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InterviewSolution
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. |
The sum of the intercepts made on the axes of coordinates by any tangent to the curve ` sqrt(x)+sqrt(y)=sqrt(a)` is equal toA. 2aB. aC. `(a)/(2)`D. none of these |
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Answer» Correct Answer - B Let P`(x_(1),y_(1))` be a point on the curve ` sqrt(x)+sqrt(y)=sqrt(a). ` Then, ` sqrt(x_(1))+sqrt(y_(1))=sqrt(a) " "...(i)` Now, ` sqrt(x)+sqrt(y)=sqrt(a)` ` rArr (1)/(2sqrt(x))+(1)/(2sqrt(y)) (dy)/(dx)=0 rArr (dy)/(dx) = -(sqrt(y))/(sqrt(x)) rArr ((dy)/(dx))_(P)=-sqrt((y_(1))/(x_(1)))` The equation of the tangent to the given curve at point `P(x_(1),y_(1))` is ` y-y_(1)=-sqrt((y_(1))/(x_(1)))(x-x_(1)) ` ` rArr (x)/(sqrt(x_(1)))+(y)/(sqrt(y_(1))) = sqrt(x_(1))+ sqrt(y_(1)) rArr (x)/(sqrt(x_(1))) + (y)/(sqrt(y_(1)))=sqrt(a) ` [Using (i)] This cuts the coordinate axes at `A(sqrt(ax_(1)),0)` and `B(0, sqrt(ay_(1)))` ` therefore OA+OB= sqrt(ax_(1))+sqrt(ay_(1))=sqrt(a)(sqrt(x_(1))+sqrt(y_(1)))= sqrt(a) xx sqrt(a)=a. ` |
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| 52. |
If the tangent at each point of the curve `y=(2)/(3) x^(3)-2ax^(2)+2x+5 ` makes an acute angle with the positive direction of x-axis, thenA. ` a ge 1 `B. `-1 le a le 1 `C. ` a le -1 `D. none of these |
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Answer» Correct Answer - B It is given that the tangent at each point of the curve ` y=(2)/(3)x^(3)-2ax^(2)+2x+5 ` makes an acute angle with the positive direction of x-axis. ` therefore (dy)/(dx) ge 0` for all x ` rArr 2x^(2)-4ax+2 ge 0 ` for all x `rArr x^(2)-2ax+1 ge 0 ` for all x ` rArr 4a^(2)-4 le 0 rArr a^(2)-1 le 0 rArr -1 le a le 1. ` |
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| 53. |
The equation of the normal to the curve `y=x+sin x cos x " at " x=(pi)/(2),` isA. `x=2`B. `x=pi`C. `x+pi=0`D. `2x=pi` |
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Answer» Correct Answer - D When `x=(pi)/(2)`, we have `y=(pi)/(2) + "sin" (pi)/(2)"cos"(pi)/(2)=(pi)/(2)` So, the coordinates of the point are `(pi//2,pi//2)` Now, `y=x+sin x cos x ` `rArr (dy)/(dx)=1+cos^(2)x-sin^(2)x rArr ((dy)/(dx))_(x=(n)/(2))=1+0-1=0` This means that the tangent at `(pi//2,pi//2)` is parallel to x-axis. So, the normal is parallel to y-axis and hence its equation is `x=(pi)/(2)`. |
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| 54. |
A tangent to the curve `y= int_(0)^(x)|x|dt,` which is parallel to the line `y=x,` cuts off an intercept from the y-axis is equal toA. 1B. `(-1)/(2),(1)/(2)`C. `(1)/(2),1`D. -1 |
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Answer» Correct Answer - B We have, `y= int_(0)^(x)|x|dt " "...(i)` Differentiating w.r.t. x, we get ` (dy)/(dx)=|x| ` Let `P (x_(1),y_(1))` be a point on the curve (i) such that the tangent at P is parallel to the line `y=x.` Then, Slope of the tangent at `(x_(1),y_(1))=1` ` rArr ((dy)/(dx))_((x_(1)","y_(1)))=1 rArr |x_(1)|=1 rArr x_(1)= pm 1 ` Now, `y=int_(0)^(x)|t|dt rArr y={(int_(0)^(x)tdt=(x^(2))/(2) ",", "if " x ge 0),(int_(0)^(x) -tdt=-(x^(2))/(2) ",", "if " x lt 0):} ` ` therefore x_(1)=1 rArr y_(1)=(1)/(2) " "["Putting " x_(1)=1 " in " y_(1)=(x_(1)^(2))/(2)] ` and, ` x_(1)=-1 rArr y_(1)=-(1)/(2) " "[" Putting " x_(1)=-1 " in " y_(1)=-(x_(1)^(2))/(2)] ` Thus, the two points on the curve are `(1,1//2)` and `(-1,-1//2)` The equations of the tangents at these two points are `y-(1)/(2)=1(x-1) " and " y+(1)/(2)=1(x+1) ` respectively. ` rArr 2x-2y-1=0 " and " 2x-2y+1=0 ` Clearly, these tangents cut off intercepts ` -(1)/(2) " and " (1)/(2) ` respectively on y-axis. |
