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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. |
Evaluate `lim_(xrarr0)((3^(2x)-1)/(2^(3x)-1)).` |
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Answer» `lim_(xto0)((3^(2x)-1)/(2^(3x)-1))=lim_(xto0)({((3^(2x)-1)/(2x)).2x})/({((2^(3x)-1)/(3x)).3x})` `=2/3.(lim_(2xto0)((3^(2x)-1)/(2x)))/(lim_(3xto0)((2^(3x)-1)/(3x)))` `=2/3.(log3)/(log2)" "[becauselim_(yto0)((a^(y)-1)/(y))=loga]` `=(2log3)/(3log2)=(log(3^(2)))/(log(2^(3)))=(log9)/(log8).` |
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| 2. |
Show that `lim_(xrarr2)(x)/([x])` does not exist. |
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Answer» Let `f(x)=(x)/({x}).` Then, `lim_(xto2^(+))f(x)=lim_(hto0)f(2+h)=lim_(hto0)(2+h)/([2+h])=1" "[becausehto0,[2+h]=2]` `lim_(xto2^(-))f(x)=lim_(hto0)f(2-h)=lim_(hto0)(2-h)/([2-h])=lim_(hto0)(2-h)/(1)=2" "[because[2-h]=1]` `thereforelim_(xto2^(+))f(x)nelim_(xto2^(-))f(x)and solim_(xto2)f(x)` does not exist. |
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| 3. |
If `y(x)=|x|-3, "find"lim_(xrarr3)f(x).` |
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Answer» `lim_(xto3^(+))f(x)=limf(3+h)=lim_(hto0){|3+h|-3}=lim_(hto0)(3+h-3)=lim_(hto0)h=0.` `lim_(xto3^(-))f(x)=lim_(hto0)f(3-h)=lim_(hto0){|3-h|-3}=lim_(hto0)(3-h-3)=lim_(hto0)(-h)=0.` `thereforelim_(xto3^(+))f(x)=lim_(xto3^(-))f(x)=0implieslim_(xto3)f(x)=0.` |
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| 4. |
Evaluate the following limits: `lim_(xrarr0)((1-cosx))/(sin^(2)x)` |
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Answer» Correct Answer - `1/2` Given limit `=lim_(xto0)((1-cosx))/((1-cos^(2)x))=lim_(xto0)(1)/((1+cosx))=1/2.` |
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| 5. |
Evaluate the following limits: `lim_(xrarr0)(cos2x-1)/(cosx-1)` |
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Answer» Correct Answer - 4 `lim_(xto0)(cos2x-1)/(cosx-1)=lim_(xto0)(2cos^(2)x-2)/(cosx-1)=lim_(xto0)(2(cos^(2)x-1))/((cosx-1))=2lim_(xto0)(cosx+1)=4.` |
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| 6. |
Evaluate the following limits: `lim_(xrarr0)(ax+x cosx)/(b sinx)` |
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Answer» Correct Answer - (a+1)/(b)` `lim_(xto0)(ax+xcosx)/(bsinx)=lim_(xto0)(a+cosx)/(b((sinx)/(x)))" "["dividing num. and denom. by x"]` `=((a+1))/((bxx1))=((a+1))/(b).` |
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| 7. |
Evaluate `(i)lim_(xrarr0)((1-cos 4x)/(1-cos5x))` `(ii) lim_(xrarr0)((1-cosmx)/(1-cosnx)).` |
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Answer» `(i)lim_(xto0)((1-cos4x)/(1-cos5x))=lim_(xto0)(2sin^(2)2x)/(2sin^(2)(5x//2))` `=lim_(xto0)(sin^(2)2x)/(sin^(2)(5xx//2))=lim_(xto0){({(sin2x)/(2x)}^(2).4x^(2))/({(sin(5x//2))/((5x//2))}^(2).(25x^(2))/(4))}` `16/25.(lim_(2xto0)((sin2x)/(2x))^(2))/(lim_((5x)/(x)to0){(sin(5x//2))/((5x//2))}^(2))=((16)/(25)xx(1^(2))/(1^(2)))=16/25.` `(ii)lim_(xto0)((1-cosmx)/(1-cosnx))` `=lim_(xto0){(2sin^(2)(mx//s))/(2sin(nx//2))}=lim_(xto0){(sin^(2)(mx//2))/(sin^(2)(nx//2))}` `=lim_(xto0){((sin^(2)(mx//2))/((mx//2)^(2))xx(m^(2)x^(2))/(4))/((sin^(2)(nx//2))/((nx//2)^(2))xx(n^(2)x^(2))/(4))}=(m^(2))/(n^(2)).(lim_((mx)/(2)to0){(sin(mx//2))/((mx//2))}^(2))/(lim_((nx)/(2)to0){(sin(nx//2))/((nx//2))}^(2))` `(m^(2)/(n^(2))xx(1^(2))/(1^(2)))=(m^(2))/(n^(2)).` |
