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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.
| 301. |
`x^(2) + xy + y^(2) =100` |
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Answer» `x^(2) + xy + y^(2) =100` Differentiate both sides w.r.t. x `2x + (x (dy)/(dx)+ y .1)+ 2y(dy)/(dx) = 0` `rArr (x + 2y)(dy)/(dx)= - (2x + y)` `rArr (dy)/(dx) = - (2x + y)/(x + 2y)` |
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| 302. |
If `y=sqrt(x+sqrt(x+sqrt(+…..oo)))` then prove that `(2y-1)dy/dx=1.` |
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Answer» `y=sqrt(x+sqrt(x+sqrt(+…..oo)))` `y=sqrt(x+y)` `rArr y^2x+y` `rArr 2dy dy/dx =1 + dy/dx` `rArr (2y -1) dy/dx =1` ` rArr dy/dx =(1)/(2y -1)` |
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| 303. |
If `x^(y)+y^(x)=a^(b)` then find `(dy)/(dx)`. |
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Answer» Let `u=x^(y)andv=y^(x)` `u=x^(y)` `rArrlogu=logx^(y)=ylogx` `rArr1/ucdot(du)/(dx)=y/x+logxcdot(dy)/(dx)` `rArr(du/(dx)=u[y/x+logxcdot(dy)/(dx)]` `=x^(y)[y/x+logxcdot(dy)/(dx)]` `=ycdotx^(y-1)+x^(y)cdotlogxcdot(dy)/(dx)` and `v=y^(x)` `rArrlogv=logy^(x)=xlogy` `rArr1/v(dv)/(dx)=x/ycdot(dy)/(dx)+logycdot1` `rArr(dv)/(dx)=v(x/y(dy)/(dx)+logy)` `=y^(x)(x/y(dy)/(dx)+logy)` `=xcdoty^(x-1)cdot(dy)/(dx)+y^(x)logy` Now `x^(y)+y^(x)=a^(b)` `rArru+v=a^(b)rArr(du)/(dx)+(dv)/(dx)=0` `rArrycdotx^(y-1)+x^(y)logxcdot(dy)/(dx)` `+xcdoty^(x-1)(dy)/(dx)+y^(x)logy=0` `rArr(dy)/(dx)(x^(y)logx+xcdoty^(x-1))=-(y^(x)logy+ycdotx^(y-1))` `rArr(dy)/(dx)=(-(y^(x)logy+ycdotx^(y-1))/((x^(y)logx+xcdoty^(x-1))` Ans. |
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| 304. |
If `y= tan ^-1 ((x)/sqrt(a^2-x^2))` then `dy/dx =? `A. `1/(sqrt(a^2-x^2))`B. `1/(sqrt(x^2-a^2))`C. `1/(sqrt(a^2+x^2))`D. `1/(sqrt(1 +x^2))` |
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Answer» Correct Answer - A |
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| 305. |
If `y=tan^(-1){(sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))}`, `-1 |
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Answer» Correct Answer - `(-x)/(sqrt(1-x^4))` |
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| 306. |
`sin^(3)(ax+b)` |
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Answer» Correct Answer - `3a sin ^2 (ax +b) cos (ax +b)` |
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| 307. |
If `x^2+y^2=25 " then find " dy/dx.` |
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Answer» `x^2+y^2=25` Differenatiate both sides with respect to x `2x+2y dy/dx =0 ` `rArr dy/dx =-x` |
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| 308. |
`sin^(-1)(ax)` |
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Answer» Correct Answer - `(a)/(sqrt(1-a^2x^2))` |
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| 309. |
`sqrt(ax^2+bx+c)` |
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Answer» Correct Answer - `(2ax+b)/(2 sqrt(ax^2+bx +c))` |
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| 310. |
`xy + y^(2) = tan x + y ` |
