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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.
| 1. |
33 The position vector of a particle moving in x-y plane is given by r (t 4) i (t 4) j. Find (a) Equation of trajectory of the particle Albo in a (b) Time when it crosses x-axis and y-axis |
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Answer» To find the trajectory,`vecr =xhati+ yhatj=(t^2-4)hati+ (t-4)hatj` So, `x=t^2 -4` and `y=t-4` On solving,`x=(y+4)^2-4` `y^2-8y-x+12=0` Time when it crosses x-axis,`t-4=0``t=4sec` Time when it crosses y-axis,`t^2-4=0``t=2sec` |
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| 2. |
A particle is moving with uniform acceleration. Its displacement at any instant t is given by `s=10t+49t^2` .find (i) initial velocity (ii) velocity at `t=3 `second and(iii) the uniform acceleration. |
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Answer» acceleration is constant `S=10t+49t^2` `V=ds/dt=d/dt(10t+49t^2)` `V=10+98t` 1)Initial velocity `V=10+98t` `V=10`m/s 2)Velocity at t=3sec `V=10+98t` `V=10+98(3)` `V=304`m/s 3)Constant acceleration `a=(dV)/dt` `a=d/dt(10+98t)` `a=98m/s^2`. |
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| 3. |
If `y=(log)_(sinx)(tanx),t h e n(((dy)/(dx)))_(pi/4)"is equal to"`(a)`4/(log2)`(b) `-4log2`(c)`(-4)/(log2)`(d) none of these |
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Answer» `y=ln(tanx)/ln(sinx)` `dy/dx=(lnsinx xx sec^(2)x/tanx - lntanx xx cosx/sinx)/((lnsinx)^2)` `dy/dx at x=pi/4=``(ln1/sqrt2xx 2-0)/(ln(1/sqrt2)^2`=`-4/lnsqrt2` |
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| 4. |
Find the derivative of the following functions from first principles: (i) ` x` (ii) `(-x)^(-1)` (iii) `s in (x + 1)` (iv) `cos(x-pi/8)`Find derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s a |
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Answer» (i) Let `f(x)=-x` From first principle. `f(x)=underset(hrarr0)"lim"(f(x+h)-f(x))/(h)` `=underset(hrarr0)"lim"(-(x+h)-(-x))/(h)` `=underset(hrarr0)"lim"(-h)/(h)=underset(hrarr0)"lim"(-1)=-1` (ii) Let `f(x)=(-x)^(-1)=(1)/(-x)=-(1)/(x)` From first principle, `f(x)=underset(hrarr0)"lim"(f(x+h)-f(x))/(h)` `=underset(hrarr0)"lim"((-(1)/(x+h)-(-(1)/(x)))/(h)` `=underset(hrarr0)"lim"(-(1)/(x+h)+(1)/(x))/(h)=underset(hrarr0)"lim"(-x+(x+h))/(hx(x+h))` `=underset(hrarr0)"lim"(h)/(hx(x+h))=underset(hrarr0)"lim"(1)/(x(x+h))` `=(1)/(x(x+0))=(1)/(x^(2))` (iii) Let `f(x)=sin (x+1)` From first principle, `f(x)=underset(hrarr0)"lim"(sin(x+h+1)-sin(x+1))/(h)` `=underset(hrarr0)"lim"(2cos(x+1+(h)/(2))sin(h)/(2))/(h)` `=underset(hrarr0)"lim"(cos(x+1+(h)/(2))sin(h)/(2))/((h)/(2))` `=cos(X+1+0).1=cos(x+1)` (iv) Let `f(x)=cos(x-(pi)/(8))` From first principle, `f(x)=underset(hrarr0)"lim"(f(x+h)-f(x))/(h)` `=underset(hrarr0)"lim"(cos(x+h-(pi)/(8))-cos(x-(pi)/(8)))/(h)` `=underset(hrarr0)"lim"(-2sin(x-(pi)/(8)+(h)/(2)).sin(h)/(2))/(h)` `=underset(hrarr0)"lim"(-sin(x-(pi)/(8)+(h)/(2)).sin(h)/(2))/((h)/(2))` `=-sin(x-(pi)/(8)+0).1` `=-sin(x-(pi)/(8))` |
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| 5. |
`(3x+5)(1+tanx)` |
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Answer» Let `y=(3x+5)(1+tanx)` `(dy)/(dx) = d/(dx)[(3x+5)(1+tanx)]` `=(3x+5)(sec^(2)x)+(1+tanx).3` `=(3x+5)sec^(2)x+3(1+tanx)` `=3xsec^(2)x+5sec^(2)x+3tanx+3` |
