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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.
| 401. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `y = x^(3) - 2x + 7 " at " (1, 6)` |
| Answer» Correct Answer - `x - y + 5 = 0, x + y - 7 = 0` | |
| 402. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `y = cot^(2) x - 2 cot x + 2 " at " x = (pi)/(4)` |
| Answer» Correct Answer - `y = 1, x = (pi)/(4)` | |
| 403. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `16x^(2) + 9y^(2) = 144 " at " (2, y_(1)), " where " y_(1) gt 0` |
| Answer» Correct Answer - `8x + 3 sqrt5 y - 36 = 0, 9 sqrt5x - 24y + 14 sqrt5 = 0` | |
| 404. |
If V denotes the volume and S is the surface area of a sphere. If radius of sphere is 2 cm, then the rate of change of V w.r.t. S isA. 2 cmB. 1 cmC. 2 cm/secD. 1 cm/sec |
| Answer» Correct Answer - B | |
| 405. |
Acute angle between two curve `x^(2)+y^(2)=a^(2)sqrt2` and `x^(2)-y^(2)=a^(2)` isA. `(pi)/(6)`B. `(pi)/(3)`C. `(pi)/(4)`D. none of these |
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Answer» Correct Answer - C `x^(2)+y^(2)=a^(2)sqrt2 and x^(2)-y^(2)=a^(2)` `therefore" "(dy)/(dx)=(-x)/(y),(dy)/(dx)=(x)/(y)` Angle between curves is given by `theta=|tan^(-1).((x)/(y)+(x)/(y))/(1-(x^(2))/(y^(2)))|` `=|tan^(-1).(2xy)/(y^(2)-x^(2))|` Squaring and subtracting the equations of given curves, `4x^(2)y^(2)=a^(4)` `therefore" "2xy= pm a^(2)` `therefore" "theta=|tan^(-1).(a^(2))/(a^(2))|` `=tan^(-1)(1)` `=(pi)/(4)` |
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| 406. |
The rate of change of volume of a sphere is equalto the rate of change of its radius, then its radius is equal to(a) 1 unit(b)units(c)unit(d)unitA. 1B. 2C. 0.5D. none of these |
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Answer» Correct Answer - B `V=(4)/(3)pir^(3)` `therefore" "(dV)/(dt)=(4)/(3)pi.3r^(2)(dr)/(dt)` Given `(dV)/(dt)=(dr)/(dt)` `therefore" "r^(2)=(1)/(4pi) rArr r=(1)/(2sqrtpi)` `therefore" "k=2` |
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| 407. |
The length x of a rectangle is decreasing at therate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute.When x = 8cm and y = 6cm, find the rates of change of (a) the perimeter, and(b) the area of the rectangle |
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Answer» Given that `(dx)/(dt) = -5 cm`/minute and `(dy)/(dt)` = 4 cm/minute. (i) Let P be the perimeter of the rectangle at any instant. Then, `P = 2(x +y) rArr (dP)/(dt) = 2 ((dx)/(dt) + (dy)/(dt)) = 2 (-5 + 4)` cm/minute Hence, the perimeter of the rectangle is decreasing at the rate of 2 cm/minute. (ii) Let A the area of the rectangle at any instant. Then, `A = x.y rArr (dA)/(dt) = (dx)/(dt) .y + x.(dy)/(dt)` `= [(-5) xx 6 + 8 xx4] cm^(2)`/minute `= 2 cm^(2)`/minute Hence the area of the rectangle is increasing at the rate of `2 cm^(2)`/minute. |
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| 408. |
The length x of a rectangle is increasing at the rate of 3 cm/sec. and the width y is increasing at the rate of 2 cm/sec. If x=10 cm and y=6 cm, then the rate of change of its area isA. `2 cm^(2)//sec`B. `-2 cm^(2)//sec`C. `38 cm^(2)//sec`D. `-38 cm^(2)//sec` |
| Answer» Correct Answer - C | |
| 409. |
