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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.
| 201. |
Number points of discontinuity of `f(x)=tan^2x- sec^2 x` in `(0,2pi)` is |
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Answer» `f(x) = tan^2x - sec^2x = sin^2x/cos^2x - 1/cos^2x` We know, at `x = pi/2` and `x = (3pi)/2`, value of `cos x` will be `0`. So, given function will not be defined and not continuous at these points. So, number of points of discontinuity will be `2`. |
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| 202. |
Let y be an implicit function of x defined by `x^(2x) -2x^xcoty-1=0`.Then y (1) equals` |
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Answer» at x= 1`1-2 xx1 cot y - 1 = 0` `cot y =(1-1)/-2 = 0` `y= pi/2` `y = x^(2x) ` `ln y = 2x ln x ` `d(v v )/dx = v (dv)/dx + v (dv)/dx` so,`1/y dy/dx = 2 ln x + 2` `dy/dx= 2y(ln x+1) ` `= 2x^(2x) [ln x +1]` `x^(2x) - 2x^xcot y - 1 = 0` `2x^(2x) (ln x+1) - 2[ x^x(ln x + 1)cot y + x^x(-cosec^2 y)dy/dx]= 0` x=1; `2 xx 1(0+1) - 2[ 1 xx1xx0 + 1(-1) dy/dx] = 0` `2 + 2 dy/dx = 0` `dy/dx = -2/2 = -1` option 1 is correct |
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| 203. |
If the function `f(x)`defined as `f(x)`defined as `f(x)={3,x=0(1+(a x+b x^3)/(x^2)),x >0`is continuous at `x=0,`then`a=0`b. `b=e^3`c. `a=1`d. `b=(log)_e3`A. a = 0B. `b=e^(3)`C. `a=1`D. `b=log_(e)3` |
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Answer» Correct Answer - A::D `underset(hrarr0)(lim)f(0+h)=underset(hrarr0)(lim)(1+(ah+bh^(3))/(h^(2)))^(1//h)` `" "=underset(hrarr0)(lim)e^((1)/(h)ln(1+(ah+bh^(3))/(h^(2))))` For limit to exist, we must have `underset(hrarr0)(lim)(ah+bh^(3))/(h^(2))=0` `rArr" "underset(hrarr0)(lim)(a+bh^(2))/(h)=0` `therefore" "a=0` So, we have `underset(hrarr0)(lim)f(0+h)=underset(hrarr0)(lim)(1+bh)^(1//h)` `" "=underset(hrarr0)(lim)(1+bh)^((1//bh)b)=e^(b)` For f(x) to be continuous at x = 0, we must have `underset(xrarr0^(+))(lim)f(x)=f(0)` `rArr" "e^(b)=3` `therefore" "b=log_(e)3` |
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| 204. |
Find the value of k so that the function f defined by`f(x)={(kcosx)/(pi-2x),3 , "if" x !=pi/2"if" x=pi/2`is continuous at `x=pi/2` |
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Answer» Correct Answer - k =6 ` f( pi/2-0) = lim_( h to 0) (k cos ( pi /2 -h))/ ( pi -2 ( pi -h)) = lim_( h to 0) ( k sin h )/( 2h) = k / 2 lim_(h to 0) (sin h)/( h ( k/2 xx 1) = k/2` ` f( pi/2 +0) = lim_( h to 0) ( k cos ( pi/2 +h))/( pi-2( pi/2 +h)) = lim_( h to 0) ( - k sin h)/( -2h) = k /2 lim_( h to0) ( sinh )/ h= (k/2 xx 1) = k/2` ` k/2 =3 Rightarrow k=6 ` |
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| 205. |
show that ` f(x)=x^(3)` is contiuous at x=2 |
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Answer» we have `f(2) =2^(3)=8` `lim_(xto2+)f(x) = lim_(hto0)(2+h)^(3) =lim_(hto0) (8+h^(3) +12h +6h^(2))=8,` ` lim_(x to 2-) f(x) = lim_(hto0) (2 -h)^(3) = lim_(h to 0) ( 8-h^(3) -12h +6h^(2)) =8 ` `lim_(x to 2x) f(x) = lim_(x to 2-) f(x) = f(2)` Hence , f(x) is continous at x =2 . |
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| 206. |
Let f(x) = [ n + p sin x], `x in (0,pi), n in Z`, p a prime number and [x] = the greatest integer less than or equal to x. The number of points at which f(x) is not not differentiable is :A. pB. p-1C. 2p+1D. 2p-1 |
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Answer» Correct Answer - D We know that the greatest integer function [x] is neither continuous nor differentiable at integer points. Therefore, f(x)=[n+p sin x] is discontinuous and non-differentiable at those points where `n+p sinx` is an integer. Clearly, n+p sin x will be an integer when p sin x is an integer Now, p sin x is an integer, if `sin x=1,(1)/(p),(2)/(p),(3)/(p),........,(p-1)/(p) [therefore x in (0, pi) therefore sin x gt 0]` `Rightarrow x=sin^(-1)(1), sin^(-1)((2)/(p)),...... sin^(-1)((p-1)/(p))` at points `pi-"sin"^(-1)(1)/(p),....mpi-sin^(-1)((p-1)/(p))"also"` Hence, the total number of points of discontiniuity of f(x) is (2p-1). |
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| 207. |