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| 8. |
Evaluate `lim_(xrarr0)((1-cosx))/(x^(2)).` |
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Answer» `lim_(xto0)((1-cosx))/(x^(2))=lim_(xto0)(2sin^(2)(x//2))/(((x)/(2))^(2)xx4)` `=1/2lim_(xto0){(sin(x//2))/((x//2))}^(2)=((1)/(2)xx1^(2))=1/2.` |
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| 9. |
Evaluate the following limits: `lim_(xrarr0)(1-cos2mx)/(1-cos2nx)` |
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Answer» Correct Answer - `(m^(2))/(n^(2))` Given limit `=lim_(xto0)(2sin^(2)mx)/(2sin^(2)nx)=lim_(xto0)(sin^(2)mx)/(sin^(2)nx)=lim_(xto0){((sinmx)/(mx)xxmx(sinmx)/(mx)xxmx)/((sinnx)/(nx)xxnx(sinnx)/(nx)xx nx)}` `=(m^(2))/(n^(2)).{(lim_(xto0)(sinmx)/(mx))/(lim_(xto0)(sinnx)/(nx))}xx{(lim_(xto0)(sinmx)/(mx))/(lim_(xto0)(sinnx)/(nx))}=((m^(2))/(n^(2))xx1/1xx1/1)=(m^(2))/(n^(2)).` |
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| 10. |
For ` x in R , lim_( x to infty) ((x -3)/(x + 2))^x ` is equal toA. eB. `e^(-1)`C. ` e^(-5)`D. `e^(5)` |
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Answer» Correct Answer - C For ` x in R, underset(x to infty) lim ((x -3)/(x+2))^(x) = underset( x to infty) lim ((1-3//x)^(x))/((1+2//x)^(x)) = e^(-3)/e^(2) = e^(-5)` |
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| 11. |
` lim_(x to - infty) [((x^(4) sin (1/x) + x^(2)))/((1+|x|^(3)))]=...` |
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Answer» Correct Answer - `-1` `underset( x to - infty) lim ((x^(4) sin. 1/x + x^(2))/(1+|x|^(3))) = underset(x to - infty) lim (x^(4)sin.1/x + x^(2))/(1-x^(3))` On dividing by `x^(3)`, we get `underset(x to - infty) lim ((sin (1//x))/(1/x) + 1/x )/(1/x^(3)-1) = (1+0)/(0-1) =- 1` |
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| 12. |
Let `y=e^x s in x^3+(t a n x)^xdotF in d(dy)/(dx)dot` |
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Answer» Correct Answer - `e^(x sin x^(3))(3x^(3) cos x^(3) + sin x^(3))+(tan x) ^(x) [2x " cosec " 2x + log (tan x)]` Since, ` y = e^(x sin x^(3))+(tan x)^(x),` then ` y = u + v," where " u=e^(x sin x ^(3)) and v = (tan x ) ^(x) ` ` rArr" " (dy)/(dx) = ((du)/(dx)+(dv)/(dx))` …(i) Here, ` u = e^(x sin x^(3)) ` and log v = x log (tan x) On differentiating both sides w.r.t. x, we get `(du)/(dx) = e^(x sin x^(3))* (3x^(3) cos x^(3) + sin x^(3))` ..(ii) and ` 1/v*(dv)/(dx) = (x*sec^(2) x)/(tan x) + log (tan x) ` `(dv)/(dx) = (tan x)^(x) [2x* cosec (2x)+ log (tan x)}`...(iii) From Eqs. (i), (ii)and (iii), we get ` (dy)/(dx) = e^(x sin x^(3))(3x^(3)*cos x^(3) + sin x^(3)) + (tan x)^(x) ` [2x cosec 2x + log (tan x)] |
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| 13. |
If `f(x) = {{:(sin x"," x ne npi"," n = 0"," pm1"," pm2","...),(" 2, ""otherwise"):}}` `and g (x) ={{:(x^(2)+1"," x ne 0","2),(" 4, "x=0),(" 5, "x=2):}},"then" lim_(x to 0) g[f(x)]` is ……… |
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Answer» Correct Answer - 1 Given, `f(x) ={{:(sin x", "x ne npi", n=0,"pm"1," pm"2,"...),(" 2, otherwise"):}` `g[f(x)]={{:({f(x)}^(2)+1" , "f(x) ne 0","2),(" 4 , "f(x) = 0),(" 5 , "f(x) = 2):}` `:. g[f(x)]={{:((sin^(2) x) +1"," x ne"," n pi = 0"," pm 1","...),(" 5,"x=npi):}` Now, `underset( x to 0) lim [f(x)] = underset( x to 0) lim (sin^(2)x) + 1 = 1` |
