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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.
| 251. |
`cosec^(-1)((1+tan ^2x)/(2tanx))` |
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Answer» Correct Answer - 2 |
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| 252. |
`cosec^(-1)3x` |
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Answer» Correct Answer - `(-1)/(xsqrt(9x^2-1))` |
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| 253. |
If y=x sin y then prove that `xdy/dx=(y)/(1-x cos y)` |
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Answer» y=sin x `rArr dy/dx(1-x cos y ) =sin y` `rArr dy/dx = (sin y )/(1-x cos y )=(y)/(x(1-x cos y)).` `rArr x.(dy)/(dx)=(y)/((1-x cos y )).` |
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| 254. |
`sqrt(secx)` |
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Answer» Correct Answer - `1/2 sqrt(sec x ). tan x ` |
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| 255. |
`y=(sinx)^(cosx)+(cosx)^(sinx)` |
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Answer» Correct Answer - `(sinx)^(cosx)[cotxcdotcosx-sinxlog(sinx)]+)cosx)^(sinx)[cosxcdotlog(cosx)-sinxcdottanx]` |
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| 256. |
If `y=secx+tanx` then prove that `(d^2y)/(dx^(2))=cosx/((1-sinx)^(2))`. |
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Answer» `y=secx+tanx` `=1/cosx+sinx/cosx=(1+sinx)/cosx` ,brgt `rArr(dy)/(dx)=d/(dx)((1+sinx)/cosx)` `(cosxd/(dx)(1+sinx)-(1+sinx)d/(dx)cosx)/(cos^(2)x)` `=(cosx.cosx+(1+sinx).sinx)/(cos^(2)x)` `=(1+sinx)/(1-sin^2x)=1/(1-sinx)` `rArr (d^(2)y)/(dx^2)=d/(dx)(1/(1-sinx))` `=((1-sinx).d/(dx)(1)-1d/(dx)(1-sinx))/((1-sinx)^(2))` `=(0+cosx)/((1-sinx)^(2))=cosx/((1-six)^(2))` |
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| 257. |
`tan (e^x+5)` |
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Answer» Correct Answer - `e^(x).sec^2(e^x +5)` |
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| 258. |
`y=sqrt((x^(2)+x+1)/(x^(2)-x+1))` |
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Answer» Correct Answer - `sqrt((x^(2)+x+1)/(x^(2)-x+1))cdot((1-x^(2)))/(x^(4)+x^(2)+1)` |
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| 259. |
cosec`sqrt(x)` |
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Answer» Correct Answer - `(-cosec sqrt(x)cot sqrt(x))/(2sqrt(x))` |
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| 260. |
The number of points in `(1,3)`, where `f(x) = a(([x^2]),a>1` is not differential is |
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Answer» Correct Answer - D Let `g(x)=x^(2)`. Then g(x) is an increasing function on (1,3) such that g(1)=1 and g(3)=9 on (1,3) such that g(1)=1 and g(3)=9 Clearly, `[g(x)]=[x^(2)]` is discontinuous and hence non-differentiable at `x=sqrt(2),sqrt(3),sqrt(4),sqrt(5),sqrt(6),sqrt(7) and sqrt(8)`. Hence, `f(x)=a^([x^(2)]` is non differentiable at x=0, then |
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| 261. |
If `sqrt(1-x^(2)) + sqrt(1-y^(2))=a(x-y)`, then prove that `(dy)/(dx) = sqrt((1-y^(2))/(1-x^(2)))` |