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Answer» `xy + y^(2) = tan x + y ` Differentiate both sides w.r.t. x `(x(dy)/(dx)+y.1)+ 2y(dy)/(dx)= sec^(2)x + (dy)/(dx)` `rArr (dy)/(dx) (x + 2y - 1)=sec^(2)x - y` `rArr (dy)/(dx) =(sec^(2) x - y)/(x + 2y-1)` |
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| 311. |
`2sqrt(cot(x^2))` |
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Answer» Let y `2sqrt(cot(x^(2)) ` `rArr (dy)/(dx) = 2 (d)/(dx)sqrtcot (x^(2))` `= 2(d)/(dx)[cot (x^(2))] ^(1//2).` `= 2(1)/(2)[cot (x^(2))] ^(1//2).(d)/(dx) cot(x^(2))` `= (-cosec^(2)(x^(2)))/ sqrt(cot(x^(2)))(d)/(dx)(x^(2))` `= (2xcosec^(2)(x^(2)))/ sqrt(cot(x^(2)))` |
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| 312. |
If `y=(sin x) ^(sin x ^sin x ....oo)` then prove that `dy/dx=(y^2 cot x)/(1-y log (sin x))` |
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Answer» `y=(sin x) ^(sin x ^sin x ....oo)` `rArr y=(sin x)^y ` `rArr log y =log (sin x)^y=y log (sin x)` Differentiate both sides with respect to x `(1)/(y)dy/dx =y.1/sin x .d/dx sin x +log (sin x) .dy/dx ` `rArr dy/dx (1/y-log sin x )=y/sin x .cos x` `rArr dy/dx (1-y log sin x)/(y)=y cot x ` `rArr dy/dx =(y^2 cot x)/1- y log sin x` |
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| 313. |
If ` y = tan ^-1[x/(1+sqrt(1-x^2))] "then find" dy/dx` |
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Answer» Let `x=sin theta` ` therefore y=tan ^-1[(sin theta)/(1+sqrt(1-sintheta))]` `=tan^-1[(sin theta)/(1+cos theta)]=tan ^-1[(2sin ""theta/2cos ""theta/2)/(2cos^2""theta/2)]` `tan^-1(tan""theta/2)=theta/2=1/2sin^-1x` `rArr dy/dx=1/2d/dxsin^-1x=1/(2sqrt(1-x^2))` |
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| 314. |
` sqrt(cot^(-1)sqrt(x))` |
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Answer» Correct Answer - `(-1)/(4sqrt(xcot^(-)1+cos^2 sqrt(x)).(1+x))` |
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| 315. |
`ax + by^(2) = cos y ` |
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Answer» `ax + by^(2) = cos y ` Differentiate both sides w.r.t. x `a + 2 by (dy)/(dx)= - sin y (dy)/(dx)` `rArr (2by + sin y) (dy)/(dx) = - a rArr (dy)/(dx) = (-a)/(2by + sin y) ` |
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| 316. |
If `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `f(x)` isA. continuous at `x=pi//2`B. `underset(x to pi//2)lim f(x)` but f(x) is not continuous at `x=pi//2`C. `underset(x to pi//2)lim f(x)` does not existD. `underset(x to pi//2^(-))lim f(x) =-1` |
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Answer» Correct Answer - B We have `underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim f(pi//2-h)` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin {cos (pi//2-h)})/(-h)` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin{sin h})/(-h)2` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin(sin h))/(-h)` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin(sin h))/(sin h)xx(sin h)/(h)=-1` and `underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim f(pi//2+h)` `Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin {cos (pi//2+h)})/((pi)/(2)+h-(pi)/(2))` `Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin{-sin h})/(h)` `Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin{sin h})/(h)xx(sin h)/(h)=-1` `"Clearly", underset(x to pi//2^(-))lim f(x)=underset(x to pi//2^(+))lim f(x) ne f((pi)/(2))` `"So", underset(x to pi//2)lim` f(x) exists but f(x) is not continuous at `x=pi//2`. |
