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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.
| 51. |
A function f(x) is defined in the interval [1,4) as follows `f(x)={{:(,log_(e)[x],1 le x lt 3),(,|log_(e)x|,3 le x lt 4):}`. Then, the curve y=f(x)A. is broken at two pointsB. is broken at exactly one pointC. does not have a definite tangent at two pointsD. does not have a definite tangent at more than two points |
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Answer» Correct Answer - A::C We have `f(x)={{:(,log_(e)[x],1 le x lt 3),(,|log_(e)x|,3 le x lt 4):}` `Rightarrow f(x)={{:(,0,1 le x lt 2),(,log_(e)2,2 le x lt 3),(,|log_(e)x|,3 le x lt 4):}` Clearly, f(x) is everywhere continuous and differentiable except possibility at x=2,3 We observe that `underset(x to 2^(-))lim f(x)=underset(x to2^(-))lim 0=0` `and underset(x to 2^(+))lim f(x)=underset(x to2^(+))lim log_(e)2=log_(e)2` Clearly, f(x) is not continuous at x=2 It can be easily seen that f(x) is not continuous at x=3 Hence, f(x) is neither continuous nor differenitable at x=2,3 Thus, the curve y=f(x) is broken at two points and it does not have a definite tangent at these points |
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| 52. |
Let f(x) be a function defined as `f(x)={{:(,int_(0)^(x)(3+|t-2|),"if "x gt 4),(,2x+8,"if "x le 4):}` Then, f(x) isA. continuous at x=4B. neither continuous nor differentiable at x=4C. everywhere continuous but not differentiable at x=4D. everywhere continuous and differentiable |
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Answer» Correct Answer - C For `x gt 4`, we have `f(x)=underset(0)overset(x)int (3+|t-1|)dt` `Rightarrow f(x)=underset(0)overset(2)int (3-|t-2|)dt+underset(0)overset(x)int (3+(t-2))dt` `Rightarrow f(x)=underset(0)overset(2)int (5-t)dt+underset(2)overset(x)int (1+t)dt` `Rightarrow f(x)=[5x-(t^(2))/(2)]^(2)+[t+(t^(2))/(2)]_(2)^(x)` `Rightarrow f(x)=(x^(2))/(2)+x+4` Thus, we have `f(x)={{:(,(x^(2))/(2)+x+4,"if "x gt 4),(,2x+8,"if "x le 4):}` Clearly, f(x) is continuous at x=4 We have `("LHD of f(x) at x=4")={(d)/(dx)(2x+8)}_(x=4)=2` `("RHD of f(x) at x=4")={(d)/(dx)(x^(2)/(2)+x+4)}_(x=4)=9` Clearly, `"Clearly", ("LHD of f(x) x=4") ne ("RHD of f(x) at x=4")` So, f(x) is not differentiable at x=4. Thus, f(x) is everywhere continuous but not differentiable at x=4. |
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| 53. |
If the function `f(x) = {(kx + 5 ", when " x le 2),(x -1 ", when " x gt 2):}` is continuous at x = 2 then k = ? |
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Answer» we have `f(2) = ( k xx 2 +5) = ( 2 k +5)` `lim_(x to 2x) f(x)= lim_(h to 0) f( 2 +h)` ` lim_( h to 0) {( 2 +h) -1} = lim_(h to 0) ( 1+h) =1` ` lim_(x to 2-) f(x) = lim_( h to 0) f ( 2-h) ` ` lim_(h to 0) {" k ( 2-h) +5} = lim_( h to 0) {( 2 k + 5) -kh } = ( 2k +5)` Now ` lim_(x to 2) f(x) " exists only when " 2k + 5 =1 ` i.e when k = -2 . when k = -2 we have ` lim_( x to 2) f (x) = f( 2) =1 ` Hence f (x) is continuous at x=2 when k =-2 |
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| 54. |
