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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.
| 551. |
The solution of the differential equation `(xy^4 + y) dx-x dy = 0,` isA. `4x^(4)y^(3)+3x^(3)=Cy^(3)`B. `3x^(3)y^(4)+4y^(3)=Cx^(3)`C. `3x^(4)y^(3)+4x^(3)=Cy^(3)`D. none of these |
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Answer» Correct Answer - C We have, `(xy^(4)+y)dx-xdy=0` `rArr" "(dy)/(dx)=(xy^(4)+y)/(x)` `rArr" "(dy)/(dx)-(y)/(x)=y^(4)` `rArr" "(1)/(y^(4))(dy)/(dx)+((-1)/(y^(3)))(1)/(x)=1` Let `-y^(-3)=v." Then, "3y^(-4)(dy)/(dx)=(dv)/(dx)` `therefore" "(1)/(3)(dv)/(dx)+(v)/(x)=1rArr (dv)/(dx)+(3)/(x)v=3" ...(i)"` This is a linear differential equaiton with integrating factor `x^(3)`. Multiplying both sides of (i) by `x^(3)` and integrating, we get `vx^(3)=(3x^(4))/(4)+C` `rArr" "(-x^(3))/(y^(3))=(3x^(4))/(4)+C` `rArr" "-4x^(3)=3x^(4)y^(3)+4y^(3)C` `rArr" "3x^(4)y^(3)+4x^(3)=-4Cy^(3)rArr 3x^(4)y^(3)+4x^(3)=lambday^(3)` |
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| 552. |
`sqrt(1-x^(2))dy+ydx=0`A. `y(x+sqrt(1-x^(2)))=c`B. `logy+sin^(-1)x=c`C. `ysin^(-1)x=c`D. `y^(2)=2sin^(-1)x+c` |
| Answer» Correct Answer - B | |
| 553. |
The solution of the differential equation `(x+y)(dx-dy)=dx+dy`, isA. `x-y=ke^(x-y)`B. `x+y=ke^(x+y)`C. `x+y=k(x-y)`D. `x+y=ke^(x-y)` |
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Answer» Correct Answer - D We have, `(x+y)(dx-dy)=dx+dy` `rArr" "dx-dy=(dx+dy)/(x+y)` `rArr" "d(x-y)=(d(x+y))/(x+y)` `rArr" "x-y=log(x+y)+logC" [On integrating]"` `rArr" "c(x+y)=e^(x-y)` `rArr" "x+y=ke^(x-y)," where k"=(1)/(C)` |
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| 554. |
The solution of the differential equation `y dx - x dy = xy dx` isA. `y=cye^(x)`B. `x=ye^(x)`C. `y=xe^(x)`D. `y=cxe^(x)` |
| Answer» Correct Answer - A | |
| 555. |
A line is drawn from a point `P(x, y)` on the curve `y = f(x),` making an angle with the x-axis which is supplementary to the one made by the tangent to the curve at `P(x, y).` The line meets the x-axis at A. Another line perpendicular to it drawn from `P(x, y)` meeting the y-axis at B. If `OA = OB,` where `O` is the origin, theequation of all curves which pass through (`1, 1)` isA. `x^(2) - y^(2) + 2xy + 2 = 0`B. `x^(2) - y^(2) + 2xy - 2 = 0`C. `x^(2) - y^(2) + 2xy + 1 = 0`D. `x^(2) - y^(2) + 2xy - 1 = 0` |
| Answer» `x^(2) + 2xy - y^(2) = c` | |
| 556. |
Solution of the differential equation `x((dy)/(dx))^(2)+2sqrt(xy)(dy)/(dx)+y=0`,isA. `x+y=a`B. `sqrtx-sqrty=a`C. `x^(2)+y^(2)=a^(2)`D. `sqrtx+sqrty=sqrta` |
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Answer» Correct Answer - D We have, `x((dy)/(dx))^(2)+2sqrt(xy)(dy)/(dx)+y=0` `rArr" "(sqrtx(dy)/(dx)+sqrty)^(2)=0` `rArr" "sqrtx(dy)/(dx)+dqrty=0` `rArr" "(1)/(sqrtx)dx+(1)/(sqrty)dy=0` `2sqrtx+2sqrty=C rArr sqrtx+sqrty=sqrta, " where "sqrta=2C`. |
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| 557. |
Find the family of curves, the subtangent at any point of which is the arithmetic mean of the co-ordinate point of tangency.A. `(x-y)^(2) = cy`B. `(y-x)^(2) = cx`C. `(x-y)^(2) = cxy`D. `(x-y)^(2) = cx^(2)y^(2)` |
| Answer» Correct Answer - `(x-y)^(2) = cy` | |
| 558. |
If a an arbitrary constant, then solution of the differential equation `(dy)/(dx)+sqrt((1-y^(2))/(1-x^(2)))=0` isA. `xsqrt(1-y^(2))+ysqrt(1-x^(2))=a`B. `ysqrt(1-y^(2))+xsqrt(1-x^(2))=a`C. `xsqrt(1-y^(2))-ysqrt(1-x^(2))=a`D. `ysqrt(1-y^(2))-xsqrt(1-x^(2))=a` |
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Answer» Correct Answer - A We have, `(dy)/(dx)+sqrt((1-y^(2))/(1-x^(2)))=0` `rArr" "sqrt(1-x^(2))dy+sqrt(1-y^(2))dx=0` `rArr" "(1)/(sqrt(1-x)^(2))dx+(1)/(sqrt(1-y^(2)))dy=0` `rArr" "sin^(-1)x+sin^(-1)y=sin^(-1)a" [On integrating]"` `rArr" "sin^(-1)(xsqrt(1-y^(2))+ysqrt(1-x^(2)))=sin^(-1)a` `rArr" "xsqrt(1-y^(2))+ysqrt(1-x^(2))-a` |