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| 6. |
Find the derivative of `sqrt(tanx)` using first principles. |
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Answer» `f(x) = sqrt(tanx)` `Lim_(h->0) (f(x+h)-f(x))/h = Lim_(h->0) (sqrt(tan(x+h))-sqrt(tanx))/h` Multiplying nunerator and denominator by `(sqrt(tan(x+h))+sqrt(tanx))`, ` =Lim_(h->0) (tan(x+h) - (tanx))/ ((sqrt(tan(x+h))+sqrt(tanx))h)` `=Lim_(h->0) ((sin h)/(cos(x+h)cosx))/ ((sqrt(tan(x+h))+sqrt(tanx))h)` `= Lim_(h->0) sin h/h xx Lim_(h->0)(1/((cos(x+h)cosx)(sqrt(tan(x+h))+sqrt(tanx))))` `=1 xx 1/(2cos^2xsqrt(tanx))` `=sec^2x/(2sqrttan x)` |
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| 7. |
If f(x) = { `sin[x] /[x],[x] != 0 ; 0, [x] = 0}` , Where[.] denotes the greatest integer function, then `lim_(x rarr 0) f(x)` is equal toA. 1B. 0C. `-1`D. Does not exist |
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Answer» Correct Answer - d Given, `f(x) = {{:(sin[x]/[x]",",[x]ne0), (0,[x]=0):},` `therefore` LHL = `lim_(xto0^(-))(sin[x])/([x])=lim(xto0)(sin[0-h])/([0-h])` `=lim_(hto0)(-sin[-h])/([-h]) =-1` RHL `=lim_(xto0)f(x) = lim_(xto0^(+))(sin[x])/([x])` `=lim_(xto0^(+)) (sin[0+h])/([0+h])=lim_(hto0)(sin[h])/([h])=1` `therefore LHL ne RHL` So, limit does not exist. |
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| 8. |
`lim_(x->1)[(2x-3)(sqrtx-1)]/[2x^2+x-3]`A. `1/10`B. `-1/10`C. 1D. None of these |
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Answer» Correct Answer - B Given, `lim_(xto1)((sqrt(x-1))(2x-3))/(2x^(2)+x-3) = lim_(xto1)((sqrt(x-1))(2x-3))/((2x+3)(sqrt(x)-1)(sqrt(x)+1))` `=lim_(xto1)(2x-3)/((2x+3)(sqrt(x)+1))=-1/(5 xx 2)=-1/10` |
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| 9. |
`lim_(y to 0) ((x+y)sec(x+y)-xsecx)/(y)` |
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Answer» Given, `underset(yto0)"lim"((x+y)sec(x+y)-xsecx)/(y)` `=underset(yto0)"lim"((x+y)/(cos(x+y))-x/(cosx))/(y)` `=underset(yto0)"lim"[(xcosx+ycosx(x+y)+ycosx)/(ycosxcos(x+y))]` `=underset(yto0)"lim"[(xcosx-xcos(x+y)+ycosx)/(ycosxcos(x+y)]]` `=underset(yto0)"lim"(x[-2sin(x+y/2)sin(-y/2)]+ycosx)/(ycosxcos(x+y))` `[therefore cosC-cosD=-2sin(C+D)/(2).sin(C-D)/(2)]` `=underset(yto0)"lim"[[(x{2sin(x+y/2)siny/2}+ycosx))/(ycosxcos(x+y)]]` `=underset(yto0)"lim"(2xsin(x+y/2))/(cosxcos(x+y)).underset(yto0)"lim"(siny/2)/(y/2).1/2+underset(hto0)"lim"cos(x+h)` `therefore underset(hto0)"lim"(sinx)/(x)=1` and `xto0 rArr kx to0]` `=underset(yto0)"lim"(2xsin(x+y/2))/(cosxcos(x+y)).1/2+underset(yto0)"lim"sec(x+y)` `(2xsinx)/(cosxcosx).1/2+secx` `xtanxsecx+secx` `=secx(xtanx+1)` |
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| 10. |
If `y=(1)/(3x^(3))` then prove that `3y+x(dy)/(dx)=0`. |
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Answer» `y=(1)/(3x^(3))=(1)/(3).x^(-3)` `rArr (dy)/(Dx)=(d)/(dx)((1)/(3)x^(-3))` `=(1)/(3)(d)/(dx)(x^(-3))=(1)/(3)(-3)x^(-4)=-x^(-4)` Now L.H.S. `=3y+x(dy)/(dx)` `=3.(1)/(3x^(3))+x(-x^(-4))` `=(1)/(x^(3))-(1)/(x^(3))=0=` R.H.S. |
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| 11. |
Find the derivative of (4 sec x sin x+cos ec x cos x-5 tan x cot x) with respect to x. |