If `f(x)=[x](sin kx)^(p)` is continuous for real x, then (where [.] represents the greatest integer function)A. `k in [npi, n in I], p gt 0`B. `k in {2npi, n in I}, p gt0`C. `k in {npi, n in I}, p in R-{0}`D. `k in {npi, n I, n ne 0}, p in R-{0}` |
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Answer» Correct Answer - A `f(x)=[x](sin kx)^(p)` `(sin kx)^(p)` is continuous function `AA x in R, k in R` and `p gt0.[x]` is discontinuous at `x in I` For `k=npi, n in I` `f(x)=[x](sin(n pix))^(p)` `underset(xrarra)(lim)f(x)=0, a in I` and f(a) = 0 So, f(x) becomes continuous for all `x in R` |
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| 55. |
For what value of k, the function `f(x) ={:{(Kx^2", " x le 2 ),(" "5", " xgt2):},` is continuous at x=2. |
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Answer» Correct Answer - `5/4` |
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| 56. |
If `f(x)=[{:(mx+1,if x le (pi)/(2)),(sinx+n,ifxgt(pi)/(2)):}` is continuous at `x = (pi)/(2)`, thenA. `m = 1, n = 0`B. `m = (n pi)/(2)+1`C. `n = (mpi)/(2)`D. `m = n = (pi)/(2)` |
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Answer» Correct Answer - C We have, `f(x)=[{:(mx+1,if x le (pi)/(2)),(sinx+n,ifxgt(pi)/(2)):}` is continuous at `x = (pi)/(2)`, `:. LHL = underset(xrarr(pi^(-))/(2))(lim)(mx+1)=underset(hrarr0)(lim)[m(pi/2-h)+1]=(mpi)/(2)+1` and `RHL = underset(xrarr(pi^(+))/(2))(lim)(sinx+n)=underset(hrarr0)(lim)[sin(pi/2+h)+n]` `= underset(hrarr0)(lim)cosh+n=1+n` `:. LHL = RHL` , [to be continuous at `x = (pi)/(2)`] `rArr m.(pi)/(2) + 1 = n + 1` `:. n = m.(pi)/(2)` |
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| 57. |
Let `f(x)=|x|+|sin x|, x in (-pi//2,pi//2).` Then, f isA. nowhere continuousB. continuous and differentiable everywhereC. nowhere differentiableD. differentiable everywhere except at x=0 |
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Answer» Correct Answer - D We have `f(x)=|x|+|sin x|"for all x"in (-pi//2,pi//2)` `Rightarrow f(x)={{:(,-x-sin x,"for"-pi//2 lt x lt 0),(,x+sin x,"for "0 le x lt x//2):}` Clearly, `underset(x to 0^(-))lim f(x)=f(0)=underset(x to 0^(+))lim f(x)` So, f(x) is continuous at x=0 Also, f(x) is continuous on `(-pi//2,0) uu (0,pi//2)` Thus, f(x) is continuous on `(-pi//2, pi//2)` Now, (LHD at x=0)`={(d)/(dx)(-x-sinx)}_("at x=0")=(-1-cos x)_("at x =0")=-1-1=-2` and (RHD at x=0) `={(d)/(dx)(x+sinx)}_("at x=0")=(1+cos x)_("at x =0")=1+1=2` Clearly, (LHD at x=0) `ne (RHD at x=0)` So, f(x) is not differentiable at x=0 Clearly f(x) is differentiable on `(-pi//2,0) uu (0,pi//2)` Hence, f(x) is everywhere differentiable except at x=0. |
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| 58. |
The set of all points where the function `f(x)=3sqrt(x^2|x|)` is differentiable, isA. `[0,oo)`B. `(0,oo)`C. `(-oo,oo)`D. `(-oo0) uu (0,oo)` |
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Answer» Correct Answer - D We have `f(x)=(x^(2)|x|)^(1//3)` `Rightarrow f(x)={{:(,(-x^(3))^(1//3)=-x, x lt 0),(,(x^(3))^(1//3)=x,x ge0):}` `Rightarrow f(x)=|x|` Clearly , it is everywhere differentiable except at x=0. Hence, the set of all points where f(x) is differentiable is `R-(0)=(-oo, 0) uu (0,oo)` |
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| 59. |
Differentiate w.r.t. x the function`(cos^(-1)x/2)/(sqrt(2x+7)),-2 |
