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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. |
The number of points of discontinuity of `f(x)=[2x]^(2)-{2x}^(2)` (where [ ] denotes the greatest integer function and { } is fractional part of x) in the interval `(-2,2)`, isA. 1B. 6C. 2D. 5 |
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Answer» Correct Answer - B Given, `f(x)=([2x]+{2x})([2x]-{2x})=4x-4x{2x}` `2x in (-4,4)` Hence f(x) is discontinuous when `2x=-3, -2, -1, 1,2,3.` At `x=0,f(x)` is continuous |
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| 2. |
The function `f(x)=(x^3)/8-s inpix+4in[-4,4]`does not take the value`-4`b. `10`c. `18`d. `12`A. `-4`B. 10C. 18D. 12 |
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Answer» Correct Answer - C `f(x)=(x^(3))/(8)-sinpix+4` `f(-4)=-4,f(4)=12` `rArr f(x)` can take value 10 as f(x) is continuous function `(x^(3))/(8)+4lt 12 and |sin pix|le 1 AA x in [-4,4]` `therefore `f(x) cannot take the value 18 |
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| 3. |
Let `f(x)={{:(-3+|x|",",-ooltxlt1),(a+|2-x|",",1lexltoo):}` and `g(x)={{:(2-|-x|",",-ooltxlt2),(-b+"sgn(x),",2lexltoo):}` where sgn(x) denotes signum function of x. If `h(x)=f(x)+g(x)` is discontinuous at exactly one point, then which of the following is not possible?A. `a=-3,b=0`B. `a=0,b=1`C. `a=2,b=1`D. `a=-3,b=1` |
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Answer» Correct Answer - D `h(x)=f(x)+g(x)` `={{:(-1",",-ooltxlt1),(a+4-2x",",1lexlt2),(a-b-1+x",",2lexltoo):}` `therefore" We must have either "a=-3, b ne 1 or b=1, a ne -3` |
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| 4. |
If `f(x) = sgn(x^5)`, then which of the following is/are false (where sgn denotes signum function)A. continuous and differentiableB. continuous but not differentiableC. differentiable but not continuousD. neither continuous nor differentiable |
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Answer» Correct Answer - A We have `sgn(x)={{:(,1,x gt0),(,0,x=0),(,-1,x lt 0):}` `therefore f(x)=x^(5)sgn(x)={{:(,x^(5),x gt 0),(,0,x=0),(,-x^(5),x lt 0):}` Clearly f(x) is continuous and differentiable at x=0. |
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| 5. |
If x = `a cos^(3) theta, y = a sin ^(3) theta` then `sqrt(1+((dy)/(dx))^(2))`=?A. `tan^(2) theta`B. `sec ^(2) theta`C. `sec theta`D. `|sec theta|` |
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Answer» Correct Answer - D |
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| 6. |
If `y=[x+sqrt(x^2+a^2)]^n` then prove that `(dy)/dx=(ny)/sqrt(x^2+a^2)` |
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Answer» `y= [x+ sqrt(x^2 + a^2)]^n` `dy/dx = n(x+ sqrt(x^2 + a^2))^(n-1) (1 + (2x)/(2sqrt(x^2+a^2)))` `= n(x+ sqrt(x^2 + a^2))^(n-1) ((sqrt(a^2+x^2) + x)/(sqrt(a^2+x^2)))` `= (n(x+ sqrt(a^2+x^2))^n)/(sqrt(a^2+x^2))` `= (ny)/sqrt(a^2+x^2)` Answer |
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| 7. |
`x=2 cos^2 t, y= 6 sin ^2 t ` |
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Answer» Correct Answer - -3 |
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| 8. |
`x=sqrt(sin 2t),y=sqrt(cos 2 t)` |
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Answer» Correct Answer - `-(tan 2 t )^3/2` |
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| 9. |
If `x=sin^-1""(2t)/(1+t^2)` and `y=tan ^-1""(2t)/(1-t) " the n find " dy/dx.` |