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| 55. |
The equation of the tangent to the curve `y""=""x""+4/(x^2)`, that is parallel to thex-axis, is(1) `y""=""1`(2) `y""=""2`(3) `y""=""3`(4) `y""=""0`A. `y=2`B. `y=3`C. `y=0`D. `y=1` |
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Answer» Correct Answer - B Let `(x_(1),y_(1))` be a point on the curve `y=x+(4)/(x^(2))` where tangent is parallel to x-axis. Then, `((dy)/(dx))_((x_(1)","y_(1)))=0 rArr 1-(8)/(x_(1)^(3)) =0 rArr x_(1)=2 ` ` therefore y_(1)=x_(1)+(4)/(x_(1)^(2))rArr y_(1)=3 ` The equation of the required tangent is `y-3=0(x-2) " or, " y=3 ` |
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| 56. |
Show that the curves `x y=a^2a n dx^2+y^2=2a^2`touch each other |
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Answer» curves are `xy= a^2` `x^2 + y^2 = 2a^2` `x^2 + (a^2/x)^2 = 2a^2` `x^2 + a^4/x^2 = 2a^2` `x^4 + a^4 = 2a^2x^2` `x^4 + a^4 = 2a^2x^2` `x^4 + a^4 - 2a^2x^2 = 0` `(x^2 - a^2)^2 = 0` `x^2 = a^2` `x= +- a` now,`xy= a^2` `xdy/dx + y(1) = 0` `dy/dx = -y/x = -a/a= m_1` now, `x^2 + y^T2 = 2a^2` `2x + 2y dy/dx= 0` `dy/dx= -x/y=-a/a= -1= m_2` as,`m_1=m_2` so, curves are touching each other hence proved |
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| 57. |
The point(s) on thecurve `y^3+ 3x^2=12 y`where the tangent isvertical, is(are) ??`(+-4/(sqrt(3)), -2)`(b) `(+- sqrt((11)/3, ) 1)``(0, 0)`(d) `(+-4/(sqrt(3)), 2)`A. `(pm(4)/(sqrt(3)),-2)`B. `(pmsqrt((pi)/(3)),1)`C. (0, 0)D. `(pm(4)/(sqrt(3)),2)` |
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Answer» Correct Answer - D We have, `y^(3)+3x^(2)=12y " …(i)" ` `rArr 3y^(2) (dy)/(dx)+6x=12(dy)/(dx) " " `[Diff. w.r.t. x] `rArr 3(y^(2)-4) (dy)/(dx)= -6x rArr (dy)/(dx) = -(2x)/(y^(2)-4)` At point(s) where the tangent(s) is (are) vertical, `(dy)/(dx)` is not defined . `therefore y^(2) -4=0 rArr y= pm 2`. From (i), we find that `y=2 rArr x=pm(4)/(sqrt(3))` and, `y = -2 rArr x^(2) =(-16)/(3)`, which is not possible. Hence, the required points are `(pm 4//sqrt(3), 2)` |
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| 58. |
The curve given by `x + y = e^(xy)` has a tangent parallel to the y-axis at the pointA. (0, 1)B. (1, 0)C. (1, 1)D. none of these |
| Answer» Correct Answer - B | |
| 59. |
If the line ` ax+by+c=0 ` is a tangent to the curve `xy=9,` thenA. ` a gt 0, b gt 0 `B. ` a gt 0, b lt 0 `C. ` a lt 0, b gt 0 `D. ` a lt 0, b lt 0 ` |
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Answer» Correct Answer - A::D We have, ` xy=9 rArr y=(9)/(x) rArr (dy)/(dx) =-(9)/(x^(2)) " "...(i) ` Let ` ax+by+c=0 ` be a tangent to the curve xy=9 at ` (x_(1),y_(1)). ` Then, ` ((dy)/(dx))_((x_(1)","y_(1))) =" Slope of the line " ax+by+c=0 ` ` rArr -(9)/(x_(1)^(2))=(-a)/(b) ` ` rArr (x_(1)^(2))/(9) =(b)/(a) ` ` rArr (b)/(a) gt 0 ` `rArr` a and b are the same sign. `rArr (a gt 0 " and " b gt 0 )" or, "(a lt 0 " and " b lt 0 ) ` |