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| 559. |
Find the curve for which the length of normal is equal to the radiusvector.A. `y^(2)pmx^(2)=k^(2)`B. `ypm x=k`C. `y^(2)=kx`D. none of these |
| Answer» Correct Answer - A | |
| 560. |
A curve has the property that area of triangle formed by the x-axis, the tangent to the curve and radius vector of the point of tangency is `k^(2)`. The equation of all such curves passing through (0, 1) is ln (ay) `= (xy^(b))/(2k^(2))` thenA. a=1B. b=1C. a=2D. b=2 |
| Answer» Correct Answer - A::B | |
| 561. |
A curve having the condition that the slope of the tangent at some point is two times the slope of the straight line joining the same point to the origin of coordinates is a/anA. circleB. ellipseC. parabolaD. hyperbola |
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Answer» Correct Answer - C Let P (x, y) be any point on the curve. Then, `(dy)/(dx)=2((y-0)/(x-0))` `rArr" "(1)/(y)dy=(2)/(x)dx` `rArr" "logy=2logx+logC rArr y-Cx^(2)" ...(i)"` Clearly, it is a parabola. |
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| 562. |
Find the curve for which the length of normal is equal to the radiusvector. |
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Answer» Correct Answer - `y^(2)+-x^(2)=c` Length of normal `=ysqrt(1+((dy)/(dx))^(2))` and radius vector `=sqrt(x^(2)+y^(2))` `therefore y^(2)[1+((dy)/(dx))^(2)]=x^(2)+y^(2)` or `y(dy)/(dx) =+-x` or `ydy+-xdx=0` or `y^(2)+-x^(2)=c` |
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| 563. |
Orthogonal trajectories of the system of curves `((dy)/(dx))^(2) = (a)/(x)` areA. `9a(y+c)^(2) = 4x^(3)`B. `y + c = (-2)/(9sqrt(a))x^(3//2)`C. `y^(2)+c = (2)/(3sqrt(a)) x^(3//2)`D. `9a (y+c)^(2) = 4x^(2)` |
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Answer» Correct Answer - A Correct option is (A) 9a (y + c)2 = 4x3 Given differential equation of curves \((\frac{dy}{dx})^2=\frac ax\) ⇒ \(\frac{dy}{dx}=\frac{\sqrt a}{\sqrt x}\) Replacing \(\frac{dy}{dx}\) with \(\frac{-dy}{dx}\), we get \(\frac{dy}{dx}=\frac{\sqrt a}{\sqrt x}\) ⇒ \(-\frac{\sqrt a}{\sqrt x}dx=dy\) ⇒ \(\frac{-1}{\sqrt a}\int\sqrt xdx=\int dy+c\) ⇒ \(\frac{-1}{\sqrt a}\times\frac23 x^{3/2}=y+c\) ⇒ \(\frac{4}{9a}x^3=(y+c)^2\) (By squaring on both sides) ⇒ 9a (y + c)2 = 4x3 |
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| 564. |
The orthogonal trajectories of the family of curves an `a^(n-1)y = x^n` are given by (A) `x^n+n^2y=constant` (B) `ny^2+x^2=constant` (C) `n^2x+y^n=constant` (D) `y=x`A. `x^(n)+n^(2)y`= constantB. `ny^(2)+x^(2)`= constantC. `n^(2)x+y^(n)`= constantD. `n^(2)x-y^(n)`= constant |
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Answer» Correct Answer - B The equaiton of the given family of curves is `a^(n-1)y=x^(n)" …(i)"` `rArr" "(n-1)loga+logy=n logx` Differentiaing w.r.t. x, we get `(1)/(y)(dy)/(dx)=(n)/(x)" …(ii)"` This is the differential equation of the family of curves given in (i). The differential equation of the orthogonal trajectories of (i) is obtained by replacing `(dy)/(dx)by-(dx)/(dy)` in (ii). Replacing `(dy)/(dx)by-(dx)/(dy)` in (ii), we get `(1)/(y)xx-(dx)/(dy)=(n)/(x)rArr xdx +ny dy=0` On integrating, we get `(x^(2))/(2)+n(y^(2))/(2)=C rArr x^(2)+ny^(2)=2C` |
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| 565. |
The orthogonal trajectories of the family of curves an `a^(n-1)y = x^n` are given by (A) `x^n+n^2y=constant` (B) `ny^2+x^2=constant` (C) `n^2x+y^n=constant` (D) `y=x`A. `x^(n)+n^(2)y` = constB. `ny^(2)+x^(2)` = constC. `n^(2)x+y^(n)` = constD. `n^(2)x - y^(n)` = const |