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| 317. |
Find `(dy)/(dx),`if `y=12(1-cost), x=10(t-sint),-pi/2 |
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Answer» `y=12(1-cost)` `implies(dy)/(dt)=12sint` `x=10(t-sint)` `implies(dx)/(dt)=10(1-cos t)` ` :. (dy)/(dx)=(dy//dt)/(dx//dt)=(12sint)/(10(1-cost))` `=(6)/(5)*(2"sin"(t)/(2)"cos"(t)/(2))/(2"sin"^(2)(t)/(2))=(6)/(5)"cot"(t)/(2)` |
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| 318. |
It is given that forthe function `f`given by `f(x)=x^3+b x^2+a x`, `x in [1, 3]`. Rolles theorem holdswith `c=2+1/(sqrt(3))`. Find the values of `a`and `b`. |
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Answer» Correct Answer - a=11 , b=-6 |
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| 319. |
`cos (sqrt(x))` |
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Answer» Let y = `cos (sqrt(x))` `rArr (dy)/(dx)= (d)/(dx) cos (sqrt(x))` `= -sin (sqrt(x)). (d)/(dx) sqrt(x) = - (sin (sqrt(x)))/(2sqrt(x))` |
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| 320. |
If `y= cot^(-1)(sqrt(1+x^2+1)/x) " then find " dy/dx` |
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Answer» Let x=tan theta therefore `y=cot ^(-1)(sqrt((1+tan ^2 theta +1)/(tan theta)))` `=cot ^(-1)((sec theta+1)/(tan theta))` `=cot ^(-1)((1+cos theta)/(sin theta))` `cot ^-1 ((2cos^2""theta/(2))/(2sin""theta/2cos""(theta)/(2)))=cot^(-1)(cot""theta/2)` `=theta/2=1/2 tan ^(-1)x` `rArr dy /dx=1/2 d/dx tan ^(-1)=(1)/(2(1+x^2))` |
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| 321. |
If `alpha, beta(alpha,beta)` are the points of discontinuity of the function `f(f(x))`, where `f(x)=1/(1-x)`, then the set of values of a foe which the points `(alpha,beta)` and `(a,a^2)` lie on the same side of the line `x+2y-3=0` , isA. `(-3//2,1)`B. `[-3//2,1]`C. `[1,oo)`D. `(-oo,-3//2]` |
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Answer» Correct Answer - A We have `f(x)=(1)/(1-x)` Clearly, f(x) is defined for all `x in R -(1)}` Now, `f(f(x))=f((1)/(1-x))=(1)/(1-(1)/(1-x))=(x-1)/(x)` We find that f(f(x)) is defined for all `x ne 0,1` Now, `f(f(x))=f((x-1)/(x))=(1)/(1-(x-1)/(x))="x for all x"in 0,1` Thus, the set of points of discontinuity is `{0,1}` the line x+2y-3=0. Therefore, `(0+2-3)(a+2a^(2)-3)gt0` `Rightarrow 2a^(2)+a-3 lt 0` `Rightarrow (2a+3) (a-1) lt 0 Rightarrow -3//2 lt a lt 1` `Rightarrow a in (-3//2,1)` |
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| 322. |
If `y=log(sinx)`, prove that `(d^3y)/(dx^3)=2cosx cos e c^3x`. |
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Answer» Correct Answer - `2cosec^2x . Cotx` |
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| 323. |
Find `(dy)/(dx)` in the following `2x + 3y = sin x ` |
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Answer» `2x + 3y = sin x ` Differentiate both sides w.r.t. x `2+ 3(dy)/(dx)= cos x ` `rArr 3(dy)/(dx) = cos x -2 rArr (dt)/(dx)= (cos x -2)/(3)` |