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| 60. |
The intercepts made by the tangent to the curve `y=int_(0)^(x) |t| dt,` which is parallel to the line `y=2x` on y-axis are equal toA. 1, -1B. `-2, 2`C. 3D. -3 |
| Answer» Correct Answer - B | |
| 61. |
The angle between the tangents to the curve `y=x^(2)-5x+6 ` at the point (2, 0) and (3, 0), isA. `pi//3`B. `pi//2`C. `pi//6`D. `pi//4` |
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Answer» Correct Answer - B The equation of the curve is `y=x^(2)-5x+6` `therefore (dy)/(dx) = 2x-5` Let `m_(1) and m_(2)` be the slopes of the tangents at P(2, 0) and Q(3, 0) to the given curve. Then, `m_(1)=((dy)/(dx))_(p)=2xx2-5= -1, m_(2)=((dy)/(dx))_(Q)=2xx3 -5=1` Clearly, `m_(1)m_(2)= -1`. So, required angle is `(pi)/(2)` |
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| 62. |
The angle between the tangents to the curve `y^(2)=2ax` at the point where `x=(a)/(2)`, isA. `pi//6`B. `pi//4`C. `pi//3`D. `pi//2` |
| Answer» Correct Answer - D | |
| 63. |
The angle between the tangents to the curve `y=x^(2)-5x+6` at the point (2, 0) and (3, 0), isA. `pi//3`B. `pi//4`C. `pi//2`D. `pi//6` |
| Answer» Correct Answer - C | |
| 64. |
The two tangents to the curve `ax^(2)+2h x y+by^(2) = 1, a gt 0` at the points where it crosses x-axis, areA. parallelB. perpendicularC. inclined at an angle of `pi//4`D. none of these |
| Answer» Correct Answer - A | |
| 65. |
The number of points on the curve `y=x^(3)-2x^(2)+x-2` where tangents are prarllel to x-axis, is |
| Answer» Correct Answer - C | |
| 66. |
Find the point ofintersection of the tangents drawn to the curve `x^2y=1-y`at the points whereit is intersected by the curve `x y=1-ydot`A. `(0,-1)`B. `(1,1)`C. `(0,1)`D. none of these |
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Answer» Correct Answer - C Solving the two equations, we get ` x^(2)y =xy rArr xy(x-1)=0 rArr x=0, y=0, x=1 ` Since y = 0 does not satisfy the two equations. So, we neglect it. Putting x = 0 in the either equation, we get x = 1. Now, putting x=1 in one of the two equations we obtain ` y= 1//2. ` Thus, the two curves intersect at (0,1) and `(1,1//2).` Now, ` x^(2)y=1-y rArr x^(2)(dy)/(dx) + 2xy = -(dy)/(dx) rArr (dy)/(dx) =-(2xy)/(x^(2)+1) ` ` rArr ((dy)/(dx))_((0","1)) =0 " and " ((dy)/(dx))_((1"," 1//2))=-(1)/(2)` The equations of the required tangents are ` y-1 =0(x-0) " and " y-(1)/(2)=(-1)/(2)(x-1) ` ` rArr y=1 " and " x+2y-2=0 ` These two tangents intersect at (0,1) |
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| 67. |
Find the points on the curve `y=x^3-2x^2-x`at which the tangent lines are parallel to the line `y=3x-2` |
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Answer» `y=mx+c,`slope of this line=m `:. y=3x-2`, slope of line `y=3x-2 `is 3 slope of tangent of any curve is `dy/dx` at that point `y= x^3 -2x^2 - x` `dy/dx = d/dx(x^3 - 2x^2 - x) ` `= 3x^2 - 4x-1` `:. dy/dx = 3` `=> 3x^2 - 4x - 1 = 3` `=> 3x^2 + 2x - 6x - 4= 0` `:. x(3x+2) - 2(3x+2) => (x-2)(3x+2) = 0` `:. x=2 or -2/3` for x=2`=> y= (2)^3 - 2(2)^2 - 2 = 8-8-2= -2` for x=-2/3 `=> y= (-2/3)^3 - 2(2/3)^2- (-2/3) ` `= -8/27 - 8/9 + 2/3` `= (-8+24+18)/27 = -14/27` `:. ` points are `(2,-2); (-2/3.-14/27)` Answer |