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Answer» Correct Answer - B Differentiating, we have `a^(n-1)(dy)/(dx) = nx^(n-1)` `rArr" "a^(n-1)= n x^(n-1) (dx)/(dy)` Putting this value in the given equation, we have `nx^(n-1)(dx)/(dy) y = x^(n)` Replacing `(dy)/(dx)` by `-(dx)/(dy)`, we have `ny = -x (dx)/(dy)` `rArr" "ny dy + x dx = 0` `rArr" "ny^(2) + x^(2) =` const. Which is the required family of orthogonal trajectories. |
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| 566. |
Find the orthogonal trajectories of family of curves `x^2+y^2=c x` |
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Answer» Correct Answer - `x^(2)+y^(2)=k^(2)` `x^(2)+y^(2)=cx`…………..(1) Differentiating w.r.t x, we get `2x+2y(dy)/(dx)=c`……………(2) Elliminating c between (1) and (2), we get `2x+2y(dy)/(dx)=(x^(2)+y^(2))/(x)` or `(dy)/(dx) =(y^(2)-x^(2))/(2xy)` Replacing `(dy)/(dx)by-(dx)/(dy)`, we get `(dy)/(dx)=(2xy)/(x^(2)-y^(2))` This equation is homogeneous, and its solution gives the orthogonal trajectories as `x^(2)+y^(2)=ky`. |
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| 567. |
The orthogonal trajectories of the family of circles given by `x^2 + y^2 - 2ay = 0`, isA. `x^(2)+y^(2)-2kx=0`B. `x^(2)+y^(2)-2ky=0`C. `x^(2)+y^(2)-2k_(1)x-2k_(2)y=0`D. none of these |
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Answer» Correct Answer - A The equation of the family of circles is `x^(2)+y^(2)-2ay=0" …(i)"` Differentiating w.r.t. x, we get `2x+2y(dy)/(Dx)-2a(dy)/(dx)=0rArra=(x+y(dy)/(dx))/((dy)/(dx))` Putting this value of a in (i), we get `x^(2)+y^(2)-2y{(x+y(dy)/(dx))/((dy)/(dx))}=0rArrx^(2)-y^(2)-2xy(dx)/(dy)=0" ...(ii)"` This is the differential equation of the family of circles givne by (i). The differential equation representing the family of orthogonal trajectories of (i) is obtained by replacing `(dy)/(dx)by-(dx)/(dy)` in (ii). So, the differential equation of the orthogoanl trajectories is `x^(2)-y^(2)+2xy(dy)/(dx)=0` `rArr" "(x^(2)-y^(2))dx+2xydy=0` `rArr" "(xd(y^(2))-y^(2))/(x^(2))=-dx` `rArr" "d((y^(2))/(x))=-dx` `rArr" "(y^(2))/(x)=-x+2krArrx^(2)+y^(2)-2kx=0` This is the required family of orthogonal trajectories. |
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| 568. |
If the solution of the differential equation `(dy)/(dx)-y=1-e^(-x)`and `y(0)=y_0`has afinite value, when `xvecoo,`then thevalue of `|2/(y_0)|`is__ |
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Answer» Correct Answer - 4 `(dy)/(dx) -y=1-e^(-x)` `P=-1,Q=1-e^(-x)` I.F`. E^(intPdx)=e^(int-1dx)=e^(-x)` `therefore y.e^(-x)=inte^(-x)(1-e^(-x))dx+C` `=e^(-x)+1/2e^(-2x)+C` When, `x=0, y=y_(0)+1/2` So, `C=y_(0)+1/2` `y=-1+1/2e^(-x) +(y_(0)+1//2)e^(x)` When `x to infty y` takes finite value. So, `y_(0)+1//2=0` or `y_(0)=-1//2` |
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| 569. |
Tangent is drawn at the point `(x_i ,y_i)`on thecurve `y=f(x),`whichintersects the x-axis at `(x_(i+1),0)`. Now, again a tangent is drawn at `(x_(i+1,)y_(i+1))`on thecurve which intersect the x-axis at `(x_(i+2,)0)`and theprocess is repeated `n`times, i.e.`i=1,2,3dot,ndot`If `x_1,x_2,x_3,ddot,x_n`from anarithmetic progression with common difference equal to `(log)_2e`and curvepasses through `(0,2)dot`Now ifcurve passes through the point `(-2, k),`then thevalue of `k`is____ |
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Answer» Correct Answer - 8 Equation of tangent at `P(x_(1),y_(1))` of `y=f(x)` is `y-y_(1) =(dy)/(dx)(x-x_(1))` This tangent cuts the x-axis. So, `x_(2)=x_(1)-y_(1)/((dy)/(dx))` `x_(1),x_(2),x_(3),………..,x-(n)` are in A.P. `x_(2)-x_(1)=-y_(1)/((dy)/(dx))=log_(2)e` (Given) or `-y=log_(2)e(dy)/(dx)` or `(dy)/ylog_(2)e=-dx` Integrating both sides, we get `log_(e)y=-xloge^(2)+c` Since `y=f(x)` passes through (0,2), `k=2` `therefore y=2.e^(-xlog_(e)2)` `therefore y=e^(1-x)` |
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| 570. |