Let [x] denote the greatest integer less than or equal to x and g (x) be given by`g(x)={{:(,[f(x)],x in (0","pi//2) uu (pi//2","pi)),(,3,x=(pi)/(2)):}` `"where", f(x)=(2(sin x-sin^(n)x)+|sinx-sin^(n)x|)/(2(sinx-sin^(n)x)-|sinx-sin^(n)x|),n in R^(+)` then at `x=(pi)/(2),g(x)`, isA. continuous and differentiable when `n gt 1`B. continuous and differentiable when `0 lt n lt 1`C. continuous but not differentiable when `n gt 1`D. continuous but not differentiable when `0 lt n lt 1` |
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Answer» Correct Answer - A Clearly, `0 lt sin x lt 1"for all "x in (0,pi//2)uu (pi//2,pi)` `CASE-I, "When n" gt1` In this case, we have `sin x gt sin^(n)x"for all "x in (0,pi//2) uu (pi//2,pi)` `Rightarrow sinx-sin^(n)x gt 0"for all "x in (0,pi//2)uu(pi//2,pi)` `Rightarrow |sin x -sin^(n)x|=sin x-sin^(n)x"for all "x in(0,pi//2) uu (pi//2,pi)` `therefore f(x)=(2(sin x sin^(n)x)+(sinx-sin^(n)x))/(2(sin x-sin^(n)x)-(sin x-sin^(n)x))=3` `Rightarrow [f(x)]=3"for all "x in (0,pi//2) uu (pi//2,pi)` ltrbgt Thus, we have `g(x)=3"for all "x in (0,pi)` Clearly, it is continuous and differential at `x=pi//2` CASE-II When `0 lt n lt 1` In this case, we have `sin x lt sin^(n) x "for all "x in (0,pi//2) uu (pi//2,pi)` `Rightarrow sin x-sin^(n)x lt 0"for all "x in(0,pi//2) uu (pi//2,pi)` `Rightarrow |sin x-sin^(n)x|=-(sinx-sin^(n)x)"for all "x in (0,pi//2)uu (pi//2,pi)` `therefore f(x)=(2(sin x-sin^(n)x)-(sin x-sin^(n)x))/(2(sin x-sin^(n)x)+(sinx-sin^(n)x))=(1)/(3)` `Rightarrow [f(x)]=0"for all "x in (0,pi//2)uu (pi//2,pi)` Thus, we have `g(x)={{:(,0,"for all"x in (0","pi//2)uu (pi//","pi)),(,3,"for "x =pi//2):}` Clearly, it is discontinuous and hence non-differerntiable also at `x=pi//2` |
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| 208. |
If `f:R to R` is defined by `f(x)={{:(,(x-2)/(x^(2)-3x+2),"if "x in R-(1,2)),(,2,"if "x=1),(,1,"if "x=2):}` `"them " lim_(x to 2) (f(x)-f(2))/(x-2)=` |
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Answer» Correct Answer - B We have `f(x)={{:(,(1)/(x-1),"if "x ne 1","2),(,2,"if "x=1),(,1,"if "x=2):}` `therefore underset(x to 2)lim (f(x)-f(2))/(x-2)` `underset(x to 2)lim ((1)/(x-1)-1)/(x-2)=underset(x to 2)lim-(x-2)/((x-1)(x-2))=-1` |
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| 209. |
If `f(x)={{:(,(e^(x[x])-1)/(x+[x]),x ne 0),(,1,x=0):}` thenA. `underset(x to 0^(+))lim f(x)=-1`B. `underset(x to 0^(-))lim f(x)=(1)/(e)-1`C. f(x) is continuous at x=0D. f(x) is discontinuous at x=0 |
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Answer» Correct Answer - D We have `f(x)={{:(,(e^(x[x])-1)/(x+[x]),x ne 0),(,1,x=0):}` `f(x)={{:(,(e^(x[x])-1)/(x+[x]),-1 le x lt 0),(,1,x=0):}` `f(x)={{:(,1,x=0),(,(e^(x)-1)/(x),0 lt x le1):}` Clearly, we have `underset(x to 0^(-))lim f(x)=underset(x to 0^(-))lim (e^(x-1)-1)/(x-1)=(e^(-1)-1)/(-1)=1-(1)/(e) and underset(x to 0^(+))lim f(x)=underset(x to 0^(+))lim (e^(x)-1)/(x)=1` So, f(x) is discontinuous at x=0 |
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| 210. |
Let `f : [0, 1] rarr [0, 1] `be a continuous function such that `f (f (x))=1 for all x in[0,1]`then:A. `f(x)=x` for at least one `x in (0,1)`B. f(x) will be differential in [0,1]C. f(x)+x=0 for at least one x such that `0 le xle 1`D. none of these |
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Answer» Correct Answer - A Since `f:[0,1] to [0,1]` is a continuous function. Therefore, Range `(f) subset [0,1] and f(x)` attains every value between f(0) and f(1). Let g(x)=f(x)-x for all `x in [0,1]` Clearly, g(x) is also continuous on [0,1] Also, `g(0)=f(0)-0=f(0) gt 0` and `g(1)=f(1)-1=0" "[therefore f(1) lt 1]` Thus, g(x) is continuous on [0,1] such that g(0) g(1)`lt 0 ` . Therefore there exists a point `c in (0,1)` such that `g(c)=0 Rightarrow f(c )=c` Hence, f(x)=x for at least one `x in (0,1)` |
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| 211. |
Let f(x) be given that `f(x)={{:(,x,"if x is rational"),(,1-x,"if x is irrational"):}` The number of points at which f(x) is continuous, isA. `oo`B. 1C. 0D. none of these |
