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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.
| 401. |
The degree of the differential equation of the curve `(x-a)^(2) + y^(2) =16` will be |
| Answer» Correct Answer - B | |
| 402. |
Write degree of the differential equation `(1+(dy)/(dx))^3=((d^2y)/(dx^2))^2dot`A. 4B. `(3)/(2)`C. Not definedD. 2 |
| Answer» Correct Answer - D | |
| 403. |
Let `f(x)` be a non-positive continuous function and `F(x)=int_(0)^(x)f(t)dt AA x ge0` and `f(x) ge cF(x)` where `c lt 0` and let `g:[0, infty) to R` be a function such that `(dg(x))/(dx) lt g(x) AA x gt 0` and `g(0)=0` The number of solution(s) of the equation `|x^(2)+x-6|=f(x)+g(x)` is/areA. 2B. 1C. 0D. 3 |
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Answer» Correct Answer - C `|x^(2)+x-6|=f(x) + g(x)` or `|x^(2)+x-6|=g(x)` Thus, no solution exists. |
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| 404. |
Let `f(x)` be a non-positive continuous function and `F(x)=int_(0)^(x)f(t)dt AA x ge0` and `f(x) ge cF(x)` where `c lt 0` and let `g:[0, infty) to R` be a function such that `(dg(x))/(dx) lt g(x) AA x gt 0` and `g(0)=0` The solution set of inequation `g(x)(cos^(-1)x-sin^(-1)) le0`A. `[-1,1/sqrt(2)]`B. `[1/sqrt(2),1]`C. `[0,1/sqrt(2)]`D. `[0,1/sqrt(2)]` |
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Answer» Correct Answer - A `g(x) (cos^(-1)x-sin^(-1)x) le0` or `(cos^(-1)x-sin^(-1)x) ge0` or `x in [-1, 1/sqrt(2)]` |
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| 405. |
If `x(dy)/(dx)=x^(2)+y-2, y(1)=1`, then `y(2)` equals ……………….. |
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Answer» Correct Answer - 2 Given, `(dy)/(dx)-1/xy=(x-2/x)` I.F. `=e^(-int1/xdx)=e^(-"ln x"=1/x)` Now, general solution is given by `y/x= int(x-2/x)1/xdx` or `y=x^(2)-2x+2` Hence, `y(2) =(2)^(2)-2x+2` Hence, `y(2)=(2)^(2)-2(2)+2=2` |
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| 406. |
In the above question, the amount of salt in the tank at the moment of overflow isA. 20 lbB. 50 lbC. 30 lbD. None of these |
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Answer» Correct Answer - D `(dQ)/(dt)+2/(10+2t)Q=4` This is a linear equation, It solution is `Q=(40t+4t^(2)+c)/(10+2t)`……….(1) At `t=0, Q=a=0`. Substituting these values into equation (1), we find that c=0. We require Q at the moment of overflow which from part (a) is t=20. Thus, `Q=(40(20)+4(20)^(2))/(10+2(20))=40 lb` |
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| 407. |
Find the general solution of the differential equations:`(dx)/(dy)+y/x=x^2` |
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Answer» In this question, `P = 1/x and Q = x^2` Now, we have to compute integrating factor(I.F.). We know, `I.F. = e^(intPdx)`,so, `I.F. = e^(int(1/x)dx) = e^lnx=x` So, solution will be `y*(I.F.) = int(I.F.)Qdx` `yx = intx*x^2dx` `yx = intx^3dx` `yx=x^4/4+c` `y = x^3/4+c/x` that is the general solution. |
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| 408. |
A 50 L tank initailly contains 10 L of fresh water, At t=0, a brine solution containing 1 lb of salt per gallon is poured into the tank at the rate of 4 L/min, while the well-stirred mixture leaves the tank at the rate of 2 L/min. Then the amount of time required for overflow to occur inA. 30 minB. 20 minC. 10 minD. 40 min |
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Answer» Correct Answer - B Here, `a=0, b=1, e=4, f=2` and `V_(0)-10`. The volume of brine in the tank at any time t is given as `V_(0)+et-ft=10+2t` We require t when `10+2t=50`. Hence, t=20 min. |
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| 409. |