Tangent is drawn at the point `(x_(i), y_(i))` on the curve y = f(x), which intersects the x-axis at `(x_(i+1), 0)`. Now again tangent is drawn at `(x_(i+1), y_(i+1))` on the curve which intersects the x-axis at `(x_(i+2), 0)` and the process is repeated n times i.e. i = 1, 2, 3,......,n. If `x_(1), x_(2), x_(3),...,x_(n)` form a geometric progression with common ratio equal to 2 and the curve passes through (1, 2), then the curve isA. circleB. hyperbolaC. ellipseD. parabola |
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Answer» Correct Answer - B Again `x_(2) = x_(1) - (y_(1))/(((dy)/(dx)))` `because" "x_(1), x_(2), x_(3),.....x_(n)` are in G.P. `" "(x_(2))/(x_(1))=1-(y_(1))/(x_(1)((dy)/(dx)))` `rArr" "1-(y)/(x)(dx)/(dy)=2` `rArr" "-(y)/(x)(dx)/(dy)=1` `rArr" "(dx)/(x)=-(dy)/(y)` Integrating, `log x = - log y + log c` `rArr" "xy = c` Curve passing through `(1, 2)" "rArr" "c = 2` `therefore" "`Curve is xy = 2. |
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| 571. |
The equation of curve in which portion of y-axis cutoff between origin and tangent varies as cube of abscissa of point of contact isA. `y = (kx^(3))/(3) + c`B. `y = (-kx^(3))/(2) + c x`C. `y = (-kn^(3))/(2) + c`D. `y = (kx^(3))/(3) + (cx^(2))/(2)` |
| Answer» Correct Answer - B | |
| 572. |
A tangent to a curve at P(x, y) intersects x-axis and y-axis at A and B respectively. Let the point of contact divides AB in the ratio `y^2 : x^2`.A. `x^(2) + y^(2) = c^(2)`B. `x^(2) + y^(2) - 2x = c`C. `x^(2) + y^(2) = cx^(2) y^(2)`D. xy = c |
| Answer» Correct Answer - A | |
| 573. |
STATEMENT-1 : To find complete solution of a second order differential equation we need two different conditions. and STATEMENT-2 : An `n^(th)` order differential equation has n independent parameters.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 574. |
A curve is such that the mid-point of the portionof the tangent intercepted between the point where the tangent is drawn andthe point where the tangent meets the y-axis lies on the line `y=xdot`If thecurve passes through `(1,0),`then thecurve is(a)`( b ) (c)2y=( d ) x^(( e )2( f ))( g )-x (h)`(i)(b) `( j ) (k) y=( l ) x^(( m )2( n ))( o )-x (p)`(q)(c)`( d ) (e) y=x-( f ) x^(( g )2( h ))( i ) (j)`(k)(d) `( l ) (m) y=2(( n ) (o) x-( p ) x^(( q )2( r ))( s ) (t))( u )`(v)A. `2y=x^(2)-x`B. `y=x^(2)-x`C. `y=x-x^(2)`D. `y=2(x-x^(2))` |
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Answer» Correct Answer - C The point on y-axis is `(0,y-x(dy)/(dx))` According, to the given condition, `x/2=y-x/2(dy)/(dx)` or `(dy)/(dx) = 2(y/x)-1` Putting `y/x=v`, we get `x(dv)/(dx)=v-1` or `"ln"|y/x-1|="ln"|x|+c` or `1-y/x=x` [as y(1)=0] |
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| 575. |
Findthe equation of a curve passing through the origin given that the slope ofthe tangent to the curve at any point (x, y) is equal to the sum of thecoordinates of the point. |
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Answer» Let the moving point be `(x,y)` Given that, `(dy)/(dx)=x+yimplies (dy)/(dx)-y=x` Here, `P=-1`, `Q=x` `:. I.F.=e^(int-1dx)=e^(-x)` and general solution : `y(e^(-x))=intxe^(-x)dx+c` `implies ye^(-x)=-xe^(-x)-int1(-e^(-x))dx+c` `=-xe^(-x)-e^(-x)+c` `implies y=-x-1+ce^(x)` This curve passes through `(0,0)` `:. 0=0-1+cimpliesc=1` `:.`Curve is `y=-x-1+e^(x)` `implies x+y+1=e^(x)` |
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| 576. |
A tangent to a curve at P(x, y) intersects x-axis and y-axis at A and B respectively. Let the point of contact divides AB in the ratio `y^2 : x^2`.A. `((25)/(4) + (2pi)/(3))`B. `2(pi + 4)`C. `25pi`D. `(16pi + (4)/(5))` |
| Answer» Correct Answer - D | |
| 577. |