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Answer» Correct Answer - C Let a be any rational number. Then, there exists a sequence `lt a_(n) gt` of irrational points such that lim `a_(n)=a` We have, `therefore underset(n to oo)lim f(a_(n))=1-a` `Rightarrow underset(n to oo)lim f(a_(n))=1-f(a)" "[therefore f(a)=a]` Thus, lim `a_(n)`=a but lim `f(a_(n)) ne f(a)`. So,f is discontinuous at x=a Similarly, f(x) is discontinuous at all iraational poitns. Hence, f(x) is nowhere continuous |
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| 212. |
Let `f(x)` be a continuous function defined for `1 |
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Answer» Correct Answer - C We know that a continuous function defined on a closed interval attains every value between its minimum and maximum values in the interval. Therefore f(x) being continous on [1,3] will attain every value between its maximum (M) and maximum (m) values. It is given that f(x) takes rational values for all x and have there are infinitely many irrational values between m and M. Therefore, f(x) can take rational value for all x, if f(x) has a constant rational value at all points between x=1 and x=3. In each other words, f(x)=constant for all ` x in [1,3]` But f(2)=10 `therefore f(x)=10 "for all "x in[1,3]` `Rightarrow f(4)=10` |
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| 213. |
Let `f(x)={{:(,3sinx+a^(2)-10a+30,x in Q),(,4 cos x,x in Q):}"which"`one of the following statements is correct?A. f(x) is continuous for all x when a=5B. f(x) must be continuous for all, x when a=5C. f(x) is continuous for all x, `=2pix-tan^(-1)((3)/(4)),n in Z,"when a=5"`D. f(x) is continuous for all `x=2pix-tan^(-1)((4)/(3)),n in Z` when a=5 |
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Answer» Correct Answer - C If f(x) is continuous for all x, then `3 sinx+a^(2)-10a+30=4cosx "for all x"` `Rightarrow (a-5)^(2)+5 cos x-3sinx"for all x"` `Rightarrow (a-5)^(2)+5 cos (x+theta),"where " theta="tan"^(-1)(3)/(4)"for all x"` We observe that `LHS ge 5`. So, the two sides of the above equality are euqal if `a=5 and cos(x+theta)=1` `Rightarrow a=5 and x=2pix-"tan"^(-1)(3)/(4)` |
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| 214. |
For every pair of continuous functions `f,g:[0,1]->R` such that `max{f(x):x in [0,1]}= max{g(x):x in[0,1]}` then which are the correct statementsA. `(f(c))^(2)+3f(c)=(g(c))^(2)+3g(c)"for some c" in [0,1]`B. `(f(c))^(2)+f(c)=(g(c))^(2)+3g(c)"for some c" in [0,1]`C. `(f(c))^(2)+3f(c)=(g(c))^(2)+g(c)"for some c" in [0,1]`D. `(f(c))^(2)+(g(c))^(2)"for some c" in [0,1]` |
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Answer» Correct Answer - A::D Since f(x) and g(x) are continuous on [0,1]. So, they attain their maximum and minimum values in [0,1] Suppose f(x) and g(x) attains their maximum values in [0,1]. Suppose f(x) and g(x) attain their maximum values at `x_(1) and x_(2)` respectively. It is given that `f(x_(1))=g(x_(2))` Let h(x)=f(x)-g(x). Then h(x) is continuous on [0,1] such that `h(x_(1))=f(x_(1))-g(x_(1))ge 0" "[{:(,therefore,f(x_(1))=g(x_(2))ge g(x_(1))),(,therefore,f(x_(1))-g(x_(1))ge 0):}` `and h(x_(2))=f(x_(2))-g(x_(2))le 0" "[{:(,therefore,g(x_(2))=g(x_(1))ge f(x_(2))),(,therefore,f(x_(2))-g(x_(2))ge 0):}` Therefore, there exists `c in (0,1)` such that `h(c)=0 Rightarrow f(c)=g(c)` Clearly, f(c)=g(c) satisfy options (a) and (d). |
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| 215. |
show that ` f(x) ={{:( (2x-1)",",if , x lt2),( (3x)/2",,if x ge 2):} ` is continuous |
| Answer» Take three cases= (i)a gt 2 ` (ii) ` a lt 2` (iii) a=2 | |
| 216. |
Locate the point of discontinuity of the function. ` f(x)={{:((x^(3) -x^(2) +2x -2)"," ,if , x ne 1),( 4",", if, if =1 ):} ` |
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Answer» Correct Answer - discontinuous at x=1 Test at x=1 |
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| 217. |
Discuss the continuity of the function ` f(x) =|x|+|x-1|` in the interval [ -1,2] |