The differential equations , find the particular solution satisfying the given condition:`(dy)/(dx)-y/x+cose c(y/x)=0; y=0`when x = 1 |
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Answer» `(dy)/(dx)=y/x+cosecy/x=f(x,y)` let y/x=t `(dy)/(dx)=t+(xdt)/(dx)` `t+(xat)/(dx)=t+cosect` `(xdt)/(dx)=cosect` `intsintdt=ln|x|+c` `-cost=ln|x|+c` given x=1 and y=0 c=-1 `ln|x|+cosy/x=1` |
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| 410. |
A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10% of its original mass, find () and expression for the mass of the material remaining at any time t, (ii) the mass of the material after four hours and (iii) the time at which the material has decayed to one half of its initial mass.A. `50^(-0.5" ln "9)`B. `50e^(-2" ln "9)`C. `50e^(-2" ln "0.9)`D. None of these |
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Answer» Correct Answer - C We require N at t=4. Substituting t=4 into (3) and then solving for N, we find `N=50e^(2" ln "0.9)` |
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| 411. |
A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10% of its original mass, find () and expression for the mass of the material remaining at any time t, (ii) the mass of the material after four hours and (iii) the time at which the material has decayed to one half of its initial mass.A. `N=50e^((1//2)("ln "9)t)`B. `50e^((1//4)("ln "9)t)`C. `N=50e^(-("ln "0.9)t)`D. None of these |
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Answer» Correct Answer - A Let N denote the amount of material present and linear at time t. Then, `(dN)/(dt) =-kN=0` This differential equation is separable and linear. Its solution is `N=ce^(kt)` At t=0, we are given that `N=50`. Therefore, from equation (1), `50=ce^(k(0))` or `c=50`. Thus, `N=50e^(kt)`............(2) At `t=2, 10%` of the original mass of 50 mg or 5 mg has decayed. Hence, at `t=2, N=50-5=45` Sustituting these values into equation (2) and solving for k, We have `45=50e^(2k)` or `k =1/2 log(45/50)` Substituting this value into (2), we obtain the amount of mass present at any time t as `N=50e^((1//2)("ln "0.9)t)` where t is measured in hours. ...............(3) where t is measured in hours. |
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| 412. |
Consider a tank which initially holds `V_(0)` liter of brine that contains a lb of salt. Another brine solution, containing b lb of salt per liter is poured into the tank at the rate of `eL//"min"`. The problem is to find the amount of salt in the tank at any time t. Let Q denote the amount of salt in the tank at any time. The time rate of change of Q, `(dQ)/(dt)`, equals the rate at which salt enters the tank at the rate at which salt leaves the tank. Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume `V_(0)` plus the volume of brine added et minus the volume of brine removed ft. Thus, the volume of brine at any time is `V_(0)+et-ft` The concentration of salt in the tank at any time is `Q//(V_(0)+et-ft)`, from which it follows that salt leaves the tank at the rate of `f(Q/(V_(0)+et-ft))`lb/min. Thus, `(dQ)/(dt)=be-f(Q/(V_(0)+et-ft))Q=be` A tank initially holds 100 L of a brine solution containing 20 lb of salt. At t=0, fresh water is poured into the tank at the rate of 5 L/min, while the well-stirred mixture leaves the tank at the same rate. Then the amount of salt in the tank after 20 min.A. 20/eB. 10/eC. `40//e^(2)`D. 5/e L |
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Answer» Correct Answer - A Here, `V_(0)=100, a=20, b=0` and `e=f=5`. Hence `(dQ)/(dt)+1/20Q=0` The solution of this linear equation is `Q=ce^(-t//20)`…………….(1) At t=0, we are given that Q=a=20. Substituting these values into equations (1), we find that `c=20,` so that equation (1), can be rewritten as `Q=20e^(-t//20)`. For `t=20, Q=20/e` |