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Answer» `(d)/(dx)(4 sec x sin x+cos ec x cos x-5tan x cot x)` `=(d)/(dx)(4(sinx)/(cos x)+(cos x)/(sin x)-5.tanx.(1)/(tanx))` `=4(d)/(dx)(tan x)+(d)/(dx)(cot x)-(d)/(Dx)(5)` `=4sec^(2)x-cos ec^(2)x-0` `=(5 sec^(2)x-cos ec^(2)x)`. |
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| 12. |
Find the derivative of the following functions: (i) `sin x cos x` (ii) `sec x` (iii) `5sec x+4 cos x` (iv) `cosec x` (v) `3 cot x+5 cosec x` (vi) `5sinx-6cosx+7` (vii) `2tanx-7secx` |
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Answer» (i) Let `y=sinx.cosx` `rArr(dy)/(dx)=(d)/(dx)(sinx.cosx)` `=sin x.(d)/(dx)cosx+cosx(d)/(dx)sinx` `=sinx(-sinx)+cos x.(cos x)` `=-sin^(2)x+cos^(2)x= cos 2 x` (ii) Let `y=sec x=(1)/(cos x)` `rArr(dy )/(dx)=(d)/(dx)((1)/(cos x))` `=(cos x(d)/(dx)(1)-1(d)/(dx)cos x)/(cos^(2)x)` `=(0-(-sinx))/(cos^(2)x)` `=(1)/(cos x).(sinx)/(cosx)=sec x tan x`. (iii) Let `y=cosec x=(1)/(sinx)rArr(dy)/(dx)=(d)/(dx)(5secx+4cosx )` `=5(d)/(dx)(sec x)+4(d)/(dx)(cos x)` `=5sec xtan x-4sinx` (iv) Let `y=cosec x=(1)/(sinx)rArr(dy)/(dx)=(d)/(dx)((1)/(sin x))` `=(sin x(d)/(dx)(1)-1.(d)/(dx)sinx)/((sin x)^(2))` `=(0-cosx)/(sin^(2)x)=-(1)/(sinx).(cosx)/(sinx)` `-cosec x cot x` (v) Let `y =3cot x+5 cosec x` `rArr(dy)/(dx)=(d)/(dx)(3 cot x+5cosec x)` `=3(d)/(dx)cot x+5(d)/(dx) cosec x` `=-3cosec^(2)x-5 cosec x cot x` (vi) Let `y=6 sin x-6cos x+7` `rArr(dy)/(dx)=(d)/(dx)(5sin x-6 cos x+7)` `=5(d)/(dx)(sinx)-6(d)/(dx)(cos x)+(d)/(dx)(7)` `=5cos x+6 sin x+0` `=5cos x +6sin x` (vii) Let `y=2 tan x-7 sec x` `rArr(dy)/(dx)=(d)/(dx)(2 tan x-7sec x)` `=2(d)/(dx)tan x-7(d)/(dx)secx` `=2sec^(2)x-7sec x tan x`. |
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| 13. |
`sin^(3)xcos^(3)x` |
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Answer» Let `y=sin^(3)xcos^(3)x` `(dy)/(dx) = sin^(3).x.d/(dx)cos^(3)x+cos^(3)xd/(dx)sin^(3)x` [By product rule] `=sin^(3)x.3cos^(2)x(-sinx) + cos^(3)x. 3sin^(2)xcosx` [By chain rule] `=-3cos^(2)xcos^(2)x(cos^(2)x-sin^(2)x)` `=3sin^(2)xcos^(2)xcos2x` `=3/4(2sinxcosx)^(2)cos2x` `=3/4sin^(2)2xcos2x` |
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| 14. |
Differentiation of `(2x-7)^2(3x+5)^3` |
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Answer» Let `y=(2x-7)^(2)(3x+5)^(3)` `(dy)/(dx) = (2x-7)^(2)d/(dy)(3x+5)^(3)+(3x+5)^(3)d/(dx)(2x-7)^(2)` [by product rule] `=(2x-7)^(2)(3)(3x+5)^(2)(3)+(3x+5)^(3)(2)(2x-7)(2)` [by chain rule] `=9(2x-7)^(2)(3x+5)^(2)+4(3x+5)^(3)(2x-7)` `=(2x-7)(3x+5)^(2)[9(2x-7)+4(3x+5)]` `=(2x-7)(3x+5)^(2)(18x-63 + 12x+20)` `=(2x-7)(3x+5)^(2)(30x-43)` |
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| 15. |
Find the derivative of `x^(2)-2` at x=10. |
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Answer» Let `y=x^(2)-2` `rArr(dy)/(dx)=2x` at x=10 `(dy)/(dx)=2xx10=20` |
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| 16. |
Find derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): `(a+bsinx)/(c+dcosx)` |
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Answer» Let `y=(a+bsinx)/(c+dcosx)` `therefore (dy)/(dx) = ((c+dcosx)d/(dx)(a+bsinx)-(a+bsinx)d/(dx)(c+dcosx))/(c+d cosx)^(2)` ["by quotinet rule"] `=((c+dcosx)(bcosx)-(a+bsinx)-(-dsinx))/(c+dcosx)^(2)` `=((bc cosx+adsinx+bd(cos^(2)x+sin^(2)x))/(c+dcosx)^(2))` `(bc cosx + adsinx+bd)/(c+dcosx)^(2)` `(bccosx+adsinx+bd)/(c+dcosx)^(2)` |