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Answer» Correct Answer - continuous The given function may be expressed as ` f(x)={{:(( 1-2x)",","when" , 1-le x lt 0),(1",","when",0le x le 1),((2x-1)",","when",1 lt x le2):}` Since a polynomial function as weill as a constant , if follows that f(x) is continuous when ` -1 x lt x lt 0, 0 lt x lt 0: 0 lt x lt 1 and 1 lt x lt 2 ` So , test the continuity at -1 , 0, 1 and 2 At x = -1 ` lim_( xto (-1) +) . f(x) = f( -1) ` so it is continuous at x =-1 |
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| 218. |
For what value of k, the function `f(x) ={:{((x^2-4)/(x-2)", " x ne 2),(" "k", " x=2):},` is continuous at x =2. |
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Answer» Correct Answer - B It is given that f(x) is right continous at x=2 `therefore underset(xto2^(+))lim f(x)=f(2) ` `Rightarrow underset(hto0)lim f(2+h)=k ` `Rightarrow underset(hto0)lim {(2+h)^(2)+e^((1)/(2-(2-h)))}^(-1)=k` `Rightarrow (4+0)^(-1)=k Rightarrow k=(1)/(4)` |
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| 219. |
The function `f(x)=[x^(2)]+[-x]^(2)`, where [.] denotes the greatest integer function, isA. continuous and derivable at x=2B. neither continuous nor derivable at x=2C. continuous but not dervable at x=2D. none of these |
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Answer» Correct Answer - B We have `underset(x to 2^(-))lim f(x)=underset(h to 0)lim f(2-h)=underset(h to 0)lim [(2-h)^(2)]+[-2+h]^(2)` `Rightarrow underset(x to 2^(-))lim f(x)=3+(-2)^(2)=7` `underset(x to 2^(+))lim f(x)=underset(h to 0)lim f(2+h)=underset(h to 0)lim [(2+h)^(2)]+[-2-h]^(2)` `Rightarrow underset(x to 2^(-))lim f(x) ne underset(x to 2^(+))lim f(x)` So, f(x) is discontinuous at x=2 Consequently, it is non -differentiable at x=2. |
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| 220. |
Thefunction `f:""R""~""{0}vecR`given by`f(x)=1/x-2/(e^(2x)-1)`can be made continuous at x = 0 by definingf(0) as(1)2(2) `-1`(3) 0(4) 1 |
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Answer» Correct Answer - B For f(x) to be continous at x=0, we must have gtbrgt `f(0)=underset(x to 0)lim f(x)` `Rightarrow f(0)=underset(xto 0) lim ((1)/(x)-(2)/(e^(2x)-1))` `Rightarrow f(0)=underset(xto 0) lim (e^(2x)-1-2x)/(x(e^(2x)-1))` `Rightarrow f(0)=underset(xto 0) lim ((1+2x+((2x)^(2))/(3!)+....)-1-2x)/((x+(1+2x+)(2x)^(2)/(2!)+.....-1))` `Rightarrow f(0)=underset(xto 0) lim ((2x)^(2) {(1)/(2!)+(2x)/(3!)...})/(x{2x+((2x)^(2))/(2!)+...})` `Rightarrow f(0)=underset(xto 0) lim (2{(1)/(2!)+(2x)/(3!)...})/({1+(2x)/(2!)+...})=2xx(1)/(2)=1` |
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| 221. |
Let `f:R to R` be any function. Defining `g: R to R` by `g(x)=|f(x)|" for "x to R`. Then g, isA. onto if if is ontoB. one-one if f is one-oneC. continuous if f is continuousD. differentiable if f is differentiable |
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Answer» Correct Answer - C Let `h(x)=|x|. "Then " h:R to R` is continuous many-one and into function. We have `hof (x)=h(f(x))=|f(x)|=g(x)` Since composition of continuous functions is continuous. Therefore, g(x) is continuous if f is continuous Since, composition of two bijections is a bijection. Here h(x) is many-one. So, g(x) cannot be one-one even if f is one-one. Also, g(x) cannot be onto even if f is onto. We observe that (f)=sin x is everywhere differentiable but |sin x|is not differentiable at `x=n pi, n in Z`. Therefore g(x) need not be differentiable even if f is differentiable |
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| 222. |
The left hand derivative of `f(x)=[x]sin(pix)` at `x = k`, `k` is an integer, isA. `(-1)^(k) (k-1)pi`B. `(-1)^(k-1) (k-1)pi`C. `(-1)^(k)kpi`D. `(-1)^(k-1)kpi` |
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Answer» Correct Answer - A We have `f(x)=[x] sin pi x` `therefore ("LHD at x=k")=underset(x to k^(-))lim (f(x)-f(k))/(x-k)` `Rightarrow ("LHD at x=k")=underset(h to 0)lim (f(k-h)-f(k))/(-h)` `Rightarrow ("LHD at x=k")=underset(h to 0)lim ([k-h)sin (pik-pih)-[k]sin pik)/(-h)` `Rightarrow ("LHD at x=k")=underset(h to 0)lim ((-1)^(k-1)[k-h]sin pi h-0)/(-h)` `Rightarrow ("LHD at x=k")=underset(h to 0)lim (-1)^(k) [k-h] (sin pi h)/(h)` `Rightarrow ("LHD at x=k")=(-1)^(k) (k-1) underset(h to 0)lim (sin pi h)/(h)=(-1)^(k) (k-1)pi` |