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| 413. |
Find the general solution of the differential equation `(dy)/(dx)=(x+1)/(2-y),(y!=2)` |
| Answer» `(dy)/(dx)=(x+1)/(2-y)``(2-y)dy=(x+1)dx`integrating both side`2y-y^2/2=x^2/2+x+c``4y-y^2=x^2+2x+C` | |
| 414. |
The degree of the differential equation satisfying `sqrt(1-x^(2))+sqrt(1-y^(2))=a(x-y)`, isA. 1B. 2C. 3D. None of these |
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Answer» Correct Answer - A 1) Putting `x=sinA` and `y=sinB` in the given relation, we get `cosA+cosB=a(sinA-sinB)` or `A-B=2cot^(-1)a` or `sini^(-1)x-sin^(-1)y=2cot^(-1)a` Differentiating w.r.t. x, we get `1/sqrt(1-x^(2))-1/sqrt(1-y^(2))(dy)/(dx)=0` Clearly, it is differential equation of degree one. |
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| 415. |
A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10% of its original mass, find () and expression for the mass of the material remaining at any time t, (ii) the mass of the material after four hours and (iii) the time at which the material has decayed to one half of its initial mass.A. `("ln "0.25)/("ln "0.9)h`B. `("ln "0.5)/("ln "0.81)h`C. `("ln "0.25)/("ln "0.81)h`D. None of these |
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Answer» Correct Answer - C We require t when `N=50//2=25`, Substituting `N=25` into equation (3) and solving for t, we find `25 =50e^((1//2)("ln "0.9)t)` or `t=("ln "0.5)/("ln "0.81)h` |
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| 416. |
Show that the differential equation `xcos(y/x)(dy)/(dx)=ycos(y/x)+x`is homogeneous and solve it. |
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Answer» Let `y/x = v` Then, `y = vx =>dy/dx = v+x(dv)/dx` So, given equation becomes, `xcosv(dy/dx) = vxcosv+x` `=>dy/dx = (vxcosv+x)/(xcosv)` `=> v+x(dv)/dx = (vcosv+1)/(cosv)` `=> v+x(dv)/dx = v+secv` `=>x(dv)/dx = secv` `=>cosdv = dx/x` Integrating both sides, `=> sinv = ln(x)+c` `=>sin(y/x) = ln(x)+c`, which is the required equation. |
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| 417. |
Suppose that a mothball loses volume by evaporation at a rateproportional to its instantaneous area. If the diameter of the ball decreasesfrom 2cm to 1cm in 3 months, how long will it take until the ball haspractically gone?A. 4 monthsB. 3 monthsC. 2 monthsD. 1 months |
| Answer» Correct Answer - A::B::C | |
| 418. |
Find the orthogonal trajectory of `y^2=4a x`(a being the parameter). |
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Answer» `y^(2)=4ax`…………..(1) `therefore 2y(dy)/(dx)=4a`……………………..(2) Eliminating a from equation (1) and (2), we get `y^(2)=2y(dy)/(dx)x` Replacing `(dy)/(dx)dy-(dx)/(dy)`, we get `y=2(-(dx)/(dy))x` `2xdx+ydy=0` Integrating both sides, we get `x^(2)+y^(2)/(2)=c` `2x^(2)+y^(2)=2c` Which is the required orthogonal trajectory. |
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| 419. |
The population of a certain is known to increase at a rate proportional to the number of people presently living in the country. If after two years the population has doubled, and after three years the population is 20,000 estimates the number of people initially living in the country. |
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Answer» Let N denote the number of people living in the country at any time t, and let `N_(0)` denote the number of people initially living in the country. Then, from `(dN)/(dt) propto N, (dN)/(dt)-kN=0` which has the solution, `N=ce^(kt)`…………………..(1) At `t=0, N=N_(0)`, hence, equation (1) states that `N_(0)=ce^(k(0))`, or `c=N_(0)`. Thus, `N=N_(0)e^(kt)`....................(2) At `t=2, N=2N_(0)`. Substituting these values into equation (2), we have `2N_(0)=N_(0)e^(2k)`or `e^(k)=sqrt(2)`. Substituting this value into equation (1) gives `N=N_(0)(sqrt(2))^(k)` ...............(3) At `t=3, N=20,000`. Substituting these values into equation (3), we obtain `N_(0)=20,000//2sqrt(2)=7071` |