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| 17. |
Find the derivative of (i) ` 2x-(3)/(4)` (ii) `(5x^(3)+3x-1)(x-1)` (iii) `x^(-3)(5+3x)` (iv) `x^(5)(3-6x^(-9))` (v) `x^(-4)(3-4x^(-5))` (vi) `(2)/( x+1)-(x^(2))/(3x-1)` |
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Answer» (i) Let `y=2x-(3)/(4)` `rArr (dy)/(dx)=(d)/(dx)(2x-(3)/(4))` `=2(d)/(dx)(x)-(d)/(dx)((3)/( 4))` `=2(1)-0=2` (ii) Let `y=(5x^(3)+3x-1)(x-1)` `rARr (dy)/(dx)=(d)/(dx)[(5x^(3)+3x-1)(x-a)]` `=(5x^(3)+3x-1)(d)/(dx)(x-1)+(x-1)(d)/(dx)(5x^(3)+3x-1)` `(5x^(3)+3x-1)(1-0)+(x-1)(15x^(2)+3)` `=5x^(3)+3x-1+15x^(3)+3x-15x^(2)-3` `=20x^(3)-15x^(2)+6x-4` (iii) Let `y=x^(-3)(5+3x)` `=5x^(-3)+3x^(-2)` `rArr (dy)/(dx)=(d)/(dx)(5x^(-3)+3x^(-2))` `=5(d)/(dx)x^(-3)+3(d)/(dx)x^(-2)` `=5(-3)x^(-4)+ 3(-2)x^(-3)` `=-(15)/(x^(4))-(6)/(x^(3))=-(3)/(x^(3))((5)/(x)+2)` (iv) Let `y=x^(5)(3-6x^(-9))` `=3x^(5)-6x^(-4)` `rArr(dy)/(dx)=(d)/(dx)(3x^(5)-6x^(-4))` `=d(d)/(dx)x^(5)-6(d)/(dx)x^(-4)` `=3(5x^(4))-6(-4x^(-5))` `=(15x^(4)+(24)/(x^(5)))` (v) Let `y=x^(-4)(3-4x^(-5))=3x^(-4)-4x^(-9)` `rArr(dy)/(dx)=(d)/(dx)(3x^(-4)-4x^(-9))` `=3(d)/(dx)x^(-4)-4(d)/(dx)x^(-9)` `=3(-4x^(-5))-4(-9x^(-10))` `=-(12)/(x^(5)+(36)/(x^(10))=-(12)/(x^(5))(1-(3)/(x^(5)))` (vi) Let `y=(2)/(x+1)-(x^(2))/(3x-1)` `rArr(dy)/(Dx)=(d)/dx)[(2)/(x+1)-(x^(2))/(3x-1)]` `=(d)/(dx)((2)/(x+1))-(1)/(dx)((x^(2))/(3x-1))` `=((x+1)(d)/(dx)2-2(d)/(dx)(x+1))/((x+1)^(2))` `((3x-1)(d)/(dx)x^(2)-x^(2)(d)/(dx)(3x-1))/((3x-1)^(2))` `=(0-2)/((x+1)^(2))-((3x-1).2x-x^(2).3)/((3x-1)^(2))` `=(-2)/((x+1)^(2))-(3x^(2)-2x)/((3x-1)^(2))` `=-(1)/((x+1)^(2))=(x(3x-2))/((3x-1)^(2))`. |
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| 18. |
Find the derivative of x at x=1, |
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Answer» Let y=x `rArr (dy)/(dx)=(d)/(dx)(x)=1` at x=1 `(dy)/(dx)=1` |
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| 19. |
Find the derivative of `(x^(n)-a^(n))/(x-a)` for some constant a. |
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Answer» Let `y=(x^(n)-a^(n))/(x-a)` `rArr (dy)/(dx)=(d)/(dx)((x^(n)-a^(n))/(x-a))` `=((xa)(d)/(dx)(x^(n)-a^(n))-(x^(n)-a^(n))(d)/(dx)(x-d))/((x-a)^(2))` `=((x-a)(nx^(n-1)-0)-(x^(n)-a^(n))(d)/(dx)(x-a))/((x-a)^(2))` `=((x-a)(nx^(n-1)-0)-(x^(n)-a^(n))(1-0))/((x-a)^(2))` `=(nx^(n)-anx^(n-1)-x^(n)+a^(n))/((x-a)^(2))` |
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| 20. |
For some constants a and b, find the derivative of : (i) `(x-a)(x-b)` (ii) `(ax^(2)+b)^(2)` (iii) `(X-a)/(x-b)` |
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Answer» Let `y=(x-a)(x-b)` `rArr (dy)/(dx)=(d)/(dx)[(x-a)(x-b)]` `=(x-a)(d)/(dx)(x-b)+(x-b)(d)/(dx)(x-a)` `=(x-a)(1-0)+(x-b)(d)/(dx)(x-a)` `=(x-a)(1-0)+(x-b)(1-0)` `=x-a+x-b=2x- a-b` (ii) Let `y=(ax ^(2)+b)^(2)` `=a^(2)x^(4)+2abx^(2)+b^(2)` `rArr(dy)/(dx)=(d)/(dx)(a^(2)x^(4)+2abx^(2)+b^(2))` `=a^(2)(d)/(dx)x^(4)+2ab(d)/(dx)x^(2)+(d)/(dx)b^(2)` ltb rgt `=4a^(2)x^(3)+4abx` (iii) Let `y=(x-a)/(x-b)` `rArr (dy)/(dx)=(d)/(dx)((x-a)/(x-b))` `=((x-b)(d)/(dx)(x-a)-(x-a)(d)/(dx)(x-b))/((x-b)^(2))` `=((x-b).1-(x-a).1)/((x-b)^(2))=(a-b)/((x-b)^(2))` |
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| 21. |
Find the derivative of `x^(n)+ax^(n-1)+a^(2)x^(n-2)+..........+a^(n-1)x+a^(n)` for some fixed real number a. |