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Answer» `therefore sqrt(1-x^2)+sqrt(1-y^2)=a (x-y)` `rArr sqrt(1-sin^2A)+sqrt(1-sin^2 )=B=a (sinA-sin B)` `rArr cosA +cos B = a(sin A-sin B)` `rArr 2cos""(A+B)/2.cos"" (A-B)/2=a 2 cos ""(A+B)/2 sin""(A-B)/2` `rArr cos""(A-B)/2=a.sin""(A-B)/2` `rArr cos""(A-B)/2=cot^-1a` `rArr (A-B)/2=cot^-1a` `rArr A-B = 2 cot^-1a` `rArr sin^-1x sin ^1y=2 cot ^-1 a` Differentiate both sides with respect to x `1/sqrt(1-x)-1/sqrt(1-y)dy/dx=0` `rArr dy/dx=(sqrt(1-y^2))/(sqrt(1-x^2))` |
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| 262. |
`y=((x+1)^(2)cdot sqrt(x-1))/((x+3)^(3)e^(x))` |
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Answer» Correct Answer - `((x+1)^(2)sqrt(x-1))/((x+3)^(3)cdote^(x))[2/(x+1)+1/(2(x-1))-3/(x+3)-1]` |
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| 263. |
`"cos" e^(x)` |
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Answer» Correct Answer - `-e^x sin e^x` |
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| 264. |
`e^("sin x")` |
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Answer» Correct Answer - `e^(sinx) cos x` |
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| 265. |
`y=(sinx)^(tanx)+(cosx)^(secx)` |
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Answer» Correct Answer - `(sinx)^(tanx)[1+log(sinx)cdot sec^(2)x]+(cosx)^(secx)cdot secxsecxtanx[log(cosx)-1]` |
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| 266. |
If `y=x^(sin^(-1)x)` then finde `(dy)/(dx)`. |
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Answer» `y=x^(sin^(-1)x)` Taking log on both sides `logy=log(x^(sin^(-1)x)` `=sin^(-1)xcdotlogx` Differentiate both sides with respect to x. `(1/y)(dy)/(dx)=sin^(-1)xcdot1/x+logxcdot1/(sqrt(1-x^(2)))` `rArr (dy)/(dx)=ycdot[(sin^(-1)x)/x+(logx)/(sqrt(1-x^(2)))]` `rArr(dy)/(dx)=x^(sin^(-1)x)[(sin^(-1)x)/x+(logx)/(sqrt(1-x^(2)))]`cdotAns |
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| 267. |
`sqrt((1-tanx)/(1+tanx))` |
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Answer» Correct Answer - `(-sec^2x)/((1+ tan x )sqrt(1-tan ^2 x ))` |
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| 268. |
Differentiate `tan^-1""((2x)/(1-x^2))` with respect to `cos ^-1 ((1-x^2)/(1+x^2))` |
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Answer» Let `y_(1)=tan^-1""(2x)/(1-x^2)=2 tan^-1 x` `rArr dy_(1)/dx =2 d/dx tan^-1 x=(2)/(1+x^2)` ` Let y_(2)=cos ^-1((1-x^2)/(1+x^2))` and x tan `theta ` `therefore y_(2)=cos ^-1""(1-tan^2 theta)/(1+tan^2theta)` `rArr (dy_2)/(dx)=2 d/dx . tan ^-1x=2/(1+x^2)` `now dy_1/dy_2=(dy_1//dx)/(dy_2//dx)=(2//(1+x^2))/(2//(1+x^2))=1.` |
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| 269. |
Let `f(x)={2/(1+x^2),xi sr a t ion a l b ,xi sr a t ion a l`has exactly two points of continuity then the value of `b`are`(0,3]`b. `[0,1]`c. `(0,2]`d.`varphi`A. (0,3]B. [0,1]C. (0,2]D. `phi` |
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Answer» Correct Answer - C `f(x)={{:((2)/(1+x^(2))",","x is irrational"),(b",","x is rational"):}` has exactly two points of continuity then equation `(2)/(1+x^(2))=b` must have two distinct roots. Now `(2)/(1+x^(2)) in(0,2]` `rArr" "b in {0,2]` |
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| 270. |
`e^(tanx)` |
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Answer» Correct Answer - `e^(tan x).sec^2x` |
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| 271. |
Find the 2nd derivative of `x^(6).e^(6x)` with respect to x. |