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Answer» `x=sin^-1""(2t)/(1+t^2)=sin^-1""(2 tan theta)/(1+tan ^2)` `=sin ^-1(sin 2 theta)` `2 theta=2tan^-1 t` Let t= tan `theta ` `rArr dxdy=(2)/(1+t^2)` and `y =tan ^-1""(2t)/(1-t^2)=2 tan ^-1 t ` `rArr dy/dx =2/(1+t^2)` `therefore (dx)/(dy)=(dy//dt)/(dx//dt)=(2//(1+t^2))/(2//(1+t^2))=1.` |
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| 10. |
`x = "sin" t, y = "cos" 2t` |
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Answer» `{:(x = "sin"t,|,y = "cos" 2t),(rArr (dx)/(dt) = "cos"t,|,rArr (dy)/(dt) = -2"sin" 2t):}` `therefore (dy)/(dx) = (dy//dt)/(dx//dt) = -(2"sin"2t)/("cos" t)` `= - (2 * 2 "sin"t "cos"t)/("cos"t) = -4"sin" t` |
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| 11. |
`x=a (t-sin t) , y =a (1-cos t) ` |
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Answer» Correct Answer - `cot t/2` |
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| 12. |
`x= 2 cos t -cos 2t ,y=2 sin t-sin 2t ` Find dy/dx |
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Answer» Correct Answer - `tan (3t)/(2)` |
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| 13. |
If `x=(3at)/ (1+t^3), y=(3at^2)/(1+t^3), then dy/dx` is |
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Answer» Correct Answer - `(t(2- t^3))/(1-2t^3)` |
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| 14. |
Find the values of a and b such that the function defined by `f(x)={{:(5, if xle2), (a x+b , if 2 lt x lt10 ),(21 , ifx ge10):}` is a continuous function. |
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Answer» Here, value of `f(x)` is changing at `x = 2` and `x = 10`. So, `f(x)` to be continuous, `f(2^-) = f(2^+)` `=>5 = a(2)+b` `=>2a+b = 5->(1)` Also, `f(x)` to be continuous, `f((10^-) = f(10^+)` `=>a(10) + b = 21` `=>10a+b = 21->(2)` Subtracting (2) -(1), `=>10a+b-2a-b = 21-5` `=>8a = 16` `=> a= 2` `:. 2(2)+b = 5` `=>b = 1` `:.` For, `a = 2, b =1, f(x)` will be continuous. |
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| 15. |
The number of points at which `g(x)=1/(1+2/(f(x)))`is not differentiable, where `f(x)=1/(1+1/x)`, is`1`b. `2`c. `3`d. `4`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `f(x)=(1)/(1+(1)/(x))=(x)/(x+1)` is not differentiable at `x=0,-1.` Also, `g(x)=(1)/(1+(2)/(f(x)))=(1)/(1+(2(x+1))/(x))=(x)/(3x+2)` Thus, the point where g(x) is not differentiable are x = 0, `-1, -2//3`. |
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| 16. |
Statement-1: If f and g are differentiable at x=c, then min (f,g) is differentiable at x=c. Statement-2: min (f,g) is differentiable at `x=c if f(c ) ne g(c )`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D Let f(x)=x nd g(x)=`x^(2)` then, `min (f(x),g(x))={{:(,x,x lt 0),(,x^(2),0 le x lt 1):}` Clearly, f(x) and g(X) are differentiable at x=0. But, min [f(x),g(x)] is not differentiable at x=0. So, statement-1 is not true. However, statement-2 is true. |
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| 17. |
Find all points of discontinuity of f, where f is defined by`f(x)={(|x|)/x , ifx!=0 0, ifx=0` |
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Answer» We know that the identity function x is continuous and the modulus function |x| is continuous . ltrgt So, the quotient funcation ` x/|x|` is continuous at eath ` x ne 0` It has already been proved that f(x) is discontinuous at x=0. Hence, the given function is continuous at each point, except at x=0. |
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| 18. |