A tangent to a curve at P(x, y) intersects x-axis and y-axis at A and B respectively. Let the point of contact divides AB in the ratio `y^(2) : x^(2)`. If a member of this family passes through (3, 4) then its equation isA. `x^(2) + y^(2) = 25`B. `x^(2) + y^(2) - 2x = 19`C. `x^(2) + y^(2) = 25x^(2) y^(2)`D. `x^(2) + y^(2) = 7` |
| Answer» Correct Answer - A | |
| 578. |
At present, a firm is manufacturing 2000 items. Itis estimated that the rate of change of production P w.r.t. additional numberof workers x is given by `(d P)/(dx)=100-12sqrt(xdot)`If the firmemploys 25 more workers, then the new level of production of items is(1) 3000 (2)3500(3) 4500(4) 2500A. (a) 2500B. (b) 3000C. (c) 3500D. (d) 4500 |
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Answer» Correct Answer - (c) Given, `(dP)/dx=(100-12sqrtx) rArr dP=(100-12sqrtx)dx` On integrating both sides, we get `int dP=int (100-12sqrtx)dx` `P=100x-8x^(3//2)+C` When `x=0, then P=2000 rArr C=2000` Now, when x=25, then is `P=100xx25-8xx(25)^(3//1) +2000` `=2500-8xx125+2000` =4500-1000+3500 |
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| 579. |
lf length of tangent at any point on th curve `y=f(x)` intercepted between the point and the x-axis is of length 1. Find the equation of the curve. |
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Answer» Correct Answer - `(sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))= pm x+c)` Since, the length of tangent = `abs(ysqrt(1+(dx/dy)^(2)))=1` `rArr y^(2) (1+((dx)/dy)^2)=1` `therefore dy/dx=pm y/(sqrt(1-y^(2)))` `rArr int (sqrt(1-y^(2)))/y dy =pm int xdx ` `rArr int (sqrt(1-y^(2)))/y dy =pm x+C` Lut `y=sin theta rArr dy = cos theta d theta` `therefore int (cos theta)/(sin theta)cdot cos theta d theta = pm x+C` `therefore int (cos^(2) theta)/(sin^(2) theta)cdot sin theta d theta = pm x+C` Again put `cos theta =t rArr -sin theta d theta =dt` `therefore -int t^(2) /(1-t^(2))dt=pm x+C` `rArr int (1-1/(1-t^(2))) dt =pm x+C` `rArr t-log abs((1=t)/(1-t))=pmx+C` `rArr sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))=pm x+C` |
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| 580. |
A spherical rain drop evaporates at a rateproportional to its surface area at any instant `tdot`Thedifferential equation giving the rate of change of the radius of the raindrop is _____ |
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Answer» Correct Answer - `((dr)/dt=-lambda)` Since, Rate of change of volume `propto` surface area `rArr (dV)/dt propto SA` `rArr 4pir^(2)cdot (dr)/dt=-lambda4pir^(2)` `(dr)/dt=-lambda` is required differential equation. |
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| 581. |
The differential equation representing the familyof curves `y^2=2c(x+sqrt(c)),`where `c`is apositive parameter, is of(a) order 1 (b) order 2(c) degree 3 (d) degree 4A. order 2, degree 1B. order 1, degree 3C. order 1, degree 2D. order 1, degree 2 |
| Answer» Correct Answer - B | |
| 582. |
The differential equation of family of curves whose tangent form an angle of `pi/4` with the hyperbola `xy=C^2` isA. `(dy)/(dx)=(x^(2)+c^(2))/(x^(2)-C^(2))`B. `(dy)/(dx)=(x^(2)-C^(2))/(x^(2)+C^(2))`C. `(dy)/(dx)=-(C^(2))/(x^(2))`D. none of these |
| Answer» Correct Answer - B | |
| 583. |
Which of the following is not the differential equation of family of curves whose tangent from an angle of `pi/4` with the hyperbola `xy=c^(2)`?A. `(dy)/(dx)=(x-y)/(x+y)`B. `(dy)/(dx) = x/(x-y)`C. `(dy)/(dx) = (x+y)/(y-x)`D. None of these |
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Answer» Correct Answer - B `xy=C` or `x(dy)/(dx)+y=0` or `(dy)/(dx)=-y/x=m_(1)` By condition, `tanpi/4=|(m_(1)-m_(2))/(1+m_(1)m_(2))|` `1=|(-y/x-m_(2))/(1-y/xm_(2))|` or `y/x+m_(2)=1-y/xm_(2)` or `y/xm_(2)-1` or `(dy)/(dx) = (x-y)/(x+y)` or `(dy)/(dx) = (x+y)/(y-x)` |