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| 324. |
`2x + 3y = sin y ` |
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Answer» `2x + 3y = sin y ` Differentiate both sides w.r.r. x `2 + 3 (dt)/(dx)= cos y(dy)/(dx)` `rArr 2 = (cos y - 3 ) (dy)/(dx) rArr (dy)/(dx) = (2)/(cos y - 3)` |
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| 325. |
If `x=a(t-sint), y=a(1-cost)` then find `(d^2y)/(dx^2)`. |
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Answer» Correct Answer - `-1/(4a)cosec^4 t/2` |
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| 326. |
if `y= (sin ^(-1)x)/(sqrt(1-x^2))` then prove that `(1-x)^2) .d/dx=xy+1` |
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Answer» `y=(sin^1x)/(sqrt(1-x^(2)))` `rArr dy/dx=d/dx((sin^(-1)x)/(sqrt(1-x^(2))))` `sqrt(1-x^(2))d/dxsin ^(-1)x-sin^(-1)x` `=(" ".d/dxsqrt(1-x^(2)))/((sqrt(1-x^(2)))^2)` `sqrt(1-x^(2)).(1)/(sqrt(1-x^(2)))-sin^(-1)x.(1)/(2sqrt(1-x^(2)))` `=(" "d/dx(1-x^(2)))/((1-x^(2))` `rArr (1-x^(2))dy/dx=1-(sin^(-1)x)/(sqrt(1-x^(2))).(1)/(2).(-2x)` `rArr (1-x^(2))d/dx=1 +x.y` Hence Proverd |
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| 327. |
If `y = sin ^(2) alpha + cos ^(2) (alpha + beta) + 2 sin alpha sin betacos(alpha+beta)` then `(d^(3)y)/(da^(3))=?`A. `sin^(3) (alpha +beta)/(cosalpha)`B. `sin ( alpha + beta)`C. 0D. 1 |
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Answer» Correct Answer - C |
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| 328. |
if `x^2+y^2 = t - 1/t` and` x^4 + y^4 = t^2 + 1/t^2 `then prove that `dy/dx = 1/(x^3y)` |
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Answer» Correct Answer - B |
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| 329. |
xsin 2y = y cos 2x |
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Answer» Correct Answer - `(2ysin2x+sin 2y)/(cos^2x-2xcos^(2)2y)` |
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| 330. |
`5x^2+5y^2-7y+3x=2=0` |
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Answer» Correct Answer - `(10x+3)/(7-10y)` |
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| 331. |
If `y^(1/m)= x + sqrt (1 + x^(2)) "then" (1 + x^(2))y_(2)+ xy _(1) = ?`A. myB. `m^(2)y`C. `m^(2)y^(2)`D. None of these |
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Answer» Correct Answer - B |
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| 332. |
`x^2+y^2=log(xy)` |
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Answer» Correct Answer - `(y(1-2x^2))/(x(2y^2-1))` |
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| 333. |
`sin(xy)+x/y=x^2-y` |
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Answer» Correct Answer - `(2xy^2-y-y^3cos(xy))/(xy^2cos(xy)-x+y^2)` |
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| 334. |
`"If "xy=e^((x-y))," then find "(dy)/(dx).` |
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Answer» `xy = e^(x-y) " "....(1)` Differentiate both sides w.r.t.x ` x(dy)/(dx) + y * 1 = (d)/(dx)e^((x-y))` `rArr x(dy)/(dx) + y = e^((x-y)) * (d)/(dx) (x-y)` `rArr x(dy)/(dx) + y = xy(1-(dy)/(dx)) " From equation" (1)` `rArr x(dy)/(dx) + xy(dy)/(dx) = xy-y` `rArr (dy)/(dx) = (y(x-1))/(x(1+y))` |