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| 68. |
Show that the line `x/a+y/b=1`touches the curve `y=b e^(-x/a)`at the point where it crosses the y-axis. |
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Answer» `y= be^(-x/a)` `dy/dx = be^(-x/a) (-1/a) ` `dy/dx = -b/a e^(-x/a)` now, `x/a + y/b = 1` `ax + ay - ab = 0` slope=`-b/a` now, `-b/ae^(-x/a) = -b/a` `e^(x/a) = 1` `x=0` `y=b` Answer |
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| 69. |
The curve `y-e^(xy)+x=0` has a vertical tangent at the point :A. (1, 1)B. at no pointC. (0, 1)D. (1, 0) |
| Answer» Correct Answer - D | |
| 70. |
If m denotes the slope of the normal to the curve `y= -3 log(9+x^(2))` at the point `x ne 0`, then,A. `n in [-1,1]`B. `m in R-(-1,1)`C. `n in R -[-1,1]`D. `m in (-1,1)` |
| Answer» Correct Answer - B | |
| 71. |
The number of possible tangents which can be drawn to the curve `4x^2-9y^2=36 ,`which are perpendicular to the straight line `5x+2y-10=0`, iszero (b)1 (c) 2(d) 4A. `5(y-3)=2(x-(sqrt(117))/(2)) `B. ` 2x-5y+10-2sqrt(18)=0 `C. ` 2x-5y-10-2sqrt(18)=0 `D. none of these |
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Answer» Correct Answer - D We have, ` 4x^(2)-9y^(2)=36 rArr 8x-18y(dy)/(dx)=0 rArr (dy)/(dx)=(4x)/(9y) ` ` therefore " Slope of the tangent " =(4x)/(9y) ` For this tangent to be perpendicular to the straight line ` 5x+2y-10=0, ` we must have `(4x)/(9y)xx (-(5)/(2))=-1 rArr y=(10x)/(9). ` Putting this value of y in `4x^(2)-9y^(2)=36, ` we get `-64x^(2)=324, ` which does not have real roots. Hence, at no point on the given curve can the tangent be perpendicular to the given line. |
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| 72. |
Find the equations of the tangents drawn to the curve `y^2-2x^2-4y+8=0.` from point `(1,2)` |
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Answer» y-2=slope(x-1) `4-2-8+8!=0` Lets say y=mx+c `(mx+c)^2-2x^2-4y+8=0` `(m^2-2)x^2+(2mc-4m)x+(c^2-4c+8)=0` `D=0=b^2-4ac` `(2mc-4m)^2-4(m^2-2)(c^2-4c+8)=0` `c=2pmsqrt(2m^2-4` `y=mx+2pmsqrt(2m^2-4` `1=m+2pmsqrt(2m^2-4` `m=pm2` `c=0,m=2` `c=4,m=-2` `y=2x` `y=-2x+4`. |
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| 73. |
Prove that the tangents to the curve `y=x^2-5+6`at the points `(2,0)a n d(3,0)`are at right angles. |
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Answer» `y= x^2 - 5x + 6` at point `(2,0)` `dy/dx = 2x- 5` at (2,0) , `dy/dx = -1` at (3,0)`dy/dx= 1` `(m_1)(m_2) = -1` so, they are at right angles hence proved |
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| 74. |
Find the equation of the tangent line to the curve `y=sqrt(5x-3)-2`which is parallel to the line `4x-2y+3=0` |
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Answer» slope = `m= -4/(-2) = 2` `dy/dx|_p = 2` `=> 1/(2sqrt(5x-3)) xx 5 = 2` `x = 73/80` putting it in the curve eqn `y= -3/4 p(75/80, -3/4)` `m=2` eqn of tangent `y-(-3/4) = 2(x-73/80)` Answer |
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| 75. |
The slope of the tangent to the curve `x=t^2+3t-8,``y=2t^2-2t-5`at the point `(2, 1)`is(A) `(22)/7` (B) `6/7` (C) `7/6` (D) `(-6)/7`A. `(22)/(7)`B. `(6)/(7)`C. -6D. none of these |