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Answer» Let `y=x^(3).e^(6x)` `rArr(dy)/(dx)=d/(dx)(x^(3).e^(6x))` `=x^(3).d/(dx)e^(6x)+e^(6x).d/(dx)x^(3)` `=x^(3).6e^(6x)+e^(6x).3x^(2)` `=6.x^3.e^(6x)+3.x^2.e^(6x)` `rArr (d^(2)y)/(dx^(2))=6.d/dx(x^3.e^(6x))+3.d/dx(x^2.e^(6x))` `=6[x^3. 6.e^(6x)+e^(6x).3x^2]` `+3[x^2 . 6 e^(6x)+e^(6x).2x]` `=e^(6x)[36x^3+18x^2+18x^2+6x]` `=6x . e^(6x)(6x^2+6x+1)`. |
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| 272. |
The function f(x)=(x) where (x) denotes the smallest integer `ge x` isA. everywhere continuousB. continuous at x=n, `n in Z`C. continuous on R-ZD. none of these |
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Answer» Correct Answer - C For any `n in Z`, we have `("LHL at x=n")=underse(x to n^(-))lim f(x)` `Rightarrow ("LHL at x=n")=underse(h to 0)lim f(n-h)=underset(h to 0)lim (n-h)=n` `("RHL at x=n")=underset(x to n^(+))lim f(x)` `Rightarrow ("RHL at x=n")=underset(h to 0)lim f(n+h)=n+1` `therefore underset(x to n^(-))lim f(x)ne underset(x to n^(+))lim f(x)` Let `x=a in R-Z`. Then, there exists `n in Z` such that `n lt a lt n+1` Now, `underset(x to a^(-))lim f(x)=underset(x to 0)limf(a-h)=underset(h to 0)lim (a-h)=n+1` `underset(x to a^(+))lim f(x)=underset(h to 0)limf(a+h)=underset(h to 0)lim (a+h)=n+1` `and, f(a)=n+1` `therefore underset(x to a^(-))lim f(x)=underset(x to a^(+))lim f(x)=f(a)` There f(x) is a continuous at x=a Hence, f(x) is a continous at all points other than integer points. |
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| 273. |
Let `f(x)=[t a n x[cot x]],x [pi/(12),pi/(12)]`, (where [.] denotes the greatest integer less than orequal to`x`). Then the number of points, where `f(x)`is discontinuous isa. oneb. zero``c. threed. infiniteA. oneB. zeroC. threeD. infinite |
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Answer» Correct Answer - C We `x in I`. `rArr" "tanx[cotx]=1` `rArr" "[tanx[cotx]lt1` When, `x in I`, `[cotx]ltcotx` `rArr" "0 lt tanx[cotx]lt1` `rArr" "[tanx[cotx]]=0` `rArr" "f(x)=[tanx[cotx]]={{:(1",",cotx in I),(0",",cot x cancelinI):}` So, f(x) is discontinuous when `cot x in I` Now `(pi)/(12)le x lt(pi)/(2)` `rArr" "0 lt cot x le2+sqrt3` Hence, number of points of discontinuity are `x=cot^(-1)3, cot^(-1)2` and `cot^(-1)1=(pi)/(4)`. Thus three points of discontinuity. |
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| 274. |
Statement-1: Let `f(x)=[3+4sinx ]`, where [.] denotes the greatest integer function. The number of discontinuities of f(x) in `[pi,2pi]` is 6 Statement-2: The range of f is `[-1,0,1,2,3]`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D We have `-1 le sin x le 0 "for all "x in [pi,2pi]` `Rightarrow -4 le 4sin x le 0 "for all "x in [pi,2pi]` `Rightarrow -1 le 4sin x le 3 le 3" for all "x in [pi,2pi]` `Rightarrow f(x)=[4 sinx+3]"assumes values"-1,0,1,2 and 3` when `x in [pi,2pi]` `Rightarrow "Range f"=[-1,0,1,2,3]` So, statement-2 is true Clearly, there are eight of discontinuity of f(x) in `[pi,2pi]`. So, statment-1 is true. |
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| 275. |
`Differential `e^(tanx)`with respect to sin x. |