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Answer» Let `y=("cos"^(-1)(x)/(2))/(sqrt(2x+7))` `implies (dy)/(dx)=(sqrt(2x+7)*(d)/(dx)"cos"^(-1)(x)/(2)-"cos"^(-1)(x)/(2)*(d)/(dx)sqrt(2x+7))/((sqrt(2x+7))^(2))` `=(sqrt(2x+7)*((-1))/(sqrt(1-(x^(2))/(4)))*(d)/(dx)((x)/(2))-"cos"^(-1)(x)/(2)*(1)/(2sqrt(2x+7))*(d)/(dx)(2x+7))/((2x+7))` `=((-sqrt(2x+7))/(sqrt(4-x^(2)))-("cos"^(-1)(x)/(2))/(sqrt(2x+7)))/((2x+7))` `= -[(sqrt(2x+7))/(sqrt(4-x^(2))*(2x+7))+("cos"^(-1)(x)/(2))/((2x+7)sqrt(2x+7))]` `= -[(1)/(sqrt(4-x^(2))sqrt(2x+7))+("cos"^(-1)(x)/(2))/((2x+7)^(3//2))]` |
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| 60. |
Differentiate w.r.t. x the function`cos" "(a" "cos" "x" "+" "b" "s in" "x)`, for some constant a and b. |
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Answer» Let `y=cos(acosx+bsin x)` `(dy)/(dx)=(d)/(dx)cos(acosx+bsin x)` `= -sin(acosx+bsin x)*(d)/(dx)(acosx+bsin x)` `= -sin(acosx+bsin x)*(-asinx+bcos x)` |
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| 61. |
`2 tan ^(-1)x-log(1+x^(2))` |
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Answer» Correct Answer - `2 tan ^-1 x ` |
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| 62. |
`log (sin ^(-1)x)` |
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Answer» Correct Answer - `(1)/(sin^-1xsqrt(1-x^2))` |
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| 63. |
`e^x.log (sin 2 x)` |
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Answer» Correct Answer - `e^x.[2 cot 2x + log (sin 2x)]` |
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| 64. |
If `xlogy-ylogx=1` then `((dy)/(dx))_(x=1)=` |
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Answer» `logy+x*1/y*dy/dx-logx*dy/dx-y*1/x=0` `logy-y/x+dy/dx(x/y-logx)=0` `dy/dx=((y/x-logy)/(x/y-logx))` `(dy/dx)_(x=1,y=e)=((e/1-loge)/(1/e-log1))` `=(e-1)/(1/e)=e(e-1)` Option C is correct. |
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| 65. |
If `lim_(x->oo)((x^2+x+1)/(x+1) -ax-b)=4` then a= ? b=? |
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Answer» `lim_(x->oo)((x^2+x+1-ax^2-ax-bx-b)/(x+1))=4` `lim_(x->oo)((x+1-2ax-a-b)/1)=4` `lim_(x->oo)((x(2-2a)+1-a-b)/1)=4` 2-2a=0 a=1 1-a-b=4 a+b=-3 b=-4. |
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| 66. |
Let `f(x) = (xe)^(1/|x|+1/x); x != 0, f(0) = 0`, test the continuity & differentiability at `x = 0`A. discontinuous everywhereB. continuous as well as differential for all xC. continuous for all c but not differential at x=0D. neither differential nor continuous at x=0 |
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Answer» Correct Answer - C We have `f(x)={{:(,xe^-(-(1)/(x)+(1)/(x))=x,x lt 0),(,xe^(-2//x),x gt 0),(,0,x=0):}` `"Clearly", underset(x to 0^(-))lim f(x)=underset(x to 0)lim x=0` `underset(x to 0^(+))lim f(x)=underset(x to 0)lim xe^(-2//x)=0xx0=0 and, f(0)=0` `therefore underset(x to 0^(-))lim f(x)=underset(x to 0^(+))lim f(x)=f(0)` So, f(x) is continuous at x=0 `"Also", ("LHD at x=0")=((d)/(dx)(x))_("at x=0")=1` `("RHD at x=0")={2^(-2//x)+(2e^(-2//x))/(x)}_("at x=0")`, which does not exist Thus, f(x) is everywhere continuous but not differerntiabl at x=0 |