Find all points of discontinuity of f, where f is defined by`f(x)={{:(x^(10)-1, ifxlt=1),(x^2, ifx >1):}` |
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Answer» `f(1)= (1)^10 - 1 = 0` LHL `= lim_(x->1^-)f(x) = lim_(x->1^-) x^10 - 1 = 0` RHL `= lim_(x->1^+)f(x) = lim_(x->1^+) x^2 = 1` LHL`cancel(=) `RHL |
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| 19. |
If `y=3e^(2x)+2e^(3x)`. Prove that `(d^2y)/(dx^2)-5(dy)/(dx)+6y=0`. |
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Answer» `y = 3e^(2x)+2e^(3x)` `=> dy/dx = 3e^(2x)(2)+2e^(3x)(3)` `=> dy/dx = 6(e^(2x)+e^(3x))` `=>(d^2y)/dx^2 = 6(2e^(2x)+3e^(3x))` Now,`L.H.S. = (d^2y)/dx^2-5dy/dx+6y ` `=12e^(2x)+18e^(3x)-30e^(2x)-30e^(3x)+18e^(2x)+12e^(3x)` `=0 = R.H.S.` |
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| 20. |
The following functions are differentiable on (-1,2)A. `underset(x)overset(2x)int (log t)^(2)dt`B. `underset(x)overset(2x)int (sin t)/(t)dt`C. `underset(x)overset(2x)int (1-t+t^(2))/(1+t+t^(2))dt`D. None of these |
| Answer» Correct Answer - C | |
| 21. |
If `x^3+y^3=3a x y`, find `(dy)/(dx)` |
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Answer» Correct Answer - `(ay-x^2)/(y^2-ax)` |
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| 22. |
If `x+4|y|-6y`, then y as a function of x isA. continuous at x=0B. derivable at x=0C. `(dy)/(dx)=(1)/(2)` for all xD. none of these |
| Answer» Correct Answer - A | |
| 23. |
Find all the points of discontinuity of f defined by`f(x)" "=" "|" "x" "|" "" "|" "x" "+" "1" "|`. |
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Answer» Let `g(x) =|x| - |x +1|` We know that the modulus function `g(x) = |x|` and `h(x) = |x +1|` are continuous for all x`ne` R. `rArr f(x) ` is not discontinuous at any point. |
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| 24. |
`f(x) = {{:((k cosx )/((pi - 2x)"," if x ne (pi)/(2))),(3"," if x = (pi)/(2)):}`` at x = (pi)/(2)` . |
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Answer» `f(pi//2) = 3 ` R.H.L = `underset(x to (pi^(+))/(2))("lim") f(x) = underset(x to 0)("lim")f((pi)/(2) + h)` `= underset(h to 0)("lim") (k cos ((pi)/(2) + h))/(pi -2 ((pi)/(2) + h)) = underset(h to 0)("lim") (-k sin h)/(-2h) = (k)/(2)` `(because underset(h to 0)("lim") (sinh)/(h) =1)` and L.H.L =` underset(x to pi//2^(2))(lim)f(x)= underset(h to 0)(lim)f((pi)/(2)-h)` `underset (h to 0)(lim)(k cos((pi)/(2)-h))/(pi-2((pi)/(2)-h))` `= underset(h to 0 )(lim)(k sin h)/(2h)= (k)/(2)` `because` Function is continuous at x = (`pi)/(2)`. `:.` R.H.L. = `f ((pi)/(2)) = LI. ` `rArr (k)/(2) = 3 rArr k = 6` |
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| 25. |
If the function `f(x)` given by `f(x)={(3a x+b,"if "x >1), (11,"if "x=1), (5a x-2b ,"if "x |
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Answer» Correct Answer - a=3, b=2 |
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| 26. |
Discuss the continuity of the following functions:(a) `f(x) = s in x + cos x` (b) `f(x) = s in x cos x` (c) `f(x) = s in x dot cos x` |
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Answer» (a) sin x and cos x are always continuous functions. `:.` sin x + cos x is also a continuous function. (b) `f(x) - sin x - cos x` `= sqrt(2)((1)/(sqrt(2))sin x (1)/sqrt(2)cosx )` `= sqrt(2)("cos"(pi)/(4) sin x - "sin" (pi)/(4) cosx)` `= sqrt(2)sin(x -(pi)/(4) )` we know that sine function is always continuous. `:. sin (x - (pi)/(4)` is continuous function. `rArr sqrt (2)"sin" (x - (pi)/(4))` is continuous function. `rArr f(x) `is continuous function (c) `f(x) = sin x cos x` `=(1)/(2).2 "sin x cos x"= (1)/(2)" sin " 2x` we know that sine function is always continuous. `:. sin 2x` is a continuous. `rArr (1)/(2) "sin "2x` is a continuous function. `rArr f(x) ` is a continuous functions. |