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| 420. |
In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself? |
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Answer» `(dp)/(dt)=5/100P` `int_1000^2000 (dp)/p=int_0^t(dt)/20` `ln|2000|-ln|1000|=t/20` `ln|2|=t/20` `t=20ln|2| years` |
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| 421. |
Suppose that a mothball loses volume by evaporation at a rateproportional to its instantaneous area. If the diameter of the ball decreasesfrom 2cm to 1cm in 3 months, how long will it take until the ball haspractically gone? |
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Answer» Let at any instance (t), the radius of moth ball be r and v be its volume. Then `v=4/3pir^(3)` or `(dv)/(dt) = 4pir^(2)(dr)/(dt)` Thus, as per the information `4pir^(2)(dr)/(dt) = -k4pir^(2)`, where `k in R^(+)` or `(dr)/(dt) = -k` or `r=-kt+c` at t=0, r=2cm, t=3 month, r=1 cm `rArr c=2, k=1/3` `rArr r=-1/3t+2` now `r to 0`, `t to6` Hence, it will take six months until the ball is practically gone. |
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| 422. |
The rate at which a substance cools in moving air is proportional tothe difference between the temperatures of the substance and that of the air.If the temperature of the air is 290 K and the substance cools from 370 K to330 K in 10 min, when will the temperature be 295 K? |
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Answer» Let T be the temperature of the substance at time t. `therefore -(dT)/(dt) = k(T-290)` or `(dT)/(dt) =-(T-290)` (Negative sign because `(dT)/(dt)` is rate of cooling) or `int(dT)/(T-290)=-kint(dt)` Integrating the L.H.S between the limits `T=370` to T=330 and the R.H.S between the limits t=0 to t=10, we get `int_(370)^(330) (dT)/(t-290)=-k_(int_(0)^(10))dt` or `log40-log80 =-k xx 10` `log2=10k` or `k=(log2)/10`..............(2) Now, integrating equation (1) between `T=370` and `T=295` and t=0 and t=t, we get `int_(370)^(295)(dT)/(T-290)=-kint_(0)^(1)dt` or `log(T-290)|underset(370)overset(295)(=-kt` or `t=(log 16)/k` Hence, from equation (2), we get `t=(log16)/k` Hence, from equation (2), we get `t=(log 16)/(log2) xx 10 = 40` i.e., after 40 min |
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| 423. |
If the population of country double in 50 years, in how many years willit triple under the assumption that the rate of increase is proportional tothe number of inhabitants. |
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Answer» Let x denote the population at time t in years. Then `(dx)/(dt) propto x rArr (dx)/(dt) = kx`, Where k is constant of proportionality. Solving `(dx)/(dt) = kx`, we get `int(dx)/x= intkdt` or `logx=kt + c` or `x=e^(kt+c)` or `x=x_(0)e^(kt)`, Where `x_(0)` is the population at time t=0. Since it doubles in 50 years, at t=50, we must have `x=2x_(0)` Hence, `2x_(0)=x_(0)e^(50k)` or `50k = log2` or `k=(log2)/(50)` so that `50k =log2` or `k=(log2)/(50)` so that `x=x_(0)e^((log2)/50)t` or `t=(50 log3)/(log 2)=50 log_(2)(3)` |
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| 424. |
What constant interest rate is required if an initial deposit placedinto an account accrues interest compounded continuously is to double itsvalue in six years? `(ln|x|=0. 6930)` |