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| 67. |
Let `f(x)=lim_(n to oo) ((2 sin x)^(2n))/(3^(n)-(2 cos x)^(2n)), n in Z`. ThenA. at `x=n pm(pi)/(6)`, f(x) is discontinuousB. `f((pi)/(3))=1`C. f(0)=0D. all of the above |
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Answer» Correct Answer - D We have `,f(x)=underset(n to oo)lim ((sin^(2)x)^(n))/(((3)/(4))^(n)-(cos^(2)x)^(n))` In the neighbourhood of `x=n pi pm (pi)/(6)`, we have `f(x)=(0)/(0-0)`, which does not exist. Hence, f(x) is discontinuous at `x=n pi pm pi//6, n in Z` Also, we have `f((pi)/(3))=underset(n to oo)lim ((3)/(3))^(n)/(((3)/(4))^(n)-((1)/(4))^(n))=underset(n to oo)lim (1)/(1-((1)/(3))^(n))=1` `and f(0)=underset( n to oo)lim (0)/(((3)/(4))^(n)-1)=0` Hence, option (a),(b) and (c ) are correct. |
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| 68. |
Let `f(x)={x^2|(cos)pi/x|, x!=0 and 0,x=0,x in RR,` then `f` isA. differentiable both at x=0 and x=2B. differentiable at x=0 but not differentiable at x=2C. not differentiable at x=0 but differentiable at x=2D. differentiable neither at x=0 nor at x=2 |
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Answer» Correct Answer - B We observe that `underset(x to 0)lim (f(x)-f(0))/(x-0)=underset(x to 0)lim (x^(2)|"cos "(pi)/(x)|-0)/(x-0)=underset(x to 0)lim x|"cos "(pi)/(2)|=0` So, f(x) is differentiable at x=0 Now, `underset(x to 2^(-))lim (f(x)-f(2))/(x-2)` `underset(h to 0 )lim (f(x)-f(2))/((2-h)-2)` `underset(h to 0 )lim ((2-h)^(2)|cos((pi)/(2-h)))/(-h)` `underset(h to 0 )lim ((2-h)^(2)sin((pi)/(2)-(pi)/(2-h)))/(h)a` `underset(h to 0 )lim ((2-h)^(2))/(h)sin{(-pih)/(2(2-h))}` `underset(h to 0 )lim (sin{(pih)/(2(2-h))})/((pih)/(2(2-h)))=pi` and `underset(x to 2^(+))lim=(f(x)-f(2))/(x-2)` `underset(x to 2^(+))lim=(f(2+h)-f(2))/((2+h)-2)` `underset(x to 2^(+))lim ((2+h)^(2)cos((pi)/(2+h))-0)/(h)` `underset(x to 2^(+))lim ((2+h)^(2))/(h)sin((pi)/(2)-(pi)/(2+h))` `underset(x to 2^(+))lim (2+h)^(2)/(h) sin {(pih)/(2(2+h))}` `=underset(h to 0)lim (sin{(pih)/(2(2+h))})/((pi)/(2(2+h)))xx(pi)/(2)(2+h)=(pi)/(2)xx2=pi` `therefore underset(x to 2^(-))lim (f(x)-f(2))/(x-2) ne underset(x to 2^(+))lim (f(x)-f(2))/(x-2)` So, f(x) is not differentiable at x=2 |
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| 69. |
Prove that the function defined by `f(x) = tan x` is a continuous function.A. RB. `R-{ n pi , n in Z }`C. ` R -{ (n pi)/(2), n in Z}`D. None of these |
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Answer» Correct Answer - D |
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| 70. |
The value of k for which `f(x)={{:(,(x^(2^(32))-2^(32)x+4^(16)-1)/((x-1)^(2)),x ne 1),(,k,x=1):}` is continuous at x=1, isA. `2^(63)-2^(31)`B. `2^(65)-2^(33)`C. `2^(62)-2^(31)`D. `2^(65)-2^(31)` |
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Answer» Correct Answer - A If f(x) is continuous at x=1, then `underset(x to 1)lim f(x)=f(1)` `Rightarrow underset(x to 1)lim f(x)=k` `Rightarrow underset(x to 1)lim (x^(2^(32))-2^(32)x+4^(16)-1)/((x-1)^(2))=k,` `Rightarrow underset(x to 1)lim ((x^(n)-1)-n(x-1))/((x-1)^(2))=k` `Rightarrow underset(x to 1)lim ((x^(n)-1)/(x-1)-n)/(x-1)=k` `Rightarrow underset(x to 1)lim ((x^(n-1)+x^(n-2)+....+x+1)-n)/(x-1)=k` `Rightarrow underset(x to 1)lim ((x^(n)-1)+(x^(n-2)-1)+...+(x-1))/(x-1)=k` `Rightarrow underset(x to 1)lim ((x^(n-1))/(x-1)+(x^(n-2)-1)/(x-1)+....+(x^(2)-1)/(x-1)+(x-1)/(x-1))=k` `Rightarrow (n-1)+(n-2)+....+2+1=k` `Rightarrow k=(n(n-1))/(2)=(2^(32)(2^(32)-1))/(2)=2^(63)-2^(31)` |