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| 14. |
The integer n for which ` lim_(x to 0) ((cos x - 1)(cos x -e^(x)))/x^(2) ` is a finite non-zero number, isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `underset( x to 0)lim ((cos x - 1)(cos x - e^(x)))/x^(2) ` `underset(x to 0) lim((-2 sin^(2). x/2){(1-x^(2)/(2!) +x^(4)/(4!)-...)-(1+x+x^(2)/(2!) +x^(3)/(3!)+...)})/x^(n)` `underset(x to 0)lim((-2 sin^(2) x/2)(-x-(2x^(2))/(2!) -x^(3)/(3!)-...))/(4(x/2)^(2) x^(n-2))` `underset(x to 0) lim (sin^(2). x/2(1+x+x^(2)/(3!)+...))/(2(x/2)^(2) x^(n-3))` Above limit is finite, if n - 3 = 0, i.e., n = 3. |
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| 15. |
If `lim_(x to 0) ({(a-n)nx - tan x} sin nx)/x^(2) = 0`, where n is non-zero real number, then a is equal to |
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Answer» Correct Answer - D Given, `underset(x to 0) lim({(a-n)nx-tan x } sin nx )/x^(2) = 0` ` rArr underset( xto0)lim {(a-n)-(tan x)/x}(sin n x)/(n x ) xx n = 0 ` `rArr " "{(a-n) n = 0` ` rArr" "(a-n) n= 1` ` rArr" " a = n + 1/n` |
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| 16. |
If `y`is a function of `xa n dlog(x+y)-2x y=0,`then the value of `y^(prime)(0)`is(b) `-1`(c) 2 (d)0A. 1B. `-1`C. 2D. `0` |
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Answer» Correct Answer - A Given that, ` log(x+y) = 2xy` ..(i) ` :. " At " x = 0, rArr log (y) =0rArr y = 1` ` :. ` To find ` (dy)/(dx) ` at (0, 1) On differentiating Eq. (i) w.r.t. X, we get `1/(x+y) (1+(dy)/(dy))= 2x (dy)/(dx) + 2y* 1` ` rArr" " (dy)/(dx) = (2y(x+y)-1)/(1-2(x+y)x) ` ` rArr" " ((dy)/(dx))_("(0,1)") = 1` |
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| 17. |
Evaluate ` lim_(x to 0) sqrt((x-sin x)/(x+cos^(2) x))`. |
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Answer» `underset( x to 0) lim sqrt((x-sin x)/(x + cos^(2) x) )=(underset(x to 0) lim (x -sin x)^(1//2))/(underset(x to 0) lim (x + cos^(2) x)^(1//2))` ` =(underset(x to 0) lim [x(1-(sin x)/x)]^(1//2))/(underset(x to 0) lim (0+1)^(1//2) )` ` = (0*0)/1 = 0` |
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| 18. |
Evaluate ` lim_(x to 1) ((x-1)/(2x^(2) - 7x+ 5))`. |
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Answer» Correct Answer - `(-1)/3` ` underset(x to 1) lim{(x-1)/((x-1)(2x-5))} = underset( xto 1) lim 1/((2x - 5))` ` = - 1/3` |
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| 19. |
Evaluate `lim_(h to 0) ((a+h)^(2) sin (a+h) -a^(2) sin a)/h `. |
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Answer» Correct Answer - `a^(2) cos alpha + 2a sin alpha` Here, ` underset( h to 0) lim ((a+h)^(2) sin (a+h)-a^(2) sin a)/h` ` underset( h to 0) lim(a^(2)[sin (a+h)-sin a])/h` `+(h[2a sin (a+h)+h sin (a+h)])/h ` ` = underset( h to 0) lim (a^(2)*2 cos(a+h/2) * sin h/2)/(2*h/2) + (2a+h) sin (a+h) ` ` = a^(2) cos a + 2a sin a` |
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| 20. |
Evaluate the following limits: `lim_(xrarr0)((1-cos3x))/(x^(2))` |
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Answer» Correct Answer - `9/2` Given limit `=lim_(xto0)(2sin^(2)((3x)/(2)))/(((3x)/(2))^(2)xx4/9)=9/2.` |
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| 21. |
Evaluate the following limits: `lim_(xrarr0)(xcosecx)` |
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Answer» Correct Answer - 1 Given limit `=lim_(xto0)(x)/(sinx)=1.` |