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| 223. |
Q. For every integer n, let an and bn be real numbers. Let function `f:R->R` be given by a `f(x)={a_n+sin pix, for x in [2n,2n+1]`, `bb_n+cos pix, for x in (2n+1,2n)` for all integers n.A. `a_(n)-b_(n+1)=-1`B. `a_(n-1)-b_(n-1)=0`C. `a_(n)-b_(n)=1`D. `a_(n-1)-b_(n)=1` |
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Answer» Correct Answer - B For the continuity of f at x=2n+1, we must have `underset(x to (2n+1))lim f(x)=underset(x to (2n+1)^(+))lim f(x)=f(2n+1)` `Rightarrow underset(x to (2n+1)^(-))lim a_(n)+sin pix=underset(x to (2n+1)^(+))lim b_(n+1)+cos pi x=a_(n)+sin x(2n+1)` `Rightarrow a_(n)+sin(2n+1)pi=b_(n+1)+cos(2pi+1)pi =a_(n)+sin(2n+1)pi` `Rightarrow a_(n)=b_(n+1)=1 Rightarrow a_(b)-b_(n+1)=-1` Replacing n by n-1, we get `a_(n-1)=b_(n)=-1` For the continuity of f at x=2n, we must have `underset(x to 2n^(-))lim f(x)=underset(x to 2n^(+))lim f(x)=f(2n)` `Rightarrow underset(x to 2n^(-))lim b_(n)+cospi x=underset(x to 2n^(+))lim a_(n)+sin pix=a_(n)+sin 2npi` `Rightarrow b_(n)+cos 2npi=a_(n)+sin2nxpi` `Rightarrow b_(n)+1=a_(n) Rightarrow a_(n)-b_(n)=1` So, option (c )is correct. Hence, option (b) does not hold. |
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| 224. |
`f(x)={(x^2+e^(1/(2-x)))^(-1)k ,x=2, x!=2`is continuous from right at the point `x=2,`then `k`equals`0`b. `1//4`c. `-1//4`d. none of these |
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Answer» Correct Answer - B `k=underset(xrarr2^(+))(lim)(x^(2)+e^((1)/(2-x)))^(-1)` `therefore" "k=underset(hrarr0)(lim)[(2+h)^(2)+e^((-1)/(4))]^(-1)` `=(4+e^(-oo))^(-1)` `=1//4` |
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| 225. |
In Example 138, h(0) equalsA. 66B. 6C. 36D. 38 |
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Answer» Correct Answer - D From example 138, we obtain `h(x)=f(f(x))"for all x"in R` Also, `g(f(x))=x"for all x"in R`. Therefore, `f(g(x))=x"for all" x in R` Now, `h(g(3))=f(f(g(3)))=f(3)=27+9+2=38` |
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| 226. |
A function `f:R rarrR` is defined as `f(x)=lim_(nrarroo) (ax^(2)+bx+c+e^(nx))/(1+c.e^(nx))` where f is continuous on R, thenA. point (a, b, c) lies on line in spaceB. point (a, b) represents the 2-dimensional Cartesian planeC. Locus of point (a, c) and (c, b) intersect at one pointD. point (a, b, c) lies on the plane in space |
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Answer» Correct Answer - A::B::C `f(x)=underset(nrarroo)(lim)(ax^(2)+bx+c+e^(nx))/(1+c.e^(nx))` `={{:(underset(nrarroo)(lim)(ax^(2)+bx+c+(e^(x))^(n))/(1+c.(e^(x))^(n)),,xlt0),(underset(nrarroo)(lim)(ax^(2)+bx+c+(e^(x))^(n))/(1+c.(e^(x))^(n)),,x=0),(underset(nrarroo)(lim)((ax^(2)+bx+c)/((e^(x))^(n)))/((1)/((e^(x))^(n))+c),,xgt0):}` `={{:(ax^(2)+bx+c,,xlt0),(1,,x=0),((1)/(c),,xgt0):}` since f(x) is continuous function `AA x in R` `therefore" "underset(xrarr0^(+))(lim)f(x)=underset(xrarr0^(-))(lim)f(x)=f(0)` `rArr" "underset(xrarr0^(+))(lim)((1)/(c))=underset(xrarr0^(-))(lim)(ax^(2)+bx+c)=1` `rArr" "(1)/(c)=1=c rArr c=1` `therefore " "c=1, a, b in R` |
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| 227. |
If `f (x) = x sin (1)/(2), x ne 0` then the value of the ltbRgt function at x = 0 so that the function is continuous at x= -0 , is :A. 1B. 0C. -1D. None of these |
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Answer» Correct Answer - B |
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| 228. |
Let `a, b in RR` and `f : RR rarr RR` be defined by `f(x) = a cos(|x^3-x|) + b|x| sin(|x^3+x|).` Then `f` isA. differentiable at x=0, if a=0 and b=1B. differentiable at x=1, if a=1 and b=0C. not differentiable at x=0, if a=1 and b=0D. not differerntiable at x=1, if a=1 and b=1 |
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Answer» Correct Answer - A::B We have `f(x)=a cos(|x^(3)-x|)+b|x|sin(|x^(3)+x|)"for all"x in R` `Rightarrow f(x)=a cos(x^(3)-x)+bx sin(x^(3)+x)"for all "x in R` Clearly, f(x) is the sum of two continuous and differentiable functions for all `x in R`. Hence options (a) and (b) are correct. |