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Answer» Let `y=x^(n)+ax^(n-1)+a^(2)x^(n-2)+.....+a^(n=1)x+a^(n)` `rArr (dy)/(dx)=(d)/(dx)[x^(n)+ax^(n-1)+a^(2)x^(n-2)+.......+a^(n-1)x+a^(n)]` `=(d)/(dx)x^(n)+a(d)/(dx)x^(n-1)+a^(2)(d)/(dx)x^(n-2)+.........+a^(n-1)(d)/(dx)x+(d)/(dx)a^(n)` `=nx^(n-1)+a(n-1)x^(n- 2)+a^(2)(n-2)x^(n-3)` `+..........+a^(n-1).1+0` `=nx^(n-1)+a(n-1)x^(n-2)+a^(2)(n-2)x^(n-3)+.....+a^(n-1)`. |
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| 22. |
For the function `f(x)=(x^(100))/(100)+(x^(99))/(99)+ .......+(x^(2))/2)+x+1`. prove that f(1)=100f(0). |
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Answer» Here `f(x)=(x^(100))/(100)+(x^(99))/(99)+.....+(x^(2))/(2)+x+1` `f(x) =(d)/(dx)[(x^(100)/(100)+(x^(99))/(99)+.....+(x^(2))/(2)+x+1]` `=(1)/(100)(d)/(dx)x^(100)+(1)/(99)(d)/(dx)x^(99)` `+......+(1)/(2)(d)/(dx)x^(2)+(d)/(dx)x+(d)/(dx)` `=(1)/(100).100x^(99)+(1)/(99).99x^(98)` `+....+(1)/(2)2x+1+0` `=x^(99)+x^(98)+.....+x+1` `rArr f(1)=1+1+.....+1+1` (upto 100 terms)=100 and `f(0)=0+0.....+0+1=1` `therefore f(1)=100:f(1)` Hence proved. |
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| 23. |
`lim_(x->0)[e^sinx-1]/x` |
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Answer» We know, `e^x = 1+x+x^2/2+x^3/3+...` `:. lim_(x->0) (e^(sinx)-1)/x = lim_(x->0) ((1+sinx+(sinx)^2/2+(sinx)^3/3+...) -1)/x` `= lim_(x->0) (sinx/x+sinx/x*sinx+sinx/x*sin^2x+...)` We know, `lim_(x->0) sinx/x = 1` and `lim_(x->0) sin x = 0` `:. lim_(x->0) (e^(sinx)-1)/x = 1+1(0)+1(0)+... = 1` |
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| 24. |
`lim_(x->0) (sinx-sina)/(sqrt(x)-sqrt(a))` |
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Answer» `sinx-sina = 2sin((x-a)/2)cos((x+a)/2)` `1/(sqrtx -sqrta) = 1/(sqrtx -sqrta)**(sqrtx +sqrta)/(sqrtx + sqrta) ` `= (sqrtx + sqrta)/(x-a)` Putting these values in the given expression, `Lim_(x->a) (2sin((x-a)/2)cos((x+a)/2))/((sqrtx + sqrta)/(x-a))` `=2 Lim_(x->a) ( (sin((x-a)/2))/(2((x-a)/2))**cos((x+a)/2)**(sqrtx + sqrta))` `= 2/2 Lim_(x->a) ( (sin((x-a)/2))/(((x-a)/2))) Lim_(x->a)cos((x+a)/2)**(sqrtx + sqrta)` Applying limits, `=1*cosa*2sqrta` `=2sqrtacosa`, which is the required value. |
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| 25. |
`lim_(x->0)sin(picos^2x)/(x^2)=` |
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Answer» `Lim_(x->0) sin(picos^2x)/x^2` `=Lim_(x->0) sin(pi(1-sin^2x))/x^2` `=Lim_(x->0) sin(pi-pisin^2x)/x^2` `=Lim_(x->0) sin(pisin^2x)/x^2` `=Lim_(x->0) sin(pisin^2x)/(pisin^2x)*(pisin^2x)/x^2` We know, `Lim_(x->0) sin(f(x))/(f(x)) = 1` So, our limits becomes, `=1*pi(1) = pi` `:. Lim_(x->0) sin(picos^2x)/x^2 = pi` |
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| 26. |
`lim_(x->0) (2+2x+sin2x)/((2x+sin2x)e^sinx)` is |
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Answer» `-1<=sinx<=1` `-1<=sin2x<=1` `lim_(x->oo)(((2/x)+2+sin2x/x)/((2+sin2x/x)e^(sinx)))` `lim_(x->oo)2/2e^(sinx)` `1/eltLlte` Non exitence. |
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| 27. |
Let f be a differentiable function on the open interval(a, b). Which of the following statements must be true? (i) f is continuous on the closed interval [a,b],(ii) f is bounded on the open interval (a,b) |