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Answer» `Let y_(1)=e^(tanx) " and " y_(2)=sin x ` `therefore dy_(1)/dx=d/dx e tan x` `=e^tan x.d/dxe^tan x` `e^tan x.s ec^2 x ` and `dy_(1)/dy_(2)=(dy_1//dx)/(dy_2//dx)=(e^tan x .sec^2 x )/(cos x)` `=e^(tan x).sec ^3 x ` |
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| 276. |
Find the derivative of `(1+cosx)^(x)` with respect to x. |
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Answer» Let y=`(1+cosx)^(x)` log y=`log(1+cosx)^(x)` `=xlog(1+cosx)` Differentiate both sides with respect to x. `(1/y)(dy)/(dx)=xcdotd/(dx)log(1+cosx)+log(1+cosx)cdotd/(dx)x` `rArr(dy)/(dx)=y[x/(1+cosx)cdotd/(dx)(1+cosx)+log(1+cosx)]` `rArr(dy)/(dx)=(1+cosx)^(x)[(-xsinx)/(1+cosx)+log(1+cosx)]` Ans. |
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| 277. |
`y=(x+1)^(2)(x+2)^(3)(x+3)^(4)` |
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Answer» Correct Answer - `(x+1)^(2)(x+2)^(3)(x+3)^(4)cdot[2/(x+1)+3/(x+2)+4/(x+3)]` |
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| 278. |
If `x(x)+y^(y)=1` then find `(dy)/(dx)`. |
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Answer» Correct Answer - `-(x^(x)(1+logx))/(y^(y)(1+logy))` |
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| 279. |
If `f(x)=[x] sin ((pi)/([x+1]))`, where [.] denotes the greatest integer function, then the set of point of discontiuity of f in its domain isA. ZB. `Z-{-1,0}`C. `R-[-1,0)`D. none of these |
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Answer» Correct Answer - B Clearly , f(x) is defined for all `x in R -[-1,0)`. For any integer `k ne -1`, we have `f(x)=k sin (pi)/(k+1), k le x lt k +1` So, f(x) being a constant function is continuous at all points other than integers in its domain. Also, `underset(x to k)lim f(x)=underset(h to 0)lim f(k-h)` `Rightarrow underset(x to k^(-))lim f(x)=underset(h to 0)lim [k-h] "sin" (pi)/([k+h+1])=k sin ((pi)/(k+1)), k ne -1` `therefore underset(x to k^(-))lim f(x) ne underset(x to k^(+))(lim f(x), k ne 0` Thus, f(x) is a discontinous at all non-zero integer points in its domain. When k=0, we have `f(x)=0 sin pi=0"for all " x in [0,1)` `Rightarrow underset(x to 0^(+))lim f(x)=f(0)` So, f(x) is right continous at x=0 Hence, the set of points of discontinuity in its domain is `Z-(-1,0)` |
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| 280. |
Differentiate `sqrt((1-x)/(1+x))` with respect to x. |
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Answer» Let y=`sqrt((1-x)/(1+x))=((1/x)/(1+x))^(1//2)` `rArrlogy=log((1-x)/(1+x))^(1//2)=1/2log((1-x)/(1+x))` `=1/2[log(1-x)-log(1+x)]` Differentiate both sides with respect to x `1/y(dy)/(dx)=1/2[(-1)/(1-x)-1/(1+x)]` `=1/2[(-1-x-1+x)/(1-x^(2))]=(-1)/(1-x^(2))` `rArr(dy)/(dx)=-y/(1-x^(2))=-1/(1-x^(2))sqrt((1-x)/(1+x))` `=(-1)/((1+x)^(1//2)(1+x)^(3//2))` Ans. |
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| 281. |
Let `f(x)=p[x]+qe^(-[x])+r|x|^(2)`, where p,q and r are real constants, If f(x) is differential at x=0. Then,A. `q=0,r=0, p in R`B. `p=0,r=0, q in R`C. `p=0,q=0, r in R`D. none of these |
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Answer» Correct Answer - C We have `f(x)=p[x]+qe^(-[x])+rx^(2)` Since, [x] and [x] are not differentiable at x=0. Therefore, f(x) will be differentiable at x=0 only when p=q=0 |
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| 282. |
`y=tanxtan2xtan3xtan4x` |