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| 584. |
The equation of the family of curves which intersect the hyperbola xy-2 orthogonally isA. `y=(x^(3))/(6)+C`B. `y=(x^(2))/(4)+C`C. `y=(-x^(3))/(6)+C`D. `y=(-x^(2))/(4)+C` |
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Answer» Correct Answer - A We have, `xy=2rArry=(2)/(x)rArr(dy)/(dx)=-(2)/(x^(2))` Let y = f(x) be the required family of curves. Then, `((dy)/(dx))_(C_(1))xx((dy)/(dx))_(C_(2))=-1` `rArr" "(dy)/(dx)xx(-2)/(x^(3))=-1rArr(dy)/(dx)=(x^(2))/(2)rArry=(x^(3))/(6)+C` This is the required family of curves. |
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| 585. |
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds. |
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Answer» Here, Volume is changing at a constant rate. So, `(dV)/dt = k`, where `k` is a constant. Now, Volume of a sphere, `V = 4/3pir^3` So, `(dV)/dt = 4/3**3pir^2(dr)/dt` `=>(dV)/dt = 4pir^2(dr)/dt` As,`(dV)/dt = k`.So, `=>kdt = 4pir^2dr` Now, integrating both sides, `=>kt+c = 4/3pir^3+c_1`->(1) `=>kt+C = 4/3pir^3`, where `C= c-c_1` So, at `(r,t) = (3,0)` `=>C = 4/3pi**(3)^3 => C= 36pi` At`(r,t) = (6,3)` `=>k(3) + 36pi = 4/3pi(6)^3=>3k = 288pi-36pi=>k = 84pi` Putting values of `C` and `k` in (1), `84pit+36pi = 4/3pir^3=>84t+36 = 4/3r^3` `=>r^3 = (63t+27) => r = (63t+27)^(1/3)` So, at any time `t` radius of the baloon will be `(63t+27)^(1/3)`. |
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| 586. |
In a bank principal increases at the rate of r%per year. Find the value of r if Rs. 100 double itself in 10 years `((log)_e2=0. 6931.)` |
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Answer» `(dP)/(dt)=(r )/(100)*Pimplies(dP)/(P)=(r )/(100)dt` `implies logP=(r )/(100)t+c` Let at `t=0`, `P=P_(0)` `:. logP_(0)=0+cimpliesc=logP_(0)` Now, `logP=(rt)/(100)+logP_(0)` `implies log(P)/(P_(0))=(rt)/(100)` Given at, `t=10`, P=2P_(0)` `:. log"(2P_(0))/(P_(0))=(r(10))/(100)=(r )/(10)=log2=0.6931` `implies r=6.931 %` |
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| 587. |
In Figure AB = 36 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region. |
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Answer» Let the radius of the circle be r. and `triangle PCM` is right `triangle` `PC^2=PM^2+MC^2` `(9+r)^2=(9)^2+(16-r)^2` `81+r^218r=81+324+r^2-36r` `54r=324` `r=324/54=6cm` `Area of shaded region = (324pi)/2-81pi+36pi` `=162pi-117pi` Therefore area of the shaded region `=45picm^2` |
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| 588. |
The equation of the curve satisfying thedifferential equation `y_2(x^2+1)=2x y_1`passingthrough the point (0,1) and having slope of tangent at `x=0`as 3 (where `y_2`and `y_1`represent 2ndand 1st order derivative), then(a)`( b ) (c) y=f(( d ) x (e))( f )`(g) is astrictly increasing function(h)`( i ) (j) y=f(( k ) x (l))( m )`(n) is anon-monotonic function(o)`( p ) (q) y=f(( r ) x (s))( t )`(u) has a threedistinct real roots(v)`( w ) (x) y=f(( y ) x (z))( a a )`(bb)has only one negative root.A. `y=f(x)` is a strictly increasing functionB. `y=f(x)` is a non-monoatomic functionC. `y=f(x)` has three distinct real rootD. `y=f(x)` has only one negative root |
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Answer» Correct Answer - A::D The given differential equation is `y_(2)(x^(2)-1)=2xy_(1)`or `y_(2)/y_(1)=(2x)/(x^(2)+1)` Integrating both sides, we get `logy_(1)=log(x^(2)+1)+logC`………(1) It is given that `y_(1)=3` at x=0 Putting `x=0, y_(1)=3` at `x=0` Substituting the value of C in (1), we obtain `y_(1)=3(x^(2)+1)` ..............(2) Integrating both sides w.r.t to x, we get `y=x^(3)+3x+C_(2)` This passes through the point (0,1). Therefore, 1`=C_(2)` Hence, the required equation of the curve is `y=x^(3)+3x+1` Obviously, is it strictly increasing from equation (2). Also, `f(0)=1 gt0`. Then the only root is negative. |