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| 27. |
Determine the value of `k`for which the following function is continuous at `x=3.``f(x)={(x^2=9)/(x-3),x!=3k ,x=3` |
| Answer» Correct Answer - k = 6 | |
| 28. |
For what value of `k`is thefollowing function continuous at `x=2?``f(x)={2x+1; x2}` |
| Answer» Correct Answer - k =5 | |
| 29. |
Discuss the continuity of the function `f(x) ={:{((1+cos x)/(tan^2 x) ", "x ne pi),((1)/(2) ", " x=pi):},` `at =pi`. |
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Answer» Correct Answer - continuous |
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| 30. |
Discuss the continuity of the function `f(x) ={:{((Sinx)/(x) ", "x lt0 ),(x+1 ", " x ge0):},` at =0. |
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Answer» Correct Answer - discontinuous |
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| 31. |
Show that the function `f(x)=2x-|x|`is continuous at `x=0`. |
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Answer» we have, ` f(0) =(2xx 0) -|0|=0` `lim_( x to 0+) f(x) = lim_(h to 0) f( 0+h) ` ` lim_(h to 0) ( 2h-|h|) = lim_( h to 0) ( 2h-h) = lim_( h to 0) h =0` ` lim_(x to 0-) f(x) - lim_(hto0) f( 0-h)` `lim_(hto0) {2(-h)-|-h|}=lim_(hto0)(-2h-h) = lim_(hto0) (-3h) =0` Thus, ` lim_(xto0+) f(x) = lim_(xto0+) f(x)=0and " therefore " , lim_(xto0) f(x)=0` ` lim_(x to 0) f(x) = f(0) =0` Hence, f(x) is continuous at x =0 |
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| 32. |
Examine the following functions for continuity.(a) `f(x)=x-5` (b) `f(x)=1/(x-5)` (c) `f(x)=(x^2-25)/(x+5)`(d) `f(x)=|x-5|` |
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Answer» (a) f(x) = x - 5 is a polynomial function. `:.` f(x) is contnuous for each real value . (b) f(x) ` = (1)/(x - 5) , x ne 5 ` which is the quotient of two polynomials and is not defined at x = 5 `:.` f(x) is contnuous for each real value of x (except x = 5). (c) f(x) = `(x^(2) - 25)/(x + 5), x ne - 5` `= ((x - 5) (x + 5))/(x + 5) = x -5` which is a polynomial function . `:.` f(x) is continous for all real value x = -5 . (d) f(x) =`|x - 5| ` `{:={(x -5, if x ge 5),(5 - x, if x lt 5):}` For x `ge ` 5, f (x) = x - 5 which is a polynomial function. `:.` f(x) is continuous for x `ge `5. For x `lt ` 5, f(x) = 5 - x which is a polynomial function. `:.` f(x) is ontinuous for x `lt` 5 Therefore, f(x) is continuous for all real value of x . |
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| 33. |
Show that the funcation ` f(x) = {{:( sinx",", if , x lt 0),( x"lt", if, x ge 0):} ` is continuous. |
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Answer» Let ` at ne R` Case I When ` a gt 0 , " then ", lim_( x to a+) f(x) = lim_( x to a-) f(x) = f(a) =a` Case II When ` a gt 0 , " then " lim_( x to a+) f(x) = lim_( x to a-) f)(x) = f(a) = sin a` Case III When ` a = 0 "then " , lim_( x to 0+) f(x) = lim_( x to 0-) f(x) = f(0) =0` |
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| 34. |
If the function ` f(x) = (( 3x + 4 tan x))/x ` continuous at x=0? If not, hwo may the funcation be defined to make it continuous at this point ? |