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Answer» (i)As `f` is a differentiable function in open interval `(a,b)`, it will be continuous in closed interval `[a,b]`. So, statement `(i)` is true. (ii)Statement `(ii)` is not true as we know differentiable function can be non-bounded. Example of this will be `tan theta` between `pi/2 and -pi/2`, is differentiable but is non-bounded. (iii)We are given, `f(a_1) lt 0 ltf(b_1)` and `a_1 lt c lt b_1`, So, there will be definitely a value of `c` where `f(c) = 0` So, this statement will also be true. |
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| 28. |
`(d^2x)/(dy^2)` equals: (1) `((d^2y)/(dx^2))^-1` (2) `-((d^2y)/(dx^2))^-1 ((dy)/(dx))^-3` (3) `-((d^2y)/(dx^2))^-1 ((dy)/(dx))^-2` (4) `-((d^2y)/(dx^2))^-1 ((dy)/(dx))^3` |
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Answer» `dy/dx = 1/(dx/dy)` `(d^2 y)/dx^2 = d/dx (1/(dx/dy))dy/dx` `= d/dy(1/(dx/dy))dy/dx` `(d^2y)/(dx^2)= -1/(dx/dy)^2 (d^2x)/dy^2dy/dx` `(d^2 x)/(dy^2) = -(d^2y)/(dx^2) (dx/dy)^2(dx/dy)` `= - (d^2y)/dx^2 (dy/dx)^-3` option 4 is correct |
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| 29. |
Evaluate: `lim_(xrarroo) (5x^(2)+3x+1)/(3x^(2)+2x+4)` |
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Answer» `underset(xrarroo)"lim"(5x^(2)+3x+1)/(3x^(2)+2x+4)` `=underset(xrarroo)"lim"(5+(3)/(x)+(1)/(x^(2)))/(3+(2)/(x)+(4)/(x^(2)))` (Divide numerator and denominator by `x^(2)`) |
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| 30. |
Let m and n be two positive integers greater than 1.If `lim_(alpha->0) (e^(cos alpha^n)-e)/(alpha^m)=-(e/2)` then the value of `m/n` is |
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Answer» `lim_(alpha->0)(e(e^(cosalpha^n)-1))/(alpha^m)` `=lim_(alpha->0)(e(cosalpha^n-1)/alpha^m)` `=-elim_(alpha->0)(1-cosalpha^n)/alpha^m` `=-2elim_(alpha->0)(sin(alpha^n)/2*sin(alpha^n)/2)/(alpha^n/2*alpha^m*alpha^n/2)` `=2e*alpha^(2n)/(4alpha^m` `alpha^(2n-m)=1` `2n-m=0` `m/n=2`. |
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| 31. |
`lim_(x to 0) (cos x)/(pi-x)` |
| Answer» `underset(xrarr0)"lim"(cosx)/(pi-x)=(cos 0)/(pi-0)=(1)/(pi)` | |
| 32. |
`lim_(xrarr0)(cos 2x-1)/(cosx-1)` |
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Answer» `underset(xrarr0)(cos 2x-1)/(cosx-1) ((0)/(0))` `=underset(rarr0)"lim"((2cos^(2)x-1))/(cosx-1)` `=underset(xrarr0)"lim"(2(cosx-1)(cosx+1))/(cosx-1)` `=underset(xrarr0)"lim"2(cos x+1)=2(1+1)=4` |
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| 33. |
`lim_(xrarr0) (ax+x cos x)/(b sinx)` |
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Answer» `underset(xrarr0)"lim"(ax+x cos x)/(b sin x)((0)/(0))` `underset(xrarr0)"lim"(ax+x cos x)/(b sin x) ((0)/(0))` `=underset(xrarr0)"lim"(a+cos x)/(b(sinx)/(x))` (Divide numerator/denominator by x) `=(a+cos 0)/(bxx1)=(a+1)/( b)` |
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| 34. |
`lim_(xrarr0) x sec x` |
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Answer» `underset(xrarr0)"lim"s.secx` `=underset(Xrarr0)"lim"(x)/(cos x)` `=(0)/(cos 0)=(0)/(1)=0` |
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| 35. |
if `y=sqrt(x) + 1/sqrt(x)`, then `(dy)/(dx)` at x=1 is equal toA. 1B. `1/2`C. `1/sqrt(2)`D. 0 |