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Answer» The balance N(t) in the account at any time t, `(dN)/(dt)-kN=0`, its solution is `N(t) = ce^(kt)`………..(1) Let initial deposit be `N_(0)`. At t=0, `N(0)=N_(0)` which when substituted into equation `N_(0)=ce^(k(0))=c` and equation (1) becomes `N(t) = N_(0)e^(kt)`. We seek the value of k for which `N=2N_(0)` when t=6, substituting these values into (2) and solving for k we find `2N_(0)=N_(0)e^(k(6))`. or `e^(6k)=2` or `k=1/6In|2|=0.1155` An interest rate of `11.55` percent is required. |
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| 425. |
A person places Rs 500 in an account that interest compounded continuously. Assuming no additional deposits or withdrawals, how much will be in the account after seven years if the interest rate isa constant 8.5 percent for the first four year and a constant 9.25 percent for the last three years |
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Answer» Let N(t) denote the balance in the account at any time t. Initially, N(0)=500. For the first four years, k=0.085 N=0 Its solution is `N(t)=ce^(0.085t) (0 letle4)`………….(1) At t-0, N(0)=500. Then from (1), `5000=ce^(0.085(0))=c` and equation (1) becomes `N(t) = 5000e^(0.085t) (0 letle4)`...................(2) Substituting t=4 into equation (2), we find the balance after four years to be `N(4) = 5000e^(0.085(4))=5000(1.404948)=7024.74`. This amount also represents the beginning balance for the last three-year period. Over the last three years, the interest rate is `9.25%` `therefore (dN)(dt) -0.0925 N=0` `(4 let le7)` In solution is `N(t)=ce^(0.0925t) (4 letle7)`..........(3) At `t=4, N(4)=7024.74`. Then from equations (3), `7024.74=ce^(0.009245) =c(1.447735)` or `c=4852.23` Then from equation (3), `N(7) = 4852.23 e^(0.0925(7))=4852.23(1.910758)` |
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| 426. |
If y= y(x) is the solution of the differential equation `dy/dx=(tan x-y) sec^(2)x, x in(-pi/2,pi/2),` such that y (0)=0, than `y(-pi/4)` is equal to(a)` 1/e-2 `(b) `1/2-e `(c) `2+1/e`(d)` e-2`A. (a)` 1/e-2 `B. (b) `1/2-e `C. (c) `2+1/e`D. (d)` e-2` |
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Answer» Correct Answer - (d) Given differential equation ` dy/dx=(tan x-y)sec^(2)x` `rArr dy/dx +(sec^(2) x)y=sec^(2)x tan x,` which is linear differential equation of the form `dy/dx+Py=Q,` where `P=sec^(2) x and Q=sec^(2) x tan x` `IF=e^(int sec^(2) x dx)=e^(tan x)` So, solution of given dofferential equation is `yxxIF= int(QxxIF)d+c` `y(e^tanx)=int e^(tanx). sec^(2)xtan x dx +C` Let `tan x=t rArr sec^(2) xdx =dt` `ye^tanx =int e^(t) cdot t dt + C=te^(t)-int e ^(t) dt +C` [using integration by parts method] `= e^(t) (t-1)+C` `rArr y cdot e^tan x=e^(tan x ) (tan x-1)+C [therefore t =tan x]` `therefore y(0)=0` ` rArr 0=1(0-1)+C rArr C=1` `therefore y cdot e^(tan x)=e^(tan x)(tan x-1)+1` Now, at `x=-pi/4` `ye^(-2) =e^(-1)(-1-1)+1` ` rArr ye^(-1) =2e^(-1)+1 rArr y=e-2` |
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| 427. |
Consider the differential equation, `y^(2)dx+(x-1/y)dy=0.` If value of y is 1 when x = 1, then the value of x for which y = 2, is(a) `5/2+1/sqrt e`(b) `3/2+1/sqrt e`(c) `1/2+1/sqrt e`(d) `3/2+sqrt e`A. (a) `5/2+1/sqrt e`B. (b) `3/2+1/sqrt e`C. (c) `1/2+1/sqrt e`D. (d) `3/2+sqrt e` |
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Answer» Correct Answer - (b) Given differential equation is `y^(2) dx+(x-1/y)dy =0` `rArr dx/dy+1/y^(2)x=1/y^(3),` which is the linear differential equation of the form `dx/dy+ Px = Q.` Here, `P=1/y^(2) and Q=1/y^(3)` Now, `IF=e^(int1/y^(2)dy)=e^(-1/y)` `therefore` The solution of linear differential equation is `x.(IF)intQ(IF)dy+c` `rArr x e^(-1//y)=int 1/y^(3) e^(-1//y)dy+C` `therefore x e^(-1//y) =int(-t) e^(t) dt+C [therefore let -1/y=trArr+1/y^(2)dy=dt]` `=-te^(t)+int e^(t)dt+C` [integration by parts] `=-te^(t)+ e^(t)+C` `rArr x c^(-1//y)=1/ye^(-1//y)+e^(-1//y)+C` Now, at y = 1, the value of x = 1, so `1cdote^(1)=e^(-1) +e^(-1)+C rArrC=-1/e` On putting the value of C, in Eq. (i), we get `x=1/y+1-e^(1//y)/e` So, at Y= 2, the value of `x=1/2+1-e^(1//2)/e= 3/2 - 1/sqrte` |