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| 335. |
If `f(x)= x^2+7x+10` then `f(2) =? `A. -4B. `-5/2`C. -11D. 11 |
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Answer» Correct Answer - D |
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| 336. |
log tan x |
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Answer» Correct Answer - 2 cosec 2x |
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| 337. |
`x=a(cos t + log tan t/2), y =a sin t` |
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Answer» Correct Answer - tan t |
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| 338. |
`(cosx)^(y)=(siny)^(x)` |
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Answer» Correct Answer - `(logsiny+ytanx)/(logcosx-xcoty)` |
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| 339. |
If `(cosx)^y=(cosy)^x`find `(dy)/(dx)`. |
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Answer» `("cos" x)^(y) = ("cos" y)^(x)` `rArr "log" ("cos" x)^(y) = "log" ("cos" y)^(x)` `rArr y"log cos" x = x"log cos"y` Differentiate both sides w.r.t.x `y * (d)/(dx) "log cos"x + "log cos"x (d)/(dx)y` `= x (d)/(dx)"log cos"y + "log cos"y(d)/(dx) x` `rArr y * ((-"sin"x))/("cos"x) + "log cos"x * (dy)/(dx)` `=(x(-"sin"y))/("cos"y)(dy)/(dx) + "log cos"y * 1` `rArr -y"tan"x + "log cos"x (dy)/(dx)` `= -x"tan"y (dy)/(dx) + "log cos"y` `rArr ("log cos" x + x"tan"y)(dy)/(dx) = "log cos" y + y"tan"x` `rArr (dy)/(dx) = ("log cos"y + y"tan"x)/("log cos"x + x"tan"y)` |
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| 340. |
Differentiate `(x^2-5x+8)(x^3+7x+9)` in three ways mentioned below:(i) by using product rule(ii) by expanding the product to obtain a single polynomial.(iii) by logarithmic differentiation.Do they all give the same answer? |
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Answer» `"Let "y = (x^(2) -5x + 8) (x^(3) + 7x +9)` `(i) (dy)/(dx) = (x^(2) -5x + 8) (d)/(dx) (x^(3) + 7x +9) + (x^(3) + 7x +9) (d)/(dx) (x^(2) -5x + 8)` `=(x^(2) - 5x + 8)(3x^(2) + 7) + (x^(3) + 7x + 9)(2x - 5)` `= 3x^(4) + 7x^(2) - 15x^(3) - 35x + 24x^(2) + 56 + 2x^(4) - 5x^(3) + 14x^(2) - 35x + 18x -45` `= 5x^(4) - 20x^(3) + 45x^(2) - 52x + 11` `(ii) y = (x^(2) - 5x + 8)(x^(3) + 7x + 9)` `= (x^(5) - 5x^(4) + 15x^(3) - 26x^(2) + 11x +72)` `rArr (dy)/(dx)= (d)/(dx) (x^(5) - 5x^(4) + 15x^(3) - 26x^(2) + 11x +72)` `= 5x^(4) - 20x^(3) + 45x^(2) - 52x + 11` `(iii) y= (x^(2) - 5x + 8)(x^(3) + 7x + 9)` `rArr "log"y = "log" [(x^(2) - 5x + 8) * (x^(3) + 7x + 9)]` `= "log" (x^(2) - 5x + 8) +"log" (x^(3) + 7x + 9)` Differentiate both side w.r.t.x `(1)/(y) (dy)/(dx) = (2x - 5)/(x^(2) - 5x + 8) + (3x^(2) + 7)/(x^(3) + 7x + 9)` `rArr (dy)/(dx) = y[((2x - 5)(x^(3) + 7x + 9) + (3x^(2) + 7)(x^(2) - 5x + 8))/((x^(2) - 5x + 8)(x^(3) + 7x + 9))]` `= y[(2x^(4) - 5x^(3) + 14x^(2) - 35x + 18x - 45 + 3x^(4) + 7x^(2) - 15x^(3) - 35x + 24x^(2) + 56)/(y)]` `= 5x^(4) - 20x^(3) + 45x^(2) - 52x + 11` Thus, the results in three cases are same. |
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| 341. |