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Answer» Correct Answer - `2tanxtan2xtan3xtan4x[cosec2x+2cosec4x+3cosec6x+4cosec8x]` |
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| 283. |
(i) `y=e^(x)sin^(3)xcos^(4)x` (ii) `y=x*e^(xsinx)` |
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Answer» Correct Answer - (i) `e^(x)sin^(3)xcos^(4)x[1+3cotx-4tanx]` (ii) `e^(xsinx)(xcosx+sinx)` |
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| 284. |
`sin^(2) x + cos ^(2) y = 1` |
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Answer» `sin^(2) x + cos ^(2) y = 1` Differentiate both sides w.r.t. x ` 2 sin x cos + 2 cos y (-sin y)(dy)/(dx)= ` `rArr sin 2x - sin 2y (dy)/(dx) = 0 rArr (dy)/(dx) = (sin2x)/(sin2y)` . |
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| 285. |
If f is continuous on its domain D; then `|f|` is also continuous on D |
| Answer» Hence, `f(x) = |x-1|` in `[0,2]` is not differentiable at `x in 1 in (0,2)`. | |
| 286. |
`sin (tan ^(-1)2x)` |
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Answer» Correct Answer - `(2 cos(tan ^(-1)2x))/(1+4x^2)` |
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| 287. |
Let `f(x)={{:(,(1)/(|x|),"if "|x| gt 2,"then "f(x)is),(,a+bx^(2), , "if"|x| le2):}` is differentiable at x=-2 forA. `a=(3)/(4), b=(1)/(6)`B. `a=(3)/(4), b=(1)/(16)`C. `a=-(1)/(4), b=(1)/(16)`D. `a=(1)/(4), b=-(1)/(16)` |
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Answer» Correct Answer - B We have `f(x)={{:(,-(1)/(x),"if "x lt -2),(,a+bx^(2),"if"-2 le x lte 2),(,(1)/(x),"if"x gt 2):}` Since, f(x) is differentiable at x=-2. So, it is continuous there at `therefore underset(x to -2^(-))lim f(x)=underset(x to -2^(+))lim f(x)=f(-2)` `Rightarrow underset(x to -2^(-))lim f(x)=underset(x to -2^(+))lima+bx^(2)=a+b(-2)^(2)` `Rightarrow (1)/(2)=a+4b` As f(x) is differentiable at x=-2. Therefore `underset(x to -2^(-))lim (f(x)-f(-2))/(x-(-2))=underset(x to -2^(+))lim (f(x)-(f-2))/(x-(-2))` `underset(x to -2^(-))lim (-(1)/(x)-(a+4b))/(x+2) =underset(x to -2^(+))lim ((a+bx^(2))-(a+4b))/(x+2)` `underset(x to -2^(-))lim (-(1)/(x)-(1)/(2))/(x+2)=b underset(x to -2^(+))lim (x^(2)-4)/(x+2)` `Rightarrow underset(x to -2^(-))lim -(1)/(2x)=b underset(x to =2^(+))lim (x-2)` `Rightarrow (1)/(4)=b(-4) Rightarrow b=-(1)/(16)...(ii)` Solving (i) and (ii), we get `a=(3)/(4) and b=-(1)/(16)` |
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| 288. |
`log (x+1/x)` |
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Answer» Correct Answer - `(x^2-1)/(x(x^2+1))` |
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| 289. |
Differentiate `sqrt((x+1)(x+2)(x+3))` with respect to x. |
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Answer» Let y=`sqrt((x+1)(x+2)(x+3))` `rArrlogy=log[sqrt((x+1)(x+2)(x+3))]` `=1/2log[(x+1)(x+2)(x+3)]` `=1/2[log(x+1)+log(x+2)+log(x+3)]` Differentiate both sides with sides with respect to x `1/y(dy)/(dx)=1/2[1/(x+1)+1/(x+2)+1/(x+3)]` `rArr(dy)/(dx)=1/2y(1/(x+1)+1/(x+2)+1/(x+3))` `=1/2sqrt((x+1)(x+2)(x+3))` `(1/(x+1)+1/(x+2)+1/(x+3))`. Ans |
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| 290. |
`sin^(2)y + cos xy = k` |
