rArr (dS)/(dA) = (1)/(2) bc cos A rArr (dS)/(dA).deltaA = (1)/(2) bc cos A .deltaA` `rArr delta S = (1)/(2) bc cos A.deltaA rArr deltaS = (1)/(2) bc sin A.(cot A).deltaA` `rArr deltaS = S.(cotA).deltaA` Hence, `(deltaS)/(S) = (cot A) .deltaA` |
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| 98. |
If `f(x)=(6)/(x^(2)+2)`, thenA. f has maximum value 0B. f has minimum value 3C. f has minima at x=3D. f has maxia at x=0 |
| Answer» Correct Answer - D | |
| 99. |
If `f(x)=(6)/(x^(2)+2)`, thenA. f has maxima at x=3B. f has minima at x=0C. f has maximum value 3D. f has minimum value 0 |
| Answer» Correct Answer - C | |
| 100. |
If `f(x)=x^(2)+(16)/(x^(2))`, thenA. f has maximum value 6B. f has minimum value 8C. f has maxima `at x=pm2`D. f has minima at `x=pm4` |
| Answer» Correct Answer - B | |