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| 71. |
In Example 138, h(0) equalsA. 6B. 16C. 2D. 15 |
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Answer» Correct Answer - B From example 138, `h(x)=f(f(x))"for all"x in R` `therefore h(0)=f(f(0))=f(2)=2^(3)+3xx2+2=16" "[therefore f(x)=x^(3)+3x+12]` |
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| 72. |
If the function `f(x)=((128 a+a x)^(1//8)-2)/((32+b x)^(1//5)-2)`is continuous at `x=0`, then the value of `a//b`is`3/5f(0)`b. `2^(8//5)f(0)`c. `(64)/5f(0)`d. none of theseA. `(3)/(5)f(0)`B. `2^(8//5)f(0)`C. `(64)/(5)f(0)`D. none of these |
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Answer» Correct Answer - C If f is continuous at x = 0, then `f(0)=underset(xrarr0)(lim)((128a+ax)^(1//8)-2)/((32+bx)^(1//5)-2)` As `xrarr0`, the denominator `rarr0`. Thus, for limit to exist the numberator must also `rarr0`. Thus, we have `(128a)^(1//8)=2` or a = 2. Now, we have `f(0)=underset(xrarr0)(lim)((256+2x)^(1//8)-2)/((32+bx)^(1//5)-2)" "((0)/(0))` `rArr" "f(0)=underset(xrarr0)(lim)((2)/(8)(256+2x)^(-7//8))/((b)/(5)(32+bx)^(-4//5))=(5)/(4b).(2^(-7))/(2^(-4))=(5)/(32b)` `rArr" "b=(5)/(32f(0))` ltBrgt Hence, we have `(a)/(b)=(64)/(5)f(0)` |
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| 73. |
Let `g(x) = f(f(x))` where `f(x) = { 1 + x ; 0 |
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Answer» Correct Answer - C f(x) is discontinuous at x = 2, So, f(f(x)) may be discontinuous when `1+x=2` (when `0le xle2)` or `3-x=2` (when `2 lt xle3`). `therefore" "x=1` Hence f(f(x)) is discontinuous at x = 1 and x = 2. |
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| 74. |
Find `(dy)/(dx),`when`x=cos^(-1)1/(sqrt(1+t^2))"and"y=sin^(-1)t/(sqrt(1+t^2)),tR` |
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Answer» Correct Answer - 1 |
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| 75. |
`f (x) = {{:((|x^(2)- x|)/(x^(2) - x),xne 0"," 1),(1",", x = 0),(-1",", x = 0 ):}` is continuose for all :A. xB. x except at x = 0C. x except at x = 1D. x except at = and x = 1 |
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Answer» Correct Answer - D |
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| 76. |
`y = sin ^(-1)(2xsqrt(1 - x^(2))),-(1)/sqrt(2) lt x lt (1)/sqrt(2)` |
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Answer» `y sin ^(-1)(2xsqrt(1-x^(2)))` Let `x = sin theta` `rArr theta= sin ^(-1)x` `rArr y = sin^(-1)(2 sin thetasqrt(1 - sin^(2) theta))` =` sin ^(-1)(2 sin thetacos theta)` ` sin ^(-1)(sin 2theta)= 2theta = 2 sin^(-1)x` ` (dy)/(dx)= 2 (d)/(dx) sin^(-1)x = - ((2)/(sqrt(1-x^(2))))` |
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| 77. |
Using Lagranges mean value theorem, find a point on the curve `y=sqrt(x-2)`defined on the interval [2,3], where the tangent is parallel to the chordjoining the end points of the curve. |