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| 428. |
Consider the statements : P : There exists some x IR such that f(x) + 2x = 2(1+x2) Q : There exists some x IR such that 2f(x) +1 = 2x(1+x) Then (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false.A. (a) Both I and II are trueB. (b) I is true and II is falseC. (c) I is false and II is trueD. (d) Both I and II are false |
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Answer» Correct Answer - (c) Here, `f(x)+2x=(1-x)^(2) cdot sin ^(2) x+x^(2) +2x …(i)` where, `I: f(x)+2x=2(1+x)^(2) …(ii)` `therefore 2(1+x^(2))=(1-x)^(2) sin ^(2) x+x^(2) +2x` `rArr (1-x)^(2) sin ^(2) x=x^(2) - 2x+2` `rArr (1-x)^(2) sin ^(2) x= (1-x)^(2)+1` `rArr (1-x)^(2) cos^(2) x=-1` which is never possible. `therefore I` is false. Again, let `h(x)=2f(x)+1-2x(1+x)` where, `h(0)=2f(0)+1-0=1` `h(1)=2(1) +1-4=-3 as [h(0)h(1)lt0]` `rArr h(x)` must have a solution. `therefore II` is true. |
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| 429. |
The solution of the differential equation `(dy)/(dx)+(2x)/(1+x^2) y=1/(1+x^2)^2`, isA. `y(1-x^(2))=tan^(-1)x+C`B. `y(1+x^(2))=tan^(-1)x+C`C. `y(1+x^(2))^(2)=tan^(-1)x+C`D. `y(1-x^(2))^(2)=tan^(-1)x+C` |
| Answer» Correct Answer - B | |
| 430. |
The equation of the curve through the point (1,0) which satisfies the differential equatoin `(1+y^(2))dx-xydy=0`, isA. `x^(2)+y^(2)=4`B. `x^(2)-y^(2)-=`C. `2x^(2)+y^(2)=2`D. none of these |
| Answer» Correct Answer - B | |
| 431. |
The general solution of the differential equation `(y^(2)-x^(3))dx-xydy=0 (x ne0)` is (where, C is a constant of integration)(A) `y^(2)-2x^(2)+Cx^(3) =0`(B) `y^(2)+2x^(3)+Cx^(2) =0`(C) `y^(2)+2x^(2)+Cx^(3) =0`(D) `y^(2)-2x^(3)+Cx^(2) =0`A. (a) `y^(2)-2x^(2)+Cx^(3) =0`B. (b) `y^(2)+2x^(3)+Cx^(2) =0`C. (c) `y^(2)+2x^(2)+Cx^(3) =0`D. (d) `y^(2)-2x^(3)+Cx^(2) =0` |
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Answer» Correct Answer - (b) Given differential equation is `(y^(2)-x^(3))dx-xy dy=0, (xne0)` `rArr xydy/dx-y^(2)=-x^(3)` Now, put `y^(2)=trArr 2y dy/dx = (dt)/dx rArr ydy/dx=1/2 (dt)/dx` `therefore x/2 (dt)/dx-t=-x^(3)` `rArr (dt)/dx -2/xt =-2x^(2)` which is the linear differential equation of the form `(dt)/dx+Pt=Q.` Here, `p=-2/xand Q=-2x^(2).` Now, `IF=e^(-f2/x dx)=1/x^(2)` `therefore` solution of the linear differential equation is `(IF) t=intQ(IF)dx+ lambda` [where `lambda` is integrating constant ] `therefore t(1/x^(2))=-2int(x^(2) xx 1/x^(2))dx + lambda` `rArr t/x^(2) =-2x+lambda` `rArr y^(2) /x^(2) +2x =lambda` `[therefore t=y^(2)]` `rArr y^(2) + 2x^(3) - lambda x^(2) = 0` or `y^(2) +2x^(3) +cx^(2)=0` " [let C"=-`lambda]` |
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| 432. |
If p and q are order and degree of differential equation `y^(2)((d^(2)y)/(dx^(2)))^(2)+3x((dy)/(dx))^(1/3) + x^(2)y^(2) = sin x, then`A. `p gt q`B. `(p)/(q) = (1)/(2)`C. `p = q`D. `p lt q` |
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Answer» Correct Answer - D `y^(2)((d^(2)y)/(dx^(2)))^(2)+x^(2)y^(2)- sin x = - 3x((dy)/(dx))^(1/3)` `therefore" "(y^(2)((d^(2)y)/(dx^(2)))^(2)+x^(2)y^(2)- sin x)^(3) = - 27x^(3)((dy)/(dx))` Here order = 2 = p Degree = 6 = q `therefore" "p lt q` |
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| 433. |