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| 229. |
If `f(x)={{:(,x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)]),x ne 0),(,k,x=0):}` and `n in N`. Then the value of k for which f(x) is continuous at x=0 isA. nB. n+1C. `n(n+1)`D. `(n(n+1))/(2) |
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Answer» Correct Answer - D For any `x gt 0`, we know that `(1)/(x)-1 lt [(1)/(x)] le (1)/(x) Rightarrow 1-x lt x[(1)/(x)] le 1` `(2)/(x)-1 lt [(2)/(x)] le (2)/(x) Rightarrow 2-x le x[(2)/(x)] le 2` `(n)/(x)-1 lt [(n)/(x)] le (n)/(x) Rightarrow n-x lt x[(n)/(x)] le n` `therefore (n(n+1))/(2)-nx lt x ([(1)/(x)]+[(2)/(x)]+....+[(n)/(x)])le (n(n+1))/(2)` But `underset(x to 0)lim {(n(n+1))/(2)-nx}=(n(n+1))/(2)and, underset(x to )lim (n(n+1))/(2)=(n(n+1))/(2)` Therefore, by Sandwhich theoram, we obtain `underset(x to 0)lim x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)])=(n(n+1))/(2)` |
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| 230. |
Let `f:[-1/2,2] rarr R` and `g:[-1/2,2] rarr R` be functions defined by `f(x)=[x^2-3]` and `g(x)=|x|f(x)+|4x-7|f(x)`, where [y] denotes the greatest integer less than or equal to y for `yinR`. Then,A. f is discontinuous exactly at three points in [-1/2,2]B. f is discontinuous exactly at four points in [-1/2,2]C. g is not differentiable exactly at four points in [-1/2,2]D. g is not differentiable exactly at five points in [-1/2,2] |
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Answer» Correct Answer - B::C We have `f(x)=[x^(2)-3]=[x^(2)]-3` `and g(x)=(|x|+4x-7|)f(x)` `Now, x in [-1//2,2] Rightarrow 0 le x^(2) le 4` So, `f(x)=[x^(2)-3` is discontinuous and hence non-differentiable at `x=1,sqrt2,sqrt3 and 2 "in "[-1//2,2]` Since, `g(x)=(|x|+4x-7|)f(x)`. So, g(x) is not continuous and hence non-differentiable at `x=1,sqrt2,sqrt3 " in "(-1//2,2)` In the left neighbourhood of x=0, we find that `g(x)=(-5x+7)(-3)=15x-21` In the right neighbourhood of x=0, we have `g(x)=(-3x+7)(-3)=9x-21` Clearly, g is not differentiable at x=0 In the neighbourhood of x=7//4 `g(x)=(-3x+7)xx0=0` In the right neighbourhood f x=7/4. `g(x)=(-5x-7)xx0=0` So, g(x) is differentiable at x=0. Hence, g(s) is not differentiable at `x=0,1,sqrt2,7//4" in "(-1//2,2)` |
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| 231. |
The value of k for which `f(x)={{:(,[1+x(e^(-1//x^(2)))sin(1/(x^(4)))]^(e1//x^(2)),x ne 0),(,k,x=0):}` is continuous at x=0, isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A For f(x) to be continuous at x=0, we must have `underset(x to 0)lim f(x)=f(0)` `Rightarrow underset(x to 0)lim (1+xe^(-1//x^(2))"sin"(1)/(x^(4)))^(e^(1//x^(2)))=k` `Rightarrow underset(e^(x) to 0)lim (x" "e^(1//x^(2)) e^(1//x^(2)) "sin"(1)/(x^(4)))=k` `Rightarrow underset(e^(x) to 0)lim x "sin"(1)/(x^(4))=k` `Rightarrow e^(0)=k Rightarrow k=1` |
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| 232. |
Let `f(x)={{:(,|x|-3,x lt 1),(,|x-2|+a,x ge 1):}` `g(x)={{:(,2-|x|,x lt 1),(,Sgn(x)-b,x ge 1):}` If h(x)=f(x)+g(x) is discontinuous at exactly one point, then which of the following are correct?A. a=3,b=0B. a=-3,b=-1C. a=2,b=1D. a=0,b=3 |
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Answer» Correct Answer - B::C We have `f(x)={{:(,-x-3,x lt 0),(,x-3,0 le x lt 1),(,-x+2+a,1 le x lt 2),(,x-2+a,x ge 2):}` `and g(x)={{:(,2+x,x lt 0),(,2-x,0 le x lt 1),(,1-b,1 le x lt 2),(,1-b,x ge 2):}` `therefore h(x)=f(x)+g(x)={{:(,-1,x lt 0),(,-1, 0 le x lt 1),(,a-b+3-x,1 le x lt 2),(,a-b-1+x,x ge 2):}` We observe that h(x) is continuous for all the values of x, except at x=1. At x=1, it will be discontinuous, if `underset(x to 1^(-))lim h(x) ne underset(x to 1^(+))lim h(x)i.e. if-1 ne a-b+2 or, if a-bne -3` Clearly, values in options (b) and (c) satisfy this relation. |
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| 233. |
Let `f(x)={{:(x[(1)/(x)]+x[x],if, x ne0),(0,if,x=0):}` (where [x] denotes the greatest integer function). Then the correct statement is/areA. Limit exists for `x=-1`.B. f(x) has a removable discontinuity at x = 1.C. f(x) has a non removable discontinuity at x = 2.D. f(x) is discontinuous at all positive integers. |