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Answer» Correct Answer - `(9/4,1/2)` |
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| 78. |
At which point the slope to tangent is zero for the curvey `y=x^2-6x+8` ?A. (3,1)B. (3,-1)C. (-3 ,1)D. (-3 ,-1) |
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Answer» Correct Answer - B |
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| 79. |
`y = sec^(-1)((1)/(2x^(2) -1 )), 0 lt x lt (1)/(sqrt(2))` |
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Answer» `y = sec^(-1)((1)/(2x^(2) -1 ))` Let x = cos theta `rArr theta cos ^(-1) x ` `rArr y sec ^(-1)((1)/(2 cos^(2)theta-1))` `= sec^(-1)((1)/(cos2 theta))` = `sec^(-1)(sec2 theta)` `= 2theta = 2 cos ^(-1)x ` `rArr (dy)/(dx)=2(d)/(dx) cos ^(-1)x = - (2)/(sqrt(1 - x ^(2)))` |
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| 80. |
Find the derivative of the function given `b y f(x) = sin(x^2)dot` |
| Answer» `(d(fx))/(dx)=cos(x^2)*2x` | |
| 81. |
`(e^(x))/(sinx)` |
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Answer» Let y = `(e^(x))/(sinx)` `rArr (dy)/(dx)=(d)/(dx) ((e^(x))/(sin x))= (sinx(d)/(dx)e^(x)-e^(x)(dy)/(dx) sin x)/(sin ^(2)x)` `(sin.x.e^(x)-e^(x). cos x)/(sin^(2)x)` ` = (e^(x) (sinx - cos x))/(sin^(2) x)` |
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| 82. |
Find `(dy)/(dx)`if `x-y =pi` |
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Answer» x-y=`pi` y=x-`pi` differentiate with respect to x `(deltay)/(deltax)=delta/(deltax)(x-pi)` `(deltay)/(deltax)=1` |
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| 83. |
Discuss the continuity of the function `f(x)={{:(2x -1 "," x lt 1),(3x-2"," x ge 1):}` |
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Answer» When `x lt 1` f(x)= 2x-1 is a polynomial function `therefore` f(x) is continuous in `x lt 1` When `x gt 1` f(x) =3x -2 is a polynominal function `there` f(x) is continous in `x gt 1` At x=1 L.H.L `=underset(xrarr1^-)lim f(x)=underset(hrarr0)limf(0-h)` `=underset(xrarr0)lim2(1-h)-1=2-1=1` R. H.L `=underset(xrarr1^+)lim f(x)=underset(hrarr0)limf(1+h)` `=underset(h rarr0)lim 3(1+h) -2=3-2=1` and f(1)=`3 xx1 -2 =1` `therefore underset(x rarr 1^+)f(x) = f(1)=underset(x rarr 1^(-1))lim f(x)` `therefore` f(x) is continuous at x=1 Therefore f(x) is a continuous function . |
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| 84. |
`log(sin x^2)` |
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Answer» Correct Answer - `2x . Cot x^2` |
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| 85. |
log (sec 2x +tan 2 x) |
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Answer» Correct Answer - 2 sec 2x |
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| 86. |
`e^(x)"sin"5x` |
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Answer» Let `y=e^(x)"sin"5x` `implies(dy)/(dx)=(d)/(dx)(e^(x)"sin" 5x)` `=e^(x)*5" cos"5x+"sin"5x*e^(x)` `=e^(x)(5" cos"5x+"sin"5x) ` `implies (d^(2)y)/(dx^(2))=(d)/(dx)[e^(x)(5" cos"5x+"sin"5x)]` `=e^(x)(-25" sin"5x+5" cos"5x)+(5" cos"5x+"sin"5x)e^(x)` `=e^(x)(10" cos"5x-24" sin"5x)` `=2e^(x)(5" cos"5x-12" sin"5x)` |
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| 87. |
Evaluate:`inte^sin^((-1)x)dxdot` |