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| 589. |
The value of `y(log4)` if `y_2 -7y_1 + 12y = 0, y(0) = 2, y_1(0) = 7` is |
| Answer» Correct Answer - 326 | |
| 590. |
If `y_1`and `y_2`are twosolutions to the differential equation `(dy)/(dx)+P(x)y=Q(x)`. Then prove that `y=y_1+c(y_1-y_2)`is thegeneral solution to the equation where `c`is anyconstant. |
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Answer» `y_(1), y_(2)` are the solutions of the differential equation `(dy)/(dx) + P(x)y=Q(x)` .............(1) Then `(dy_(1))/(dt) + P(x)y_(1)=Q(x)`............(2) and `(dy_(2))/(dx) + P(x)y_(2)=Q(x)`.............(3) From equations (1) and (2), we get `(d(y-y_(1)))/(dx) + P(x)(y-y_(1))=0`..............(4) and from equation (1) and (2), we get `(d(y_(1)-y_(2)))/(dx) + P(x)(y_(1)-y_(2))=0`.............(5) Now, from equation (4) and (5), we get `(d/(dx)(y-y_(1)))/(d/(dx)(y_(1)-y_(2))) = (y-y_(1))/(y_(1)-y_(2))` or `int(d(y-y_(1))/(y-y_(1))) = int(d(y_(1)-y_(2))/(y_(1)-y_(2)))` or `"ln"(y-y_(1))="ln"(y-y_(2))+"ln"c` or `y-y_(1)=c(y_(1)-y_(2))` or `y=y_(1)+c(y_(1)-y_(2))` or `y-y_(1)=c(y_(1)-y_(2)` or `y=y_(1)+c(y_(1)-y_(2))` |
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| 591. |
The equation of the curve in which the portion of the tangent included between the coordinate axes is bisected at the point of contact, isA. a parabolaB. an ellipseC. a circleD. a hyperbola |
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Answer» Correct Answer - D The eqaution of the tangent at any point `P(x,y)` is `Y-y=(dy)/(dx)(X-x)` This cuts the coordinate axes at `A(x-y(dx)/(dy),0) and B(0,y-x(dy)/(dx))`. It is given that `P(x,y)` is the mid-point of AB. `therefore" "x-y(dx)/(dy)=2xand y -x(dy)/(dx)=2y` `rArr" "x+y(dx)/(dy)=0 and y+x(dy)/(dx)=0` `rArr" "x dy+ydx=0 and ydx+xdy =0` `rArr" "d(xy)=0 rArr xy=C` Clearly, it represents a rectangular hyperbola. |
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| 592. |
The solution of the differential equation `(d^2y)/(dx^2)=sin3x+e^x+x^2`when `y_1(0)=1`and `y(0)`is(a) `( b ) (c) (d)(( e )-sin3x)/( f )9( g ) (h)+( i ) e^(( j ) x (k))( l )+( m )(( n ) (o) x^(( p )4( q ))( r ))/( s )(( t ) 12)( u ) (v)+( w )1/( x )3( y ) (z) x-1( a a )`(bb)(cc)`( d d ) (ee) (ff)(( g g )-sin3x)/( h h )9( i i ) (jj)+( k k ) e^(( l l ) x (mm))( n n )+( o o )(( p p ) (qq) x^(( r r )4( s s ))( t t ))/( u u )(( v v ) 12)( w w ) (xx)+( y y )1/( z z )3( a a a ) (bbb) x (ccc)`(ddd)(eee)`( f f f ) (ggg) (hhh)(( i i i )-cos3x)/( j j j )3( k k k ) (lll)+( m m m ) e^(( n n n ) x (ooo))( p p p )+( q q q )(( r r r ) (sss) x^(( t t t )4( u u u ))( v v v ))/( w w w )(( x x x ) 12)( y y y ) (zzz)+( a a a a )1/( b b b b )3( c c c c ) (dddd) x+1( e e e e )`(ffff)(d)None of theseA. `y=(-sin3x)/9 + e^(x)+x^(4)/12+1/3x-1`B. `y=(-sin3x)/9 + e^(x)+x^(4)/12+1/3x`C. `y=(-cos3x)/3 + e^(x) + x^(4)/12+1/3x +1`D. None of these |
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Answer» Correct Answer - A Integrating the given differential equation, we have `(dy)/(dx) = (-cos3x)/3+e^(x)+x^(3)/3+C_(1)` But `y_(1)(0)=1` So, `1=(-1/3)+C_(1)` or `C(1)=(1//3)` Again integrating, we get `y=(-sin3x)/9 + e^(x)+x^(4)/12+1/3x+C_(2)` But y(0)=0, So, 0=0+1 `+C_(2)` or `C_(2)=-1`. Thus, `y=(-sin3x)/9 + e^(x)+x^(4)/12+1/3x-1` |
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| 593. |
The curve that passes through the point `(2, 3)` and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given byA. `y=(6)/(x)`B. `x^(2)+y^(2)=13`C. `((x)/(2))^(2)+((y)/(3))^(2)=2`D. `2y-3x=0` |