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Answer» Correct Answer - B For the point (2, -1) on the curve `x=t^(2)+3t-8,y=2t^(2)- 2t -5` we have `t^(2)+3t-8=2 and 2t^(2)-2t-5=-1` `rArr t^(2)+3t-10=0 and 2t^(2)-2t-4=0` `rArr (t+5)(t-2)=0 and (t-2)(t+1)=0` `rArr t=2` Now, `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(4t-2)/(2t+3) rArr ((dy)/dx)_(t=2)=(4xx2-2)/(2xx2+3)=(6)/(7)` |
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| 76. |
The fixed point P on the curve `y=x^(2)-4x+5` such that the tangent at P is perpendicular to the line `x+2y-7=0` is given byA. (3, 2)B. (1, 2)C. (2, 1)D. none of these |
| Answer» Correct Answer - A | |
| 77. |
If `y=4x-5` is a tangent to the curve `y^(2)=px^(3) +q` at (2, 3), thenA. p = 2, q = -7B. p = -2, q = 7C. p = -2, q = -7D. p = 2, q = 7 |
| Answer» Correct Answer - A | |
| 78. |
The slope of the tangent to the curve `x=t^(2)+3t-8,y=2t^(2) -2t -5` at the point (2, -1), isA. `22//7`B. `6//7`C. -6D. none of these |
| Answer» Correct Answer - B | |
| 79. |
If m be the slope of the tangent to the curve `e^(2y) = 1+4x^(2)`, thenA. `m lt 1`B. `|m| le 1`C. `|m| ge 1`D. none of these |
| Answer» Correct Answer - B | |
| 80. |
The point on the curve `sqrt(x) + sqrt(y) = sqrt(a)`, the normal at which is parallel to the x-axis, isA. (0, 0)B. (0, a)C. (a, 0)D. (a, a) |
| Answer» Correct Answer - B | |
| 81. |
The length of the Sub tangent at `(2,2)` to the curve `x^5 = 2y^4` isA. `5//2`B. `8//5`C. `2//5`D. `5//8` |
| Answer» Correct Answer - B | |
| 82. |
The normal to the curve `x=a(cos theta + theta sin theta), y=a(sin theta - theta cos theta)` at any `theta` is such thatA. it makes a constant angle with x-axisB. it passes through the originC. it is at a constant distance from the originD. none of these |
| Answer» Correct Answer - C | |
| 83. |
The area bounded by the coordinate axes and normal to the curve `y=log_(e)x` at the point P( 1 , 0), isA. 1 sq. unitB. 2 sq. unitsC. `(1)/(2)` sq. unitD. none of these |
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Answer» Correct Answer - C We have, `y=log_(e)x rArr (dy)/(dx) = (1)/(x) rArr ((dy)/(dx)) _(p) = 1` The equation of the normal at (1, 0) is `y-0= -1 (x-1) rArr x+y=1` Clearly, it makes a triangle of area `(1)/(2)` sq. unit with the coordinate axes. |
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| 84. |
The equation of the tangent to the curve `y=1-e^(x//2)` at the tangent to the curve `y=1-e^(x//2)` at the point of intersection with the y-axis, isA. `x+2y=0`B. `2x+y=0`C. `x-y=2`D. none of these |
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Answer» Correct Answer - A We have, `y= 1-e^(x//2) rArr (dy)/(dx) = -(1)/(2) e^(x//2)` The curves`y=1-e^(x//2)` meets y-axis at (0, 0). `therefore((dy)/(dx))_((0","0))= -(1)/(2)` The equation of the tangent at (0, 0) is `y-0 = -(1)/(2) (x-0) rArr x+2y =0` |
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| 85. |
The normal to the curve `x^(2) +2xy-3y^(2)=0,` at (1, 1):A. meets the curve again in the third quadrant.B. Meets the curve again the fourth quadrant .C. does not meet the curve again.D. meets the curve again in the second quadrant. |