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Answer» Correct Answer - d Given, `y=sqrt(x) + 1/sqrt(x)` Now, `(dy)/(dx) = 1/(2sqrt(x)) -1/(2x^(3//2))` `therefore ((dy)/(dx))_(x=1)=1/2-1/2=0` |
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| 36. |
Evaluate: `lim_(xrarr1) [(x-2)/(x^(2)-x)-(1)/(x^(3)-3x^(2)+2x)]` |
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Answer» `underset(xrarr1)"lim"[(x-2)/(x^(2)-x)-(1)/(x^(3)-3x^(2)+2x)]` `=underset(xrarr1)"lim"[(x-2)/(x(x-1))-(1)/(x(x^(2)-3x+2))]` `=underset(Xrarr1)"lim"[(x-2)/(x(x-1))-(1)/(x(X-1)(x-2))]` `=underset(xrarr1)"lim"[((x-2) ^(2)-1)/(x(x-1)(x-2))]` `underset(xrarr1)"lim"((x-3)(x-1))/(x(x-1)(X-2))` `=underset(xrarr1)"lim"(x-3)/(x(x-2))=(-2)/(1(-1))=2`. |
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| 37. |
Evaluate `lim_(xrarr5) (1-sqrt(x-4))/(x-5)` |
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Answer» `underset(xrarr5)"lim"(1-sqrt(x-4))/(x-5)` `=underset(xrarr5)"lim"(1-sqrt(x-4))/(x-5)xx(1+sqrt(X-4))/(1+sqrt(x-4))` `=underset(xrarr5)"lim"(1-(x-4))/((x-5)(1+sqrt(x-4))` `=underset(xrarr5)"lim"((5x-x))/(-(5-x)(1+sqrt(x-4)))` `=underset(xrarr5)"lim"(-1)/(1+sqrt(x-4))` `=(-1)/(1+sqrt(5-4))=(-1)/(2)`. |
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| 38. |
Evaluate : `lim_(xrarr0) (sqrt(1+x)-sqrt(1-x))/( x)` |
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Answer» `underset(xrarr0)"lim" (sqrt(1+x)-sqrt(1-x))/(x)` `=underset(xrarr0)"lim"(sqrt(1+x)-sqrt(1-x))/(x)xx(sqrt(1+x)+sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))` `=underset(xrarr0)"lim"((1+x)-(1-x))/(x(sqrt(1+x)+sqrt(1-x)))` `=underset(xrarr0)"lim"(2)/(sqrt(1+x)+sqrt(1-x))` `=(2)/(sqrt(1)+sqrt(1))=(2)/(2)=1`. |
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| 39. |
Evaluate: `lim_(xrarr0) (tan x-sin x)/(x^(3))` |
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Answer» `underset(xrarr0)"lim"(tan x-sin x)/(x^(3))` `=underset(xrarr0)"lim"((sinx)/(cosx)-sinx)/(x^(3)) ` `underset(xrarr0)"lim"(sin x-cos x sin x)/(x^(3)cos x)` `=underset(xrarr0)"lim"(sin x(1-cos x))/(x^(3)cos x).((1+cos x))/((1+cos s))` `=underset(xrarr0)"lim"(sin x(1-cos^(2)x))/(x^(3).cos x (1+cos x))` `=underset(xrarr0)"lim"(sin^(3)x)/(x^(3).cos x(1+cos x))` `=underset(xrarr0)"lim"((sin x)/(x))^(3).(1)/(cos x(1+cos x))` `=1xx(1)/(cos 0(1+cos 0))` `(because underset(xrarr0)"lim"(sinx)/(x)=1)` `=(1)/(2)` |
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| 40. |
Evaluate: `lim_(xrarr0) ((3+x)^(1//2)-(3-x)^(1//2))/(x)` |
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Answer» ` underset(xrarr0)"lim"((3+x)^(1//2)-(3-x)^(1//2))/(x)` `=underset(xrarr0)"lim"(sqrt(3+x)^(1//2)-(3-x)^(1//2))/(x)xx(sqrt(3+x)+sqrt(3-x))/(sqrt(3+x)+sqrt(3-x))` `=underset(xrarr0)"lim"(2x)/(x(sqrt(3+x)+sqrt(3-x)))` `underset(xrarr0)"lim"(2x)/(sqrt(3+x)+sqrt(3-x))` `=(2)/(sqrt(3)+sqrt(3))=(1)/sqrt(3)` |
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| 41. |
`lim_(xrarr3) (x^(4)-81)/(2x^(2)-5x- 3) ` |
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Answer» `underset(xrarr0)"lim"(x^(4)-81)/(2x^(2)-5x-3) ((0)/(0))` `=underset(Xrarr3)"lim"((x^(2)-9)(x^(2)+9))/((x-3)(2x+1))` `underset(xrarr3)"lim"((x-3)(x+3)(x^(2)+9))/((x-3)(2x+1))` `=underset(Xrarr3)"lim"((x+3)(x^(2)+9))/(2x+1)` `=((3+3)(9+9))/(2(3)+1)=(108)/(7)` |