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Answer» Correct Answer - A::B::C::D `f(1^(+))=underset(xrarr1^(+))(lim)(x[(1)/(x)]+x[x])` `=underset(xrarr1^(+))(lim)(x(0)+x(1))` `=1` `f(1^(-))=underset(xrarr1^(-))(lim)(x[(1)/(x)]+x[x])` `=underset(xrarr1^(-))(lim)(x(0)+x(1))` `=1` `f(2^(+))=underset(xrarr2^(+))(lim)(x[(1)/(x)]+x[x])` `=underset(xrarr2^(+))(lim)(x(0)+x(2))` `=4` `f(2^(-))=underset(xrarr2^(-))(lim)(x[(1)/(x)]+x[x])` `=underset(xrarr2^(-))(lim)(x(0)+x(2))` `=2` Obviously f(x) is discontinuous at all positive integers but at x = 1 it has removable discontinuity. |
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| 234. |
Statement-1: If `|f(x)| le |x|` for all `x in R` then `|f(x)|` is continuous at 0. Statement-2: If `f(x)` is continuous then `|f(x)|` is also continuous.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A Let f be continuous, at x-a, Then for every `epsilon gt 0` there exist `delta gt 0` such that `|f(x)-f(a)| lt in "whenever" |x-a| lt delta` `Rightarrow ||f(x)-|f(a)|| lt |f(x)-f(a)| lt in "whenever"|x-a| lt delta` `Rightarrow |f|(x)-|f|(a) lt in "whenever" |x-a| lt delta` `Rightarrow |f|` is continuous at x=a So, statement-2 is true. Now, `|f(x)| le |x|` `Rightarrow ,|f(0)| le 0" "["Replacing x by 0"]` `Rightarrow f(0)=0` `therefore |f(x)| le |x|` `Rightarrow |f(x)-f(0)| lt |x-0|" "[therefore f(0)=0]` `Rightarrow |f(x)-f(0)| lt in "whenever" |x-0| lt delta ( in)` `Rightarrow |f(x)|` is continuous at x=0 [Using statement-2] Hence both the statements are true and statement-2 is a correct explanation for statement-1, |
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| 235. |
Let `f(x)={{:(,sum_(r=0)^(x^(2)[(1)/(|x|)])r,x ne 0),(,k,x=0):}` where [.] denotes the greatest integer function. The value of k for which is continuous at x=0, isA. 1B. 2C. 4D. `(1)/(2)` |
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Answer» Correct Answer - A Clearly, f(x) is an even function and is given by `f(x)={{:(,(x^(2))/(2)[(1)/(|x|)] ([(1)/(|x|)]+1),x ne 0),(,k,x=0):}` If f is continous at x=0, then `underset(x to 0)lim f(x)=f(0)` `underset(x to 0)lim (x^(2))/(2)[(1)/(|x|)] ([(1)/(|x|)]+1)=k` `Rightarrow underset(x to 0)lim ([(1)/(|x|)] ([(1)/(|x|)]+1))/((1)/(|x|^(2)))=2k` `Rightarrow underset( y to oo)lim ([y]([y]+1))/(y^(2))-2k,"where "y=(1)/(|x|)` `Rightarrow underset( y to oo)lim ([y])/(y) (([y])/(y)+(1)/(y))=2k` `Rightarrow 1(1+0)=2k" "[therefore underset(x to oo)lim (x)/([x])=1]` `Rightarrow k=1` |
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| 236. |
If `f(x)=(tan(pi/4-x))/(cot2x)`for `x!=pi/4,`find the value which can beassigned to `f(x)`at `x=pi/4`so that the function `f(x)`becomes continuous every wherein `[0,pi/2]dot`A. 1B. `1//2`C. 2D. none of these |
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Answer» Correct Answer - B For f(x) to be continous at `x=(pi)/(4)`, we must have `underset(x to pi//4)limf(x)=f((pi)/(4))` `underset(x to pi//4)lim (tan(pi//4-x))/(cot 2x)=f((pi)/(4))` `underset(x to pi//4)lim (tan(pi//4-x))/(tan (pi//2-2x))=f((pi)/(4))` `Rightarrow (1)/(2) underset(x to pi//4)lim ({tan (pi//4-x)/((pi//4-x))})/({(tan2(pi//4-x))/(2(pi//4-x))})=f((pi)/(4)) Rightarrow (1)/(2)=f((pi)/(4))` |
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| 237. |
`(log x)^(log x),x gt1` |
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Answer» Let `y=(log x)^(log x)` `implieslogy=log[(log x)^(log x)]=logx*log(logx)` Differentiate both sides w.r.t. x `(1)/(y)(dy)/(dx)=logx*(d)/(dx)log(logx)+log(logx)(d)/(dx)logx` `=logx*(1)/(logx)*(d)/(dx)logx+log(logx)*(1)/(x)` `implies (dy)/(dx)=y[(1)/(x)+(1)/(x)log(logx)]` `=(logx)^(logx)[(1)/(x)+(1)/(x)log(logx)]` |
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| 238. |
The domain of the derivative of the function: `f(x)={{:(,tan^(-1)x,|x| le1),(,(1)/(2)(|x|-1),|x|gt1):}`A. `R-{0}`B. `R-{1}`C. `4-{-1}`D. `R-{-1,1}` |