The solution of the differential equation `e^(-x) (y+1) dy +(cos^(2) x - sin 2x) y (dx) = 0` subjected to the condition that y = 1 when x = 0 isA. `(y+1) + e^(x) cos^(2)x = 2`B. `y + log y = e^(x) cos^(2) x`C. `log(y+1)+e^(x) cos^(2)x = 1`D. `y = log y + e^(x) cos^(2)x = 2` |
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Answer» Correct Answer - D `e^(-x)(y+1)dy+(cos^(2)x-sin 2x) y (dx)=0` `therefore" "(1+y^(-1))dy + e^(x)(cos^(2)x-sin 2x)dx = 0` Integration we get, `y+log_(e) y + e^(x) cos^(2) x = c` y = 1 when x = 0 `therefore" "1+0+1 = c` `therefore" "c = 2` `therefore" ""Curve is y" + log y + e^(x) cos^(2) x = 2` |
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| 434. |
The solutio of the differential equation `(x^(2)-xy^(2))(dy)/(dx)+y^(2)+xy^(2) = 0` isA. `log((x)/(y))=(1)/(x)+(1)/(y)+c`B. `log((y)/(x))=(1)/(x)+(1)/(y)+c`C. `log(xy) = (1)/(x) + (1)/(y)+c`D. `log(xy)+(1)/(x)+(1)/(y)=c` |
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Answer» Correct Answer - A The given equation `(x^(2)-yx^(2))(dy)/(dx)+ y^(2)+xy^(2)=0` `rArr" "(1-y)/(y^(2))dy+(1+x)/(x^(2))dx = 0` `rArr" "((1)/(y^(2))-(1)/(y))dy+((1)/(x^(2))+(1)/(x))dx = 0` On integrating, we get the required solution `log((x)/(y))=(1)/(x)+(1)/(y)+c` |
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| 435. |
`xydx+sqrt(1+x^(2))dy=0`A. `2(1+x^(2))^(3//2)+3logy=c`B. `sqrt(1+y^(2))+logx+c`C. `4sqrt(1+x^(2))+y^(2)=c`D. `sqrt(1+x^(2))+logy=c` |
| Answer» Correct Answer - D | |
| 436. |
The number of arbitraryconstants in the particular solution of a differential equationof third order are:(A) 3 (B) 2 (C) 1 (D)0A. `3`B. `2`C. `1`D. `0` |
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Answer» Correct Answer - D The number of arbitrary constants in the particular solution of the given differential equation is zero because we put some particular values in the general solution of the differential equation to eliminate all constants. |
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| 437. |
Find the curve for which the perpendicular from the foot of theordinate to the tangent is of constant length. |
| Answer» `c^(2)e^((2x)/(k)) - 2cye^((x)/(k))+k^(2) = 0` | |
| 438. |
The solution of differential equation `(1-xy + x^(2) y^(2))dx = x^(2) dy` isA. tan xy = log |cx|B. tan (y/x) = tan log |cx|C. xy = tan log |cx|D. None of these |
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Answer» Correct Answer - C `(1-xy+x^(2)y^(2))dx = x^(2) dy` `(1-(xy)+(xy)^(2))dx = x^(2) dy` Let xy = v. `y = (v)/(x)` `therefore" "(dy)/(dx)=(x*(dv)/(dx)-v)/(x^(2))` `x^(2)dy = xdv - v dx` So,`" "(1-v+v^(2))dx = xdv - vdx` `rArr" "(1+v^(2))dx = xdv` `rArr" "int (dv)/(1+v^(2))=int(dx)/(x)` `rArr" "tan^(-1)v=log|x|+log c` `rArr" "tan^(-1)v = log |cx|` `rArr" "v = tan log |cx|` `rArr" "xy = tan log |cx|` |
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| 439. |
The solution of `((dy)/(dx))^(2)-2(x+(1)/(4x))(dy)/(dx)+1=0`A. `y = x^(2) + c`B. `y = (1)/(2) In (x) + c, x gt 0`C. `y = (x)/(2) + c`D. `y = (x^(2))/(2) + c` |
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Answer» Correct Answer - A::B `rArr" "((dy)/(dx))^(2)-3(x+(1)/(4x))(dy)/(dx)+1 = 0` `rArr" "((dy)/(dx)-2x)((dy)/(dx)-(1)/(2x))=0` `therefore" "dy = 2x dx or dy=(1)/(2x)dx` `rArr" "y=x^(2)+ c or y = (1)/(2)"In x"+c, x gt 0`. |
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| 440. |
If the solution of the equation `(d^(2)x)/(dt^(2))+4(dx)/(dt)+3x = 0` given that for `t = 0, x = 0 and (dx)/(dt) = 12` is in the form `x = Ae^(-3t) + Be^(-t)`, thenA. `A + B = 0`B. `A + B = 12`C. `|AB| = 36`D. `|AB| = 49` |
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Answer» Correct Answer - A::C `x = Ae^(-3t) + Be^(-t)` `therefore" "(dx)/(dt)= -3Ae^(-3t) - Be^(-t)` When t = 0, x = 0 `therefore" "A + B = 0" "(i)` At t = 0, `(dx)/(dt) = 12` `therefore" "12 = -3A - B" "(ii)` Solving, we get A = -6, B = 6. |