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Answer» Let y ` = e^(sin ^(-1)x)` `rArr (dy)/(dx)(d)/(dx) e^(sin ^(-1)x)` ` = e^(sin ^(-1)x) .(dy)/(dx)sin ^(-1)x` ` = e^(sin ^(-1)x) .(1)/(sqrt(1 - x^(2)))=(e^(sin ^(-1)x))/(sqrt(1 - x ^(2)))` . |
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| 88. |
If `f(x) = [x] [sinx]` in `(-1,1)` then f(x) isA. continuous on (-1,0)B. differentiable on (-1,1)C. differentiable at x=0D. none of these |
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Answer» Correct Answer - A We have `f(X)=[x sin x]={{:(,-1xx -1=1,"if "-1 lt x lt 0),(,0xx0,"if "0 le x lt 1):}` Clearly, f(x) is not continuous at x=0. Consequently it is not differentiable at x=0 Since f(x) is constant function on `(-1,0) uu (0,1).` Som it is continuous and differentiable on `(-1,0) uu (0,1).` |
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| 89. |
`e^(6x)"cos"3x` |
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Answer» Let `y=e^(6x)" cos"3x` `implies(dy)/(dx)=(d)/(dx)(e^(6x)"cos"3x)` `=e^(6x)*(-3" sin"3x)+"cos"3x*(6e^(6x))` `=e^(6x)(-3" sin"3x+6" cos"3x)` `implies (d^(2)y)/(dx^(2))=(d)/(dx)[e^(6x)(-3" sin"3x+6" cos"3x)]` `=e^(6x)(-9" cos"3x-18" sin"3x)+(-3" sin"3x+6" cos"3x)*(6e^(6x))` `=e^(6x)(-9" cos"3x-18" sin"3x-18" sin"3x+36" cos"3x)` `=e^(6x)(27" cos"3x-36" sin"3x)` `=9e^(6x)(3" cos"3x-4" sin"3x)` |
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| 90. |
` = e^(x ^(3))` |
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Answer» Let y ` = e^(x ^(3))` `rArr (dy)/(dx)=(d)/(dx) = e^(x ^(3))` ` = e^(x ^(3)) (d)/(dx)x^(3) = 3x^(2) .= e^(x ^(3))` |
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| 91. |
Find the derivative of `x^x-2^(sinx)`w.r.t. `x` |
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Answer» `"Let" y = x^(x) - 2^("sin"x)` `"Let " u=x^(x) and v=2^("sin"x)` `therefore y=u-v rArr (dy)/(dx)= (du)/(dx)-(dv)/(dx) " "....(1)` `"Now", u=x^(x)` `rArr "log " u = "log " (x^(x)) = x" log " x` Differentiate both sides w.r.t.x `(1)/(u) (du)/(dx) = x*(d)/(dx)"log"x + "log" x * (d)/(dx)x` `rArr (du)/(dx) = u[x*(1)/(x) + "log"x*1] = x^(x)(1+"log"x)` `"and "v=2^("sin"x)` `rArr (dv)/(dx) = (d)/(dx)2^("sin"x)` `=2^("sin"x) " log" 2 * (d)/(dx) sinx` `=2^("sin"x) * "log"2 * "cos"x` Now from equation (1) `(dy)/(dx) = x^(x) (1+ "log"x)- 2^("sin"x) * "log"2 * "cos"x` |
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| 92. |
If `f(x)=|2-x|+(2+x)`, where (x)=the least integer greater than or equal to x, themA. `underset(x to 2^(-))lim f(x)=f(2)=2`B. f(x) is continuous and differentiable at x=2C. f(x) is neither continuous nor differentiable at x=2D. f(x) is continuous and non-differentiable at x=2 |
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Answer» Correct Answer - C We know that the |2-x| is everywhere continuous but (2+x) is not continuous at x=2. Therefore , f(x) is not continuous at x=2. Hence f(x) is not differentiable at x=2. |
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| 93. |
` "tan"^(-1)x` |
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Answer» Let `y= "tan"^(-1)x` `implies (dy)/(dx)=(d)/(dx)"tan"^(-1)x=(1)/(1+x^(2))` `implies (d^(2)y)/(dx^(2))=(d)/(dx)((1)/(1+x^(2)))=(d)/(dx)(1+x^(2))^(-1)` `= -(1+x^(2))^(-2)*(d)/(dx)(1+x^(2))` `= -(2x)/((1+x^(2))^(2))` |