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| 42. |
` lim_(z to 1) (2^(1//3)-1)/(z^(1//6)-1)` |
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Answer» `underset(zrarr1)"lim"(z^(1//3)-1)/(z^(1//6)-1) ((0)/(0))` `=underset(Zrarr1)"lim"((z^(1//6)^(2)-(1)^(2)))/(z^(1//6)-1)` `=underset(zrarr1)"lim"((z^(1//6)-1)(z^(1//6)+1))/(z^(1//6)-1)` `=underset(xrarr1)"lim"(z^(1//6)+1)=1+1= 2`. |
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| 43. |
`lim_(xrarr0) (ax+b)/(cx+1)` |
| Answer» `underset(xrarr0)"lim"(ax+b)/(cx+1)=(0+b)/(0+1)=b` | |
| 44. |
`lim_(xrarr0) (sinax+bx)/(ax+si nbx),a,b,a+bne0` |
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Answer» `underset(xrarr0)"lim"(sin ax+bx)/(dx+sinbx)(a,b,a+bne0)((0)/(0))` `=underset(xrarr0)"lim"((sinax)/(x)+b)/(a+(sinbx)/(x))` (Divide numerator/denominator by x) |
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| 45. |
`lim_(xrarr1) (ax^(2)+bx+c)/(cx^(2)+bx+a)a+b+cne0` |
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Answer» `underset(xrarr1)"lim"(ax^(2)+bx+c)/(cx^(2)+bx+a),a+b+cne0` `=(a+b+c)/(c+b+a)=1` |
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| 46. |
`lim_(xrarr0) (sinax)/(sinbx),a,bne0` |
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Answer» `underset(xrarr0)"lim"(sinax)/(sinbx)(a,bne0) ((0)/(0))` `=underset(xrarr0)"lim"(sinax)/(ax).(bx)/(sinbx).(a)/(b)` `=1xx1xx(a)/(b)(because underset(0rarr0)"lim"(sintheta)/(theta)=1)` `=(a)/(b)` |
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| 47. |
If the function f(x) satisfies `lim_(xrarr1) (f(x)-2)/(x^(2)-1)=pi`, evaluate `lim_(Xrarr1) f(x)`. |
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Answer» According to the problem, `underset(xrarr1)"lim"(f(x)-2)/(x^(2)-1)=pi` Now, `underset(Xrarr1)"lim"[f(x)-2]=underset(Xrarr1)"lim"[(f(x)-2)/(x^(2)-1).(x^(2)-1)]` `=underset(xrarr1)"lim"(f(x)-2)/(x^(2)-1).underset(xrarr1)"lim"(x^(2)-1)` `=pixx0=0` `rArr underset(xrarr1)"lim"f(x)=2` |
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| 48. |
if `y=(1+1/x^(2))/(1-1/(x)^(2))`,then `(dy)/(dx)` is equal toA. `(-4x)/(x^(2)-1)^(2)`B. `(-4x)/(x^(2)-1)`C. `(1-x^(2))/(4x)`D. `(4x)/(x^(2)-1)` |
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Answer» Correct Answer - a Given, `y=(1+1/x^(2))/(1-1/x^(2)) rArr y=(x^(2)+1)/(x^(2)-1)` `therefore (dy)/(dx) = ((x^(2)-1)2x-(x^(2)+1)(2x))/((x^(2)-1)^(2))` [by quotient rule] `(dy)/(dx) = (2x(x^(2)-1-x^(2)-1))/(x^(2)-1)^(2)` `=(2x(-2))/(x^(2)-1)^(2)=(-4x)/(x^(2)-1)^(2)` |
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| 49. |
Evaluate: `lim_(xrarr(pi)/(4)) (sin x -cos x)/(x-(pi)/(4))` |
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Answer» `underset(xrarr(pi)/(4))"lim"(sin x-cos x)/(x-(pi)/(4)) (0)/(0)` `underset(x-(pi)/(4)rarr0)"lim"(sqrt(2)[sinx.cos(pi)/(4)-cos x.sin(pi)/(4)])/((x-(pi)/(4)))` `=underset(x-(pi)/(4)rarr0)"lim"(sqrt(2).sin(x-(pi)/(4)))/(x-(pi)/(4))` `=sqrt(2)xx1=sqrt(2)`. |
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| 50. |
`lim_(xrarr-2)((1)/(x)+(1)/(2))/(x+2)` |
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Answer» `underset(Xrarr-2)"lim"((1)/(x)+(1)/(2))/(x+2)` `=underset(Xrarr-2)"lim"((2+x)/(2x))/(x+2)` `=underset(xrarr-2)"lim"(1)/(2x)=(1)/(2(-2))=-(1)/(4)` |
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