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Answer» Since f(x) is not defined at x = 0 , it cannot be continuous at x=0 However, ` lim_(x to 0) f(x) = lim_(x to 0) (( 3x + 4 tan x)/x) = lim_( x to 0) [ 3+4 .( sin x)/ . 1/ (cos x) ]` ` = 3+4. lim_( x to 0) { ( sin x)/x } . { lim_( x to 0) 1/ ( cos x)} =7 ` So, in order to make f(x) continuous at x =0 , we define it as `f(x)= {{:(((3x + 4 tanx))/(x)",","when", x ne0),(7",","when" , x=0):}` |
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| 35. |
Examine the continuity of the funcation `f(x)={{: ((|sinx|)/x",", xne0),(1",",x=0 " at " x=0):}` |
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Answer» we have f(0) =1 `lim_(xto0+)f(x)=lim_(hto0)f(0+h)` `lim_(hto0)(|sin(0 +h)|)/(( 0+h))=lim_(h to 0) (|sin h|)/(h) = lim_(h to0) (sinh)/h=1` `lim_(xto0-)f(x)=lim_(hto0) f(0-h)` ` lim_(h to0) (|sin (-h)|)/ (-h) =lim_(hto0) (|-sin h|)/(-h)=lim_(hto0) (sinh)/(-h)=-1` `lim_(xto0+) f(x)ne lim_(xto0-) f(x) . " so" , lim_(xto0) f(x)` does not exist. Hence, f(x) is discontinuous at x=0. |
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| 36. |
Show that the funcation `f(x)={{:(3x-2",", " when ", x le0),(x+1",", " when ", x gt 0):}` is discontinuous at x =0 |
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Answer» we have ` f(0) = (3 xx 0-2) = -2` ` lim_(x to 0+) f(x) = lim_(h to 0) f( 0+h)` ` lim_(h to 0) ( h+1) =1` ` lim_(x to 0-) f(x) = lim_(h to 0) f( 0-h)` ` lim_(h to 0) [ 3 (-h) -2] = lim_( h to 0) (-3 h-2) = -2` ` lim_(x to 0 +) f(x) ne lim_( x to 0-) f(x) and " therefore, " lim_(x to 0) f(x) ` does not exist. Hence, f(x) is discontinuous at x=0 |
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| 37. |
Show that the funcation `f(x)={{:(x","," if x is an intger ") ,( 0",", " if x is not an integer "):}` is discontinuous at each integral value of x. |
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Answer» Let x= n when n is an integer. Then , f(n) =n ` lim_(x to n+) f(x) = lim_(h to 0) f( n +h) =0 ` [ ( n +h) is not an integer ` Rightarrrow ` f( n +h) =0] ` and lim_(x to n-) f(x)( = lim( h to 0) f( n -h) =0` [ ( n-h) is not an interger ` Rightarrow ` f ( n -h) =0] ` lim_(x to n +) f(x) = lim_( x to n-) f(x) =0 ` So, ` lim_( x to n) f(x) = 0 ne f( n)` Hence, f(x) is discontinuous at x = n |
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| 38. |
Statement-1: The function `f(x)=[x]+x^(2)` is discontinuous at all integer points. Statement-2: The function g(x)=[x] has Z as the set of points of its discontinuous from left.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A Let k be an integer. Then, `underset(x to k^(-))lim g(x)=k-1 and g(k)=k Rightarrow underset(x to k^(-))lim g(x) ne g(k)` So, k is a point of left discontinuity of g. Hence, the points of discontinuous from left of function g is the set Z of all integers. So, statement-2 is true. We have `f(x)={{:(,k-1+x^(2),"if "k-1 le x lt k),(,k+x^(2),"if "k le x lt k+1):}` Clearly, f(x) is discontinuous at all integers points as g(x)=[x] is discontinuous there at. |
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| 39. |
The function `f(x) = (4-x^(2))/(4x-x^(3))` isA. discontinuous at only one pointB. discontinuous at exactly two pointsC. discontinuous at exactly three pointsD. None of the above |
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Answer» Correct Answer - C We have, `f(x) = (4-x^(2))/(4x-x^(2))=((4-x^(2)))/(x(4-x^(2)))` ` = ((4-x^(2)))/(x(2^(2)-x^(2))) = (4-x^(2))/(x(2+x)(2-x))` Clearly , `f(x)` is discontinuous at exactly three points `x = 0, x = - 2` and `x = 2`. |