Which of the following pair(s) of family is/are orthogonl? where c and k are arbitrary constant.A. `16x^(2)+y^(2)=c and y^(16)=kx`B. `y=x+ce^(-x) and x+2=y+ke^(-y)`C. `y=cx^(2) and x^(2)+2y^(2)=k`D. `x^(2)-y^(2)=c and xy =k` |
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Answer» Correct Answer - A::B::C::D (a) `16x^(2)+y^(2)=c rArr(dy)/(dx)=m_(1)=-(16x)/(y)` `y^(16)=kx" "rArr " "(dy)/(dx)=m_(2)=(k)/(16y^(15))` `m_(1)m_(2)=-(16x)/(y).(k)/(16y^(15))` `=-(x)/(y^(16)).k` `=-(x)/(y^(16)).(y^(16))/(x)=-1` `rArr" Curves are orthogonal."` (b) `y=x+ce^(-x)` `rArr" "(dy)/(dx)=1-ce^(-x)=1-(y-x)=-(y-x-1)` `x+2=y+ke^(-y)` `rArr" "(dy)/(dx)-k(dy)/(dx)e^(-y)=1` `rArr" "(dy)/(dx)[1-ke^(-y)]=1` `"or "[1-(x+2-y)](dy)/(dx)=1` `rArr" "(dy)/(dx)=m_(2)=(1)/(y-x-1)` `rArr: "m_(1)m_(2)=-1` `rArr" Curves are orthogonal."` (c) `y=cx^(2)` `rArr" "(dy)/(dx)=2cx=2x(y)/(x^(2))=(2y)/(x)=m_(1)` `"and "x^(2)+2y^(2)=k` `rArr" "2x+4y(dy)/(dx)=0` `rArr" "(dy)/(dx)=-(x)/(2y)=m_(2)` `rArr" "m_(1)m_(2)=-1` `rArr" Curves are orthogonal."` (c) `x^(2)-y^(2)=c rArr (dy)/(dx)=(x)/(y)=m_(1)` `xy=k rArr (dy)/(dx)=-(y)/(x)=m_(2)` `therefore" "m_(1)m_(2)=-1` `rArr" Curves are orthogonal."` |
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| 410. |
The eccentricity of the ellipse `3x^2+4y^2=12` is decreasing at the rate of 0.1 per sec.The time at which it will coincide with auxiliary circle is:A. 2 secondsB. 3 secondsC. 5 secondsD. 6 seconds |
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Answer» Correct Answer - C Eccentricity of ellipse `=(1)/(2)` `(de)/(dt)=-0.1" (given)"` Eccentricity of auxiliary circle = 0 `rArr" "int_(1//2)^(0)de=-0.1 int_(0)^(T)dt` T is time at which it will touch the auxiliary circle. `rArr" "-(1)/(2)=-0.1 (T-0)" "thereforeT=5 " seconds."` |
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| 411. |
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is:A. `10pi`B. `12pi`C. `8pi`D. `11pi` |
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Answer» Correct Answer - B Let r be the radius and A be the area of the circle. `:. A = pi r^(2)` `rArr (dA)/(dr) = 2 pi r` at r = 6 cm ` (dA)/(dr) = 2pi xx 6 = 12 pi cm^(2)`/sec |
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| 412. |
The rate of change of `sqrt(x^2+16)` with respect to `x/(x-1)` at `x=3` isA. 1B. `(11)/(5)`C. `-(12)/(5)`D. `-3` |
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Answer» Correct Answer - C `u=sqrt(x^(2)+16)` `therefore" "(du)/(dx)=(2x)/(2sqrt(x^(2)+16))=(x)/(sqrt(x^(2)+16))` `v=(x)/(x-1)` `rArr" "(dv)/(dx)=(-1)/((x-1)^(2))` `therefore" "(du)/(dv)=(du//dx)/(dv//dx)=(-12)/(5)` |
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| 413. |
Find the rate of change of the area of a circle with respect to its radius r when(a) `r = 3 c m` (b) `r = 4 c m` |
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Answer» Let radius of circle be r and area be A. Area A = `pi r^(2)` `rArr (dA)/(dr) = pi (2r)` (a) at r = 3 cm ` (dA)/(dr) = pi xx 2 xx 3 = 6 pi cm^(2)//sec` (b) at r = 4 cm ` (dA)/(dr) = pi xx 2 xx 4 = 8 pi cm^(2) //sec` |
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| 414. |
The total revenue of selling of x units of a product is represented by `R (x) = 2x^(2)+x+5`.Find its marginal revenue for 5 units. |
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Answer» Correct Answer - Rs. 21 For x units, marginal revenue =`(dR)/(dx)` |
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