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| 94. |
Differentiate `sin ^(-1)(2x sqrt(1-x^2))` with respect to `sin^-1 x.` |
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Answer» Correct Answer - 2 |
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| 95. |
Find derivative of `sin(tan^(-1)e^(-x))`w.r.t. to `x` |
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Answer» Let y =` sin (tan ^(-1) e^(-x))` `rArr (dy)/(dx) sin (tan^(-1) e ^(-x))` ` = cos ( tan ^(-1)e^(-x)) (d)/(dx) (tan^(-1)e^(-x))` `= cos (tan ^(-1)e^(-x)). (1)/(1+ (e^(-x))^(2))(d)/(dx) e^(-x)` `= cos (tan ^(-1)e^(-x))/(1 + e^(-2x)).e ^(-x). (d)/(dx) (-x)` `=-(e ^(x).cos (tan^(1) e^(x)))/(1+e^-2x) ` |
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| 96. |
`y^(x) = x^(y)` then find dy/dx |
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Answer» `y^(x) = x^(y)` `rArr "log" (y^(x)) = "log"(x^(y))` `rArr x"log"y = y"log"x` Differentiate both sides w.r.t.x `x * (d)/(dx) "log" y + "log" y * (d)/(dx)x` `= y * (d)/(dx)"log"x + "log"x * (d)/(dx)y` `rArr (x)/(y)(dy)/(dx) + "log" y * 1 = (y)/(x) + "log"x (dy)/(dx)` `rArr ((x)/(y) - "log"x)(dy)/(dx) = (y)/(x) - "log" y` `rArr (x-y "log"x)/(y) (dy)/(dx) = (y-x "log"y)/(x)` `rArr (dy)/(dx) = (y)/(x) * (y-x"log"y)/(x-y"log"x)` |
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| 97. |
Let `f(x) =` Degree of `(u^(x^2) + u^2 +2u + 3).` Then, at` x=sqrt2, f(x)` isA. continuous but not differentiableB. differentiableC. dicontinuousD. none of these |
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Answer» Correct Answer - A We have `f(x)="Degree of "(u^(x^(2))+u^(2)+2u+3)` `Rightarrow f(x)={{:(,x^(2),x gt sqrt2),(,2,x le sqrt2):}` `therefore underset(x to sqrt2)lim f(x)=2=f(2)` and `underset(x to sqrt2^(+))limf(x)=underset(x to sqrt2)lim =(sqrt2)^(2)=2 ` `therefore underset(x to sqrt2^(-))limf(x)=underset(x to sqrt2^(+))lim f(x)=f(sqrt2) ` So, f(x) is continuous at `x=sqrt2` Now, `("LHD at x=sqrt2")=((d)/(dx)(2))_(x=sqrt2)=0` `("RHD at x=sqrt2")=((d)/(dx)(x^(2)))_(x=sqrt2)=(2x)_(x=sqrt2)=2sqrt2` So, f(x) is not differentiable at `x=sqrt2` |
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| 98. |
If `cosy=xcos(a+y),`with `cosa!=+-1,`provethat`(dy)/(dx)=(cos^2(a+y))/(sina)dot` |
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Answer» `cosy=x cos(a+y)` `impliesx=(cosy)/(cos(a+y))` Differentiate both sides w.r.t. y `(dy)/(dx)=(cos(a+y)(d)/(dx)cosy-cosy(d)/(dy)cos(a+y))/([cos(a+y)]^(2))` `=(-cos(a+y)siny+cosysin(a+y))/(cos^(2)(a+y))` `=(sin(a+y-y))/(cos^(2)(a+y))=(sina)/(cos^(2)(a+y))` `implies(dy)/(dx)=(cos^(2)(a+y))/(sina) " " `[Hence Proved. |
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| 99. |
`Cosx^@` |
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Answer» Correct Answer - `-(pi)/&180)sin ((pi x)/(180))` |
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| 100. |
`y=e^(sinx)+(tanx)^(x)` |
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Answer» Correct Answer - `e^(sinx)cdotcosx+(tanx)^(x)[log(tanx)+2xcdotcosec2x]` |
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