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| 40. |
The function `f(x) =cot x` is discontinuous on setA. `{x=npi:ninZ }`B. `{x = 2npi : n in Z }`C. ` {x =(2n+1)(pi)/(2),n in Z}`D. `{x=(npi)/(2), n in Z}` |
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Answer» Correct Answer - A We know that, `f(x) = cotx` is continuous in `R-{npi:ninZ}`. Since, `f(x) = cotx = (cosx)/(sinx)`, [since, `sinx = 0` at `n pi, n in Z`] Hence`f(x)= cotx` is discontinuous on the set ` {x=npi : n in Z}`. |
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| 41. |
If `f(x)={(x^2+1", "x ne 1),(" "3 ", "x=1):}` , then check whether the function f(x) is continuous or discontinuous at x=1 |
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Answer» Correct Answer - discontinuous |
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| 42. |
If `f(x) = 2x` and `g(x) = (x^(2))/(2)+1` , then which of the following can be a discontinuous functions?A. `f(x)+g(x)`B. `f(x)-g(x)`C. `f(x).g(x)`D. `(g(x))/(f(x))` |
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Answer» Correct Answer - D We know that, if f and g be continuous functions, then (a) `f+g` is continuous , (b) `f-g` is continuous (c ) `fg` is continuous , (d) `(f)/(g)` is continuous at these points, where ` g(x) ne 0`. Here, ` (g(x))/(f(x)) = ((x^(2))/(2)+1)/(2x)=(x^(2)+2)/(4x)` Which is discontinuous at `x = 0`. |
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| 43. |
`f:R->R` is defined by `f(x)={(cos3x-cosx)/(x^2), x!=0lambda, x=0` and `f` is continuous at `x=0;` then `lambda=`A. `-2`B. `-4C. `-6`D. `-8` |
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Answer» Correct Answer - B If f is continous x=0 then, `underset(x to 0)lim f(x)=f(8)` `Rightarrow underset(x to 0)lim (cos 3x-cosx)/(x^(2))=lambda` `Rightarrow underset(x to 0)lim (-2sin 2x sinx)/(x^(2))=lambda` `Rightarrow -4 underset(x to 0)lim ((sin 2x)/(2x))((sin x)/(2x))=lambdaRightarrow -4xx1xx1=-4` |
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| 44. |
(i) Dissusse the continuity of the function `f(x)={(|x-a|", " xne a ),(" "0 ", "x=a):}` at `x=a` (ii) Discuss the continutiy of the function `f(x)={(|x-3|/(x-3)", " xne 3 ),(" "0 ", "x=3):}` at `x=3` |
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Answer» Correct Answer - (i) continuous (ii) discontinuous |
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| 45. |
Disuss the continutiy of `f(x) ={:{((sin^2 x)/(x^2)", "x ne 0),(" "0", " x= 0):},` at x=0 |
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Answer» Correct Answer - discontinuous |
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| 46. |
Discuss the continutiy of the function `f(x)= {(3x+5", " x ge 2),(6x-1", " x lt 2):}` |
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Answer» Correct Answer - everywhere continuous |
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| 47. |
If the function f(x) is given by `f(x)={{:(,2^(1//(x-1)),x lt 1),(,ax^(2)+bx,x ge 1):}` is everywhere differentiable, thenA. a=0, b=1B. a-0, b=0C. a=1, b=0D. none of these |