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Answer» `sin^(2)y + cos xy = k` Differentiate both sides w.r.t. x ` 2sin y cos y (dy)/(dx) + (-sin xy) (d)/(dx)(xy) =0` `rArr sin 2y (dy)/(dx)-sin xy(x(dy)/(dx)+ y .1)=0` `rArr (dy)/(dx)(sin2y- x sin xy)= ysin xy` `rArr (dy)/(dx)=(y sin xy)/(sin 2y = x sin xy)` |
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| 291. |
(i) `x^(y)cdoty^(x)=1` (ii) `y=e^(x^(x))` |
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Answer» Correct Answer - (i) `(-y(y+xlogy))/(x(ylogx+x))` (ii) `e^(x^(x))cdotx^(x)(1+logx)` |
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| 292. |
`tan ^(-1)(cos sqrt(x))` |
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Answer» Correct Answer - `(-sinsqrt(x))/(2sqrt(x)(1+cos^2sqrt(x)))` |
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| 293. |
If `f.g` is continuous at `x = 0` , then f and g are separately continuous at `x = 0`. |
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Answer» Let `f(x) = sinx` and `g(x) = cotx` `:. f(x). g(x) = sinx. (cosx)/(sinx) = cosx` which is continuous at `x = 0` but `cot x` is not continuous at `x = 0`. |
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| 294. |
If `y= (x)/(x+5) " then " x dy/dx = ? `A. y(1-y)B. y(1-y)C. (1-y)D. (1+y) |
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Answer» Correct Answer - A |
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| 295. |
`x^(3) + x^(2)y+ xy^(2) + y^(3) = 81` |
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Answer» `x^(3) + x^(2)y+ xy^(2) + y^(3) = 81` Differentiate both sides w.r.t. x `3x^(2) + (x^(2)(dy)/(dx)+2xy)` ` + (2xy (dy)/(dx)+y^(2))+ 3y^(2)(dy)/(dx)=0` `rArr (x^(2) + 2xy + 3xy^(2))(dy)/(dx)=-(3x^(2)+2xy+y^(2))` `rArr (dy)/(dx)=-((3x^(2)+2xy+y^(2)))/((x^(2) + 2xy+ 3y^(2)))` |
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| 296. |
If `x^(a)cdoty^(b)=(x+y)^(a+b)` then prove that `(dy)/(dx)=y/x` |
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Answer» `x^(a)cdoty^(b)=(x+y)^(a+b)` `rArrlog(x^(a)cdoty^(b))=log(x+y)^(a+b)` `rArrlogx^(a)+logy^(b)=(a+b)log(x+y)` `rArralogx+blogy=(a+b)log(x+y)` Differentiate both sides with respect to x `a/x+b/y(dy)/(dx)=((a+b))/((x+y))cdotd/(dx)(x+y)` `=(a+b)/(x+y)(1+(dy)/(dx))` `=(a+b)/(x+y)+(a+b)/(x+y)+(a+b)/(x+y)cdot(dy)/(dx)` `rArr(b/y-(a+b)/(x+y))(dy)/(dx)=(a+b)/(x+y)-a/x` `rArr(b(x+y)-y(a+b))/(y(x+y))cdot(dy)/(dx)-(x(a+b)-a(x+y))/(x(x+y))` `rArr(bx+by-ay-by)/ycdot(dy)/(dx)=(ax+bx-ax-ay)/x` `rArr(bx-ay)/ycdot(dy)/(dx)=(bx-ay)/xrArr(dy)/(dx)=y/x` Hence proved. |
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| 297. |
`tan^-1((sqrt(1+x^2)-1)/x)` |
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Answer» Correct Answer - `(1)/(2(1+x^2))` |
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| 298. |
If `x^y = e^(x-y) " then " dy/dx ?`A. `1/log x `B. `(1)/(log ex )^2`C. `(log x )/(log ex )^2`D. `1/(log ex )^2` |
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Answer» Correct Answer - C |
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| 299. |
`logs(sqrt(x)+1/sqrt(x))` |
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Answer» Correct Answer - `(x-1)/(2x(x+1))` |
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| 300. |
`sqrt("log"x)` |
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Answer» Correct Answer - `(1)/(2xsqrt(log x))` |
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