Prove that tangent function is continous in its domain. |
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Answer» Let f(x) = tan x its domain `=R-{(2n+1)pi/2,n in Z}` `therefore` f(x)= tan x is continuous in `R-{(2n+1)pi/2:n in Z}` Hence proved |
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| 342. |
Find `(dy)/(dx)`, if ` y + siny = cosx` |
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Answer» Differentiating with respect to x, `(dy)/(dx)+cosy(dy)/(dx)=-sinx` `(dy)/(dx)=-sinx/(1+cosy)` |
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| 343. |
If u, v and w are functions of x, then show that`d/(dx)(udotvdotw)=(d u)/(dx)vdotw+udot(d v)/(dx)dotw+udotv(d w)/(dx)`in two ways - first by repeated application of product rule, second by logarithmic differentiation. |
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Answer» `(i) (dy)/(dx) = (d)/(dx) {(uv)w}` `= w * (d)/(dx) (uv) + uv * (d)/(dx)(w)` `= w[v(du)/(dx) + u(dv)/(dx)] + uv(dw)/(dx)` `=vw (du)/(dx) + uw (dv)/(dx) + uv(dw)/(dx)` (ii) y = uvw `rArr "log"y = "log (uvw)` `="log" u + "log"v + "log"w` Differentiate both sides w.r.t.x `(1)/(y) (dy)/(dx) = (1)/(u) (du)/(dx) + (1)/(v) (dv)/(dx) + (1)/(w) (dw)/(dx)` `rArr (dy)/(dx) = (y)/(u) (du)/(dx) + (y)/(v) (dv)/(dx) + (y)/(w) (dw)/(dx)` `= (uvw)/(u) (du)/(dx) + (uvw)/(v) (dv)/(dx) + (uvw)/(w) (dw)/(dx)` `= vw(du)/(dx) + uw(dv)/(dx) + uv(dw)/(dx)` Thus, the results in two cases are same. |
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| 344. |
In the curve `x= a (cos t+ log tan(t/2))`,` y =a sin t`. Show that the portion of the tangent between the point of contact and the x-axis is of constant length. |
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Answer» `x=a("cos"t + "log tan"(t)/(2)) " and "y=a"sin"t` `rArr (dx)/(dt)=a[-"sin"t + ("sec"^(2)(t)/(2))/(2"tan"(t)/(2))] "and" (dy)/(dx) = a"cos"t` `=a[-"sin"t + (1)/(2"tan"(t)/(2) "cos" (t)/(2))]` `=a(-"sin"t + (1)/("sin"t))` `=a((1-"sin"^(2)t)/("sin"t))= a("cos"^(2)t)/("sin"t)` `"Now", (dy)/(dx) = (dy//dt)/(dx//dt) = ((a"cos"t)/(a"cos"^(2)t))/("sin"t) = "tan"t` |
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| 345. |
If x and y are connected parametrically by the equations given, without eliminating the parameter, Find `(dy)/(dx)`.`x=costheta-cos2theta, y=sintheta-sin2theta` |
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Answer» `(deltay)/(deltatheta)=costheta-2cos2theta` `(deltax)/(deltatheta)=sintheta+2sintheta` `(deltay)/(deltax)=(costheta-2cos2theta)/(-sintheta+2sin2theta)` |
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| 346. |
If x and y are connected parametrically by the equations given, without eliminating the parameter, Find `(dy)/(dx)`.`x=2a t^2, y=a t^4` |
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Answer» `{:(x=2at^(2), |, y=at^(4)),(rArr (dx)/(dt) = 4at,|,rArr (dy)/(dt) = 4at^(3)):}` `therefore (dy)/(dx) = (dy//dt)/(dx//dt) = (4at^(3))/(4at) = t^(2)` |
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| 347. |
If x and y are connected parametrically by the equations given, without eliminating the parameter, Find `(dy)/(dx)`.`x=(sin^3t)/(sqrt(cos2t)), y=(cos^3t)/(sqrt(cos2t))` |