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Answer» Correct Answer - B Clearly, f(x) is everywhere continuous and differentiable except possible at x=1. At x=1, we have `underset(x to 1^(-))lim f(x)=underset(x to 1^(-))lim 2^(1//(x-1))=underset(h to 0)lim 2^(-1//h)=0` `underset(x to 1^(+))lim f(x)=underset(x to 1^(+))lim (ax^(2)+bx)=a+b` and f(1)=a+b For f(x) to be differentiable at x=1, it must be continuous there at . `therefore underset(x to 1^(-))lim f(x)=underset(x to 1^(+))lim f(x)=f(1)` `Rightarrow 0=a+b........(i)` Also, `("LHD at x=1")=underset(x to 1^(-))lim (f(x)-f(1))/(x-1)a` `Rightarrow ("LHD at x=1")=underset(h to 0)lim (f(1-h)-f(1))/(-h)` `Rightarrow ("LHD at x=1")=underset(h to 0)lim (2^(-1//h)-(a+b))/(-h)a` `Rightarrow ("LHD at x=1")=underset(h to 0)lim (2^(-1//h))/(-h)" "[therefore a+b=0]` `Rightarrow ("LHD at x=1")=-underset(h to 0)lim (1//h)/(2^(1//h))" "[(oo)/(oo)"form"]` `Rightarrow ("LHD at x=1")=-underset(h to 0)lim (-1//h^(2))/(2^(1//h)"log_(e) 2(-(1)/(h^(2))))` `Rightarrow ("LHD at x=1")=-underset(h to 0)lim (2^(-1//h))/(log_(e)2)=(0)/(log_(e)2)=0` and `("RHD at x=1")=underset(x to 1^(+))lim (f(x)-f(1))/(x-1)` `Rightarrow ("RHD at x=1")=underset(h to 0)lim (f(1+h)-f(1))/(h)` `Rightarrow ("RHD at x=1")=underset(h to 0)lim (a(1+h)^(2)+b(1+h)-(a+b))/(h)` `Rightarrow ("RHD at x=1")=underset(h to 0)lim (ah^(2)+2ah+bh)/(h)` `Rightarrow ("RHD at x=1")=underset(h to 0)lim ah+2a+b=2a+b` For f(x) to be differentiable at x=1, we must have (LHD at x=1)=(RHD at x=1) `Rightarrow 0=2a+b......(ii)` Solving (i) and (ii) we get a=b=0. |
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| 48. |
If `f(x)={{:(,e^(x),x lt 2),(,ax+b,x ge 2):}` is differentiable for all `x in R`, themA. `a=e^(2),b=-e^(2)`B. `a=-e^(2),b=e^(2)`C. `a=b=e^(2)`D. none of these |
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Answer» Correct Answer - A Clearly, f(x) is everywhere continuous and differentiable except possible at x=2. At x=2, we have `underset(x to 2^(-))lim f(x)=underset(x to 2^(-))lim e^(x)=e^(2)` `underset(x to 2^(+))lim f(x)=underset(x to 2^(+))lim ax+b=2a+b` `f(2)=2a+b` Also, `("LHD at x=2")=((d)/(dx)(e^(x)))_(x=2)=e^(2)` `("RHD at x=2")=((d)/(dx)(ax+b))_(x=2)=a` For f(x) to be differentiable at x=2, it should be both continuous and differentiable at x=2 `therefore underset(x to 2^(-))lim f(x)=underset(x to 2^(+))lim f(x)=f(2)` and (LHD at x=2)=(RHD at x=2) `Rightarrow e^(2)=2a+b and a=e^(2) Rightarrow a=e^(2) and b=-e^(2)` |
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| 49. |
Let a function f(x) defined on [3,6] be given by `f(x)={{:(,log_(e)[x],3 le x lt5),(,|log_(e)x|,5 le x lt 6):}` then f(x) isA. continuous and differentiable on [3,6]B. continuous on [3,6] but not differentiable at x=4,5C. differentiable on [3,6] but not continuous at x=4,5D. none of these |
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Answer» Correct Answer - D We have `f(x)={{:(,log_(e)3,3 le x lt 4),(,log_(e)4,4 le x lt 5),(,log_(e)x,5 le x lt 6):}` Clearly, f(x) is continuous and differentiable `[3,4) uu (4,5) uu (5,6)` At x=4, we have `underset(x to 4^(-))lim f(x)=log_(e) 3 and underset(x to 4^(+))lim f(x)=log_(e)4` `therefore underset(x to 4^(-))lim f(x)ne underset(x to 4^(+))lim f(x)` Thus, f(x) is neither continuous nor differentiable at x=4, At x=5, we have So, f(x) is neither continuous nor differentiable at x=5. |
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| 50. |
The fuction f(x) is defined in the interval [0,1] as follows: `f(x)={:{(0", "x=0),(1/2-x", " 0ltxlt1/2),(1/2", " x=1/2),(2/3-x", " 1/2 lt x lt1),(1"," " "x=1):}` Discuss the continuity of the function at `x=1/2` |
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Answer» Correct Answer - discontinuous at `x=1/2` |
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