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| 22. |
Evaluate `underset(xrarr0)"lim"(2sinx-sin2x)/(x^(3))` |
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Answer» Correct Answer - 1 Given limit `=lim_(xto0)(2sinx(1-cosx))/(x^(3))=lim_(xto0)[2((sinx)/(x)).(2sin^(2)(x//2))/(((x)/(2))^(2)xx4)]=1.` |
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| 23. |
0.4 neighbourhood of 4: |
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Answer» Here, a = 4; δ = 0.4 In Modulus form: |x – a| < δ ∴ N(4, 0.4) = |x – 4| < 0.4 In Interval form:(a – δ, a + δ) ∴ N(4, 0.4) = (4 – 0.4, 4 + 0.4) = (3.6, 4.4) |
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| 24. |
0.05 neighbourhood of 0: |
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Answer» Here, a = 0; δ = 0.05 In Modulus form : | x – a | < δ ∴ N (0, 0.05) = |x – 0| < 0.05 = | x | < 0.05 In Interval form : (a – δ, a + δ) ∴ N(0, 0.05) = (0 – 0.05, 0 + 0.05) = (- 0.05, 0.05) |
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| 25. |
What is the interval form of |x – 5| < 0.25?(a) (4.75, 5.25)(b) (- 4.75, + 5.25)(c) (- 5.25, – 4.75)(d) (- 5.25, 4.75) |
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Answer» Correct option is (a) (4.75, 5.25) |
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| 26. |
0.02 neighbourhood of 2: |
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Answer» Here. a = 2; δ = 0.02 In Modulus form: |x – a| < δ ∴ N(2. 0.02) = | x – 2| < 0.02 In Interval form: (a – 8, a + δ) ∴ N (2, 0.02) = (2 – 0.02, 2 + 0.02) = (1.98, 2.02) |
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| 27. |
Express the interval form (- 0.5, 0.5) in modulus form. |
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Answer» Interval form: {- 0.5, 0.5) ∴ a – δ = – 0.5, a + δ = 0.5 By adding, 2a = 0 ∴ a = 0 Putting a = 0 in a + δ, δ = 0.5 In modulus form: |x – a| < δ ∴ For (- 0.5, 0.5), modulus form is expressed by |x – 0| < 0.5 = |x| < 0.5 |
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| 28. |
Define the 5 neighbourhood of a. |
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Answer» If a ∈ R and δ is non-negative real number, then the open interval (a – δ, a + δ) is called δ neighbourhood of a. It is denoted by N (a, δ). Thus, N (a, δ) = {x|(a – δ) < x < (a + δ), x ∈ R} |
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| 29. |
What is the interval form of 0.02 neighbourhood of -2?(a) (1.98, 2.02)(b) (- 1.98, 2.02)(c) (- 2.02, -1.98)(d) (- 2.02, 1.98) |
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Answer» Correct option is (c) (- 2.02, – 1.98) |
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| 30. |
Express N (50, 0.8) in modulus form. |
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Answer» N (50, 0.8) in modulus form is expressed by |x – 50| < 0.8. |
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| 31. |
Express |2x| < \(\frac{1}{2}\) in interval form. |
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Answer» |2x| < \(\frac{1}{2}\) in interval form is expressed by (−\(\frac{1}4,\frac{1}4\)) |
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| 32. |
Explain the meaning of x → a. |