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| 441. |
The differential equation for which `sin^(-1) x + sin^(-1) y = c` is given byA. `sqrt(1-x^(2))dy+sqrt(1-y^(2))dx=0`B. `sqrt(1-x^(2))dx+sqrt(1-y^(2))dy=0`C. `sqrt(1-x^(2))dx-sqrt(1-y^(2))dy=0`D. `sqrt(1-x^(2))dy-sqrt(1-y^(2))dx=0` |
| Answer» Correct Answer - A | |
| 442. |
Solution of the differential equation `(dx)/(x)+(dy)/(y)=0` isA. `logx=logy`B. `(1)/(x)+(1)/(y)=c`C. `x+y=c`D. `xy=c` |
| Answer» Correct Answer - D | |
| 443. |
If `(x^2+y^2)dy=xydx` and y(1)=1 and `y(x_o)=e`, then `x_o=`A. `sqrt2e`B. `sqrt3e`C. `sqrt5e`D. `e//sqrt2` |
| Answer» Correct Answer - B | |
| 444. |
The order of the differential equation of family of circles touching two given circles externally isA. 1B. 2C. 3D. none of these |
| Answer» Correct Answer - A | |
| 445. |
Form the differentialequation of the family of circles touching the y-axis at origin. |
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Answer» If the circles touches `Y`-axis at origin then its centre will lie on `X`-axis. `:.` equation of circle `(x-h)^(2)+(y-0)^(2)=h^(2)` `implies x^(2)+y^(2)-2hx=0`……..`(1)` differentiate w.r.t.x, `2x+2y(dy)/(dx)-2h=0` `implies 2h=2x+2y(dy)/(dx)` Put this value in equation `(1)`, `x^(2)+y^(2)-x(2x+2y(dy)/(dx))=0` `implies y^(2)-x^(2)-2xy(dy)/(dx)=0` |
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| 446. |
The general solution of `(dy)/(dx) = 1 - x^(2) -y^(2) + x^(2) y^(2)` isA. `2 sin^(-1) y = x sqrt(1-y^(2))+c`B. `sin^(-1) y = (1)/(2) sin^(-1) x + c`C. `cos^(-1) y = x cos^(-1) x + c`D. `(1)/(2)"log"((1+y)/(1-y))=x-(x^(3))/(3)+c` |
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Answer» Correct Answer - D `(dy)/(dx) = 1 - x^(2) - y^(2) + x^(2)y^(2)` `therefore" "(dy)/(dx)=(1-x^(2))(1-y^(2))` `therefore" "(dy)/(1-y^(2))=(1-x^(2))dx` `therefore" "int (dy)/(1-y^(2))=int (1-x^(2))dx` `therefore" "(1)/(2) log((1+y)/(1-y))=x-(x^(3))/(3)+C` |
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| 447. |
If the differenential equation `(dx)/(3y+f)+(dy)/(px+g)=0` represents a family of circle, then p=A. gB. fC. 4D. 3 |
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Answer» Correct Answer - D `(px+g) dx + (3y + f) dy = 0` Integrating `(px^(2))/(2)+gx+(3y^(2))/(2)+fy+c = 0` `rArr" "p = 3` (for a circle) |
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| 448. |
Form differential equation for `x^(2)=4by`A. `y=2xy_(1)`B. `xy_(1)=2y`C. `xy=2y_(1)`D. `yy_(1)=2x` |
| Answer» Correct Answer - B | |
| 449. |
`(dy)/(dx)+xsin2y=x^(3)cos^(2)y` |
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Answer» Correct Answer - `e^(x^(2))tany=1/2e^(x^(2))(x^(2)-1)+c` The given equation can be expressed as `sec^(2)y(dy)/(dx)+2xtany=x^(3)` Put `tany=z`, so that `sec^(2)y(dy)/(dx)=(dz)/(dx)` Given equation transforms to `(dz)/(dx) +2xz+x^(3)`, which is linear in z. `I.F. = e^(2intxdx)=e^(x^(2))` Therefore, solution is given by `ze^(x^(2))=intx^(3)e^(x^(2))dx+c` or `tanye^(x^(2))=1/2e^(x^(2))(x^(2)-1)+c` (substitute for `x^(2)=t` and then integrate by parts) |
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| 450. |
Find the order and degree of the differential equation. `(d^(3)y)/(dx^(3))=sqrt(x+((dy)/(dx))^(3))` |
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Answer» `(d^(3)y)/(dx^(3))=sqrt(x+((dy)/(dx))^(3))` `implies ((d^(3)y)/(dx^(3)))^(2)=x+((dy)/(dx))^(2)` Here the highest order derivative is `(d^(3)y)/(dx^(3))`. Therefore the order of the differntial equation is `3`. The degree of highest derivative is `2`. Therefore the degree of the differential equation is `2`. |
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