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Answer» Correct Answer - D We have `f(x)={{:(,(1)/(2)(-x-1),x lt -1),(,tan^(-1)x,-1lexle1),(,(1)/(2)(x-1),x gt1):}` We observe that `underset(x to -1^(-))lim f(x)=underset(x to -1^(-))lim (1)/(2)(-x-1)=0` `underset(x to -1^(-))lim f(x)=underset(x to -1^(+))lim tan^(-1)x=tan^(-1)=-pi//4` Clearly, `underset(x to -1^(-))lim f(x) ne underset(x to -1^(+))lim f(x)` So, f(x) is not continuous at x=-1 Similarly, f(x) is not continuous at x=1 Consequenctly f(x) is not differentiable at `x=pm1` At all other points f(x) is differentiable. |
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| 239. |
Differentiate the following w.r.t. x:`e^x+e^x^2+. . . .+e^x^5` |
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Answer» `"Let"y = e^(x) + e^(x^(2)) + e^(x^(3)) + e^(x^(4)) + e^(x^(5))` `rArr (dy)/(dx) = (d)/(dx) (e^(x) + e^(x^(2)) + e^(x^(3)) + e^(x^(4)) + e^(x^(5)))` `=e^(x) + e^(x^(2)) (d)/(dx) x^(2) + e^(x^(3))(d)/(dx) x^(3) + e^(x^(4))(d)/(dx) x^(4) + e^(x^(5))(d)/(dx) x^(5)` `=(e^(x) + 2x* e^(x^(2)) + 3x^(2)* e^(x^(3)) + 4x^(3)* e^(x^(4)) + 5x^(4)*e^(x^(5)))` |
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| 240. |
`"log" ("log"x), x gt 1` |
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Answer» `"Let"y = "log" ("log"x)` `rArr (dy)/(dx) = (d)/(dx) "log" ("log"x)` `=(1)/("log"x)*(d)/(dx)"log"x = (1)/( x "log" x)` |
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| 241. |
Differentiate the following w.r.t. x:`sqrt(e^(sqrt(x))), x >0` |
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Answer» `" Let"y = sqrt(e^sqrt(x))` `rArr (dy)/(dx) = (d)/(dx) sqrt(e^sqrt(x)) = (1)/(2sqrt(e^(sqrt(x))))*(d)/(dx)*e^(sqrt(x)` `=(e^(sqrt(x)))/(2sqrt(e^(sqrt(x))))*(d)/(dx)sqrt(x)` `=(e^(sqrt(x)))/(2sqrt(e^(sqrt(x))))*(1)/(2sqrt(x)) = (e^(sqrt(x)))/(4sqrt(x*e^(sqrt(x))))` |
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| 242. |
Differentiate the functions given w.r.t. x:`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` |
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Answer» Let `y = sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))` Taking log both sides, `=> lny = 1/2ln[((x-1)(x-2))/((x-3)(x-4)(x-5))]` Differentiating both sides w.r.t. x, `1/ydy/dx = 1/2[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)-1/(x-5)]` `=>dy/dx = y/2[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)-1/(x-5)]` `=>dy/dx = 1/2sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)-1/(x-5)]` |
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| 243. |
`sec^(-1)(x/a)` |
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Answer» Correct Answer - `(a)/(xsqrt(x^2-a^2))` |
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| 244. |
`y=(1+x)^(x)` |
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Answer» Correct Answer - `(1+x)^(x)cdot[x/(1+x)+log(1+x)]` |
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| 245. |
(i) `y=((x-a)(x-b))/(sqrt(x-c))` (ii) `y=sqrt(((x-a)(x-b))/((x-c)(x-d)))` |
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Answer» Correct Answer - (i) `((x-a)(x-b))/(sqrt(x-c)){1/(x-a)+1/(x-b)-1/(2(x-c))}` (ii) `1/2sqrt(((x-a)(x-b))/((x-c)(x-d)))cdot[1/(x-a)+1/(x-b)-1/(x-c)-1/(x-d)]` |
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| 246. |
`sec^-1((1+tan ^2x)/(1-tan ^2 x))` |
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Answer» Correct Answer - 2 |
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| 247. |
`(tanx)^(y)=(tany)^(x)` |
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Answer» Correct Answer - `(logtany-2ycosec2x)/(logtanx-2xcosec2y)` |
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| 248. |
`y=sinxcdotsin2xcdotsin4xcdotsin8x` |
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Answer» Correct Answer - `sinxcdot sin2xsin4xcdot sin8x[cotx+2cot2x+4cot4x+8cot8x]` |
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| 249. |
`(1)/(sqrt(x+1)+sqrtx)` |
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Answer» Correct Answer - `1/2 [(1)/(sqrt(x+1))-(1)/(sqrt(x))]` |
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| 250. |
`tan sqrt(x)` |
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Answer» Correct Answer - `(sec^2sqrt(x))/(2sqrt(x))` |
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