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Answer» Correct Answer - A Let `P(x,y)` be any point on the curve. The equation of tangent at P is givne by `Y-y=(dy)/(dx)(X-x)` This cute coordinate axes at `A(x-y(dx)/(dy),0) and B(0,y-x(dy)/(dx))` It is given that `P(x,y)` is the mid-point of AB. `therefore" "2x=x-y(dx)/(dy)and 2y=y-x(dy)/(dx)` `rArr" "(dy)/(dx)=-(y)/(x)` `rArr" "xdy+ydx=0rArrd(xy)=0rArrxy=C` The curve passes through (2,3). `therefore" "2xx3=CrArr C=6.` The equation of the curve is xy = 6 or, `y=(6)/(x)`. |
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| 594. |
Find the differential equation of all the circleswhich pass through the origin and whose centres lie on y-axis.A. a homogeneous differential equaionB. a differential equation of order 1 and degree 2C. a differential equation in variable separable formD. a differential equation reducible to variable separable form |
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Answer» Correct Answer - A The differential equation representing the given family of circle is `(x^(2)-y^(2))(dy)/(dx)=2xy` `or, (dy)/(dx)=(2xy)/(x^(2)-y^(2))` Clearly, it is a homogeneious differential equation. |
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| 595. |
Determine order and degree (if defined) of differential equations given`yprimeprime+(yprime)^2+2y=0` |
| Answer» The derivative of highest order in the given differential equation is `y'` whose order is `2` and degree is `1`. So the order is `2` and degree is `1` of the differential equation. | |
| 596. |
Determine order and degree (if defined) of differential equations given`(d^2y)/(dx^2)=cos3x+sin3x` |
| Answer» The derivative of highest order in the given differential equation is `(d^(2)y)/(dx^(2))` whose order is `2` and degree is `1`. So the order is `2` and degree is `1` of the differential equation. | |
| 597. |
Determine order and degree (if defined) of differential equations given`((d s)/(dt))^4+3s(d^2s)/(dt^2)=0` |
| Answer» The derivative of highest order in the given differential equation is `(d^(2)s)/(dt^(2))` whose order is `2` and degree is `1`. Therefore, the order is `2` and degree is `1` of the differential equation. | |
| 598. |
A curve has a property that the slope of the tangent at a point (x, y) is `(y)/(x + 2y^(2))` and it passes through (1, 1). Find the equation of the curve. |
| Answer» Correct Answer - `x = y(2y-1)` | |
| 599. |
Show that the differential equation `((x-y)dy)/(dx)=x+2y ,`ishomogeneous and solve it. |
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Answer» `(x-y)(dy)/(dx)=x+2y` `implies (dy)/(dx)=(x+2y)/(x-y)`………`(1)` Let `f(x,y)=(x+2y)/(x-y)` `implies f(lambdax,lambday)=(lambdax+2lambday)/(lambdax-lambday)=(lambda(x+2y))/(lambda(x-y))` `=(x+2y)/(x-y)=f(x,y)` `:.` Given differential equation is homogenous. Let `y=vx` `implies (dy)/(dx)=v+x(dv)/(dx)` Put this value in equation `(1)` `v+x(dv)/(dx)=(x+2vx)/(x-vx)=(1+2v)/(1-v)` `implies x(dv)/(dx)=(1+2v)/(1-v)-v` `=(1+2v-v+v^(2))/(1-v)=(1+v+v^(2))/(1-v)` `implies (v-1)/(v^(2)+v+1)dv=-(dx)/(x)` `implies int(v-1)/(v^(2)+v+1)dv=-int(dx)/(x)` `implies int((2v+1)-3)/(v^(2)+v+1)dv-3int(1)/(v^(2)+v+1)dv` `=-2logx+c` `impliesl og(v^(2)+v+1)-3int(1)/((v+(1)/(2))^(2)+((sqrt(3))/(2))^(2))dv` `=-logx+c` `implies log(v^(2)+v+1)+logx^(2)` `-3*(1)/(sqrt(3)//2)tan^(-1)"(v+(1)/(2))/(sqrt(3)//2)=c` `implies log(x^(2)v^(2)+vx^(2)+x^(2))-2sqrt(3)tan^(-1)"(2v+1)/(sqrt(3))=c` `implies log(y^(2)+xy+x^(2))-2sqrt(3)tan^(-1)"(2y+x)/(xsqrt(3))=c` |
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| 600. |
Consider a curved mirror y = f(x) passing through (8, 6) having the property that all light rays emerging from origin, after reflected from the mirror becomes parallel to x-axis. The equation of the mirror is `y^(a) = b(c-x^(d))` where a, b, c, d are constants, thenA. b=4B. b = 36C. c= 9D. c = 1 |
| Answer» Correct Answer - B::C | |