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Answer» `y = cos^3t/sqrt(cos2t)` `dy/dt = 1/(cos2t)(sqrt(cos2t)(3cos^2t)(-sint) - cos^3t(1/2(1/(sqrt(cos2t))(-sin2t)(2)))` `dy/dt = 1/(cos2t)(-sqrt(cos2t)(3cos^2t)(sint) +(cos^3tsin2t)/(sqrt(cos2t)))` `dy/dt = 1/((cos2t)(sqrt(cos2t)))(cos^3tsin2t-3cos^2tsintcos2t)->(1)` `x = sin^3t/sqrt(cos2t)` `dx/dt = 1/(cos2t)(sqrt(cos2t)(3sin^2t)(cost) - sin^3t(1/2(1/(sqrt(cos2t))(-sin2t)(2)))` `dx/dt = 1/(cos2t)(sqrt(cos2t)(3sin^2t)(cost) +(sin^3tsin2t)/(sqrt(cos2t)))` `dx/dt = 1/((cos2t)(sqrt(cos2t)))(3sin^2tcostcos2t + sin^3tsin2t)->(2)` Dividing (1) by (2), `dy/dx = (cos^3tsin2t-3cos^2tsintcos2t)/(3sin^2tcostcos2t + sin^3tsin2t)` |
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| 348. |
If `x=sin^3t/(sqrtcos2t), y=cos^3t/sqrt(cos2t)` show that `dy/dx =0 at t=pi/6` |
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Answer» `x = ("sin"^(3)t)/(sqrt("cos" 2t))` `rArr (dx)/(dt) = (sqrt("cos"2t) * (d)/(dt)"sin"^(3)t - "sin"^(3)t(d)/(dt) sqrt("cos"2t))/(sqrt("cos" 2t))^(2)` `= (sqrt("cos"2t) * 3"sin"^(2)t "cos"t - "sin"^(3)t * ((-2"sin"2t)/(2sqrt("cos"2t))))/("cos" 2t)` `= ("cos"2t * 3"sin"^(2)t "cos"t + "sin"^(3)t "sin"2t)/("cos" 2tsqrt("cos"2t))` `= ("sin"^(2)t[3 "cos"t " cos"2t + "sin"t " sin"2t])/("cos" 2tsqrt("cos"2t))` `= ("sin"^(2)t[3 "cos"t (1-2"sin"^(2)t) + "sin"t * 2"sin"t " cos"t])/("cos" 2tsqrt("cos"2t))` `= ("sin"^(2)t "cos" t(3-4 "sin"^(2)t))/("cos" 2tsqrt("cos"2t))` `" and "y = ("cos"^(2)t)/(sqrt("cos"2t))` `rArr(dy)/(dt) = (sqrt("cos"2t)(d)/(dt)"cos"^(3)t-"cos"^(3)t(d)/(dt)sqrt("cos"2t))/((sqrt("cos"2t))^(2)` `= (sqrt("cos"2t) * 3"cos"^(2) t (-"sin"t)-"cos"^(3)t * ((-2"sin"2t))/(2sqrt("cos"2t)))/("cos" 2t)` `=-((3"cos"^(2) t "sin"t * "cos" 2t - "sin" 2t "cos"^(3) t))/("cos" 2tsqrt("cos"2t))` `=-("cos"^(2)t[3"sin"t(2 "cos"^(2)t-1)-2 "sin" t "cos"t * "cos"t])/("cos" 2t sqrt("cos"2t))` `=-("cos"^(2)t"sin"t(4"cos"^(2)t-3))/("cos" 2t sqrt("cos"2t))` `Now (dy)/(dx) = (dy//dt)/(dx//dt) = -("cos"^(2)t"sin"t(4"cos"^(2)t-3))/("sin"^(2)t"cos"t(3-4"sin"^(2)t))` `=- ("cos"t(4"cos"^(2)t-3))/("sin"t(3-4"sin"^(2)t))` `=- ((4"cos"^(3)t-3"cos"t))/((3"sin"t-4"sin"^(3)t))` `=-("cos"3t)/("sin"3t) = -"cos" 3t` |
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| 349. |
`x=3sin t-2 sin ^3t, y=3 cos t -2cos^3t` |
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Answer» Correct Answer - tan t |
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| 350. |
Let `f: RvecR`be any function. Also `g: RvecR`is defined by `g(x)=|f(x)|`for all `xdot`Then isOnto if `f`is ontoOne-one if `f`is one-oneContinuous if `f`is continuousNone of theseA. onto if f is ontoB. one-one if f is one-oneC. continuous if f is continuousD. None of these |
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Answer» Correct Answer - C (a) Since g(x) = `|f(x)|` `therefore" "g(x) ge 0` `therefore" Range of "g ne R". Hence, is not onto"`. (b) If we take `f(x)=x`, then f is one-one but `|f(x)|=|x|` is not one-one. (c) If f(x) is continuous then `|f(x)|` is also continuous. |
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