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Answer» Correct Answer - B The equation of the curve is `x^(2) + 2xy-3y^(2)=0 " " ` …(i) Differentiating with respect to x, we get `2x+2y+2x(dy)/(dx)-6y(dy)/(dx)=0rArr (dy)/(dx)=(x+y)/(3y-x) rArr ((dy)/(dx))_((1","1))=1` The equation of the normal at (1, 1) is `y-1= -1(x-1) or, (x+y)=2 " " ` ...(ii) The x-coordinates of the points of intersection of (i) and (ii) are given by `x^(2)+2x(2-x)-3(2-x)^(2)=0` `rArr 4x^(2)-16x+12=0` `rArr x^(2)-4x+3=0 rArr (x-3) (x-1)=0 rArr x=1, 3 ` Now, `x=1 and x+y=2 rArr y=1` `x=3 and x+y=2 rArr y = -1` Hence, the normal to the curve meets it again at (3, -1) which lies in the fourth quadrant. |
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| 86. |
The area of a triangle formed by a tangent to the curve `2xy =a^(2)` and the coordinate axes, isA. `2a^(2)`B. `a^(2)`C. `3a^(2)`D. none of these |
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Answer» Correct Answer - B Let `P(x_(1), y_(1))` be a point on the curve `2xy =a^(2) " " `…(i) Then, `2x_(1)y_(1)=a^(2) " " `…(ii) Now, `2xy=a^(2) rArr 2(x(dy)/(dx)+y) =0 rArr (dy)/(dx)= -(y)/(x) rArr ((dy)/(dx))_(p) = -(y_(1))/(x_(1))` The equation of the tangent to (i) at `P(x_(1),y_(1))` is `y-y_(1)=-(y_(1))/(x_(1))(x-x_(1))` `rArr x_(1)y-x_(1)y_(1)= -xy_(1) + x_(1)y_(1)` `rArr xy_(1)+yx_(1)=2x_(1)y_(1) rArr xy_(1)+yx_(1)=a^(2) " " `[Using (ii)] This tangent meets the coordinate axes at `A(a^(2)//y_(1),0) and B(0, a^(2)//x_(1))` `therefore " Area of " Delta OAB = (1)/(2) xx OA xx OB` `rArr " Area of " Delta OAB=(1)/(2) xx (a^(2))/(y_(1)) xx (a^(2))/(x_(1))=(a^(4))/(2x_(1)y_(1))=a^(2) " " `[Using (i)] |
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| 87. |
Find the equation of the normal to the curve `y=(1+y)^y+sin^(-1)(sin^2x)a tx=0.`A. `x+y=2`B. `x+y=1`C. `x-y=1`D. none of these |
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Answer» Correct Answer - B We have, `y=(1+x)^(y)+sin^(-1) (sin^(2)x) " " `…(i) When x = 0, we have y = 1 Differentiating (i) w.r.t. x, we get `(dy)/(dx)=(1+x)^(y){(dy)/(dx)log(1+x)+(y)/(1+x)}+(sin2x)/(sqrt(1-sin^(4)x))` `rArr ((dy)/(dx))_((0","1)) =1` So, the equation of the normal at (0, 1) is `y-1= -1 (x-0)rArr x+y=1` |
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| 88. |
If the tangent at a point on the ellipse `x^2/27+y^2/3=1` meets the coordinate axes at `A` and `B,` and the origin, then the minimum area (in sq. units) of the triangle `OAB` is:A. 9B. `(9)/(2)`C. `9sqrt(3)`D. `3sqrt(3)` |
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Answer» Correct Answer - A Let `P(3sqrt(3) cos theta, sqrt(3) sin theta)` be a point on the ellipse. Then, the equation of the tangent at P is `(x)/(3sqrt(3)) cos theta +(y)/(sqrt(3)) sin theta =1`. This meets the coordinate axes at `A(3sqrt(3) sec theta, 0) and B(0, sqrt(3) " cosec" theta)`. Let `Delta` be the area of `Delta OAB.` Then, `Delta=(1)/(2)(OAxxOB)=(1)/(2) xx (3sqrt(3))/(cos theta)xx (sqrt(3))/(sin theta)=(9)/(sin2theta)` Clearly, `Delta` is minimum when `sin 2 theta` is maximum. The maximum value of `sin 2 theta` is 1. Hence, the minimum value of `Delta` is 9. |
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| 89. |
The slope of the tangent to the curve `y =sqrt(9-x^(2))` at the point where ordinate and abscissa are equal, isA. 1B. -1C. 0D. none of these |
| Answer» Correct Answer - B | |