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Answer» When the value of a variable x is brought very close to a number ‘a’ by increasing or decreasing its value, then it can be said that x tends to a. It is denoted by x → a. x → a means that value of x is very close to a but x ≠ a. Taking a = 1, let us understand the meaning of x → 1. |
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| 33. |
What is the value of ?(a) 9(b) 10(c) 4/3(d) 8 |
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Answer» Correct option is (d) 8 |
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| 34. |
If y = 10 – 3x and x → – 3, then y tends to which value ?(a) 1(b) 9(c) 19(d) 7 |
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Answer» Correct option is (c) 19 |
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| 35. |
If N(a, 0.2) = |x – 7|< b, then find the value of a. |
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Answer» N (a, 0.2) = | x – 7| < b |
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| 36. |
What is the modulus form of 0.3 neighbourhood of 3?(a)|x – 0.3|<3(b) |x- 31 < 0.3(c) |x + 3|< 0.3(d) |x – 3| > 0.3 |
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Answer» Correct option is (b) |x- 31 < 0.3 |
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| 37. |
Define an open interval. |
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Answer» If a, b ∈ R and a < b, then the set of all real numbers between a and b not including a and b is called an open interval. It is denoted by (a, b). Thus, (a, b) = {x|a < x < b, x ∈ R}. |
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| 38. |
If |x + 4| < 0.04 = (k, – 3.96), then find the value of fc. |
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Answer» If |x + 4| < 0.04 = (k, – 3.96), then |
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| 39. |
What is the value of \(\lim \limits_{x\to-1}\,\frac{x^5+1}{x+1}\)?(a) – 5(b) 5(c) 4(d) – 4 |
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Answer» Correct option is (b) 5 |
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| 40. |
What is the value of \(\lim\limits_{x\to3}\, \frac{x^4-81}{x-3}\)?(a) 192(b) 324(c) 36(d) 108 |
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Answer» Correct option is (d) 108 |
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| 41. |
If \(\lim\limits_{x\to-1}4x+K=6\) then find the value |
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Answer» \(\lim\limits_{x\to-1}4x+K=6\) = 6 = 4 (- 1) + k = 6 |
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| 42. |
Find the value of \(\lim\limits_{x\to5}\,(3x+5)\) |
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Answer» \(\lim\limits_{x\to5}\,(3x+5)\)= 3 (5) + 5 = 20 |
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| 43. |
Evaluate the following limits: `lim_(xrarr0)((e^(4x)-1)/(x))` |
| Answer» Correct Answer - 4 | |
| 44. |
Evaluate the following limits: `lim_(xrarr0)((2^(x)-1)/(x))` |
| Answer» Correct Answer - 2 | |
| 45. |
Evaluate the following limits: `lim_(xrarr0)(tan alphax)/(tanbetax)` |
| Answer» Correct Answer - `(alpha)/(beta)` | |
| 46. |
Evaluate the following limits: `lim_(xrarr0)((3^(2+x)-9)/(x))` |
| Answer» Correct Answer - `9log3` | |
| 47. |
Evaluate the following limits: `lim_(xrarr0)(tan(x//2))/(3x)` |
| Answer» Correct Answer - `1/6` | |
| 48. |
Evaluate the following limits: `lim_(xrarr0)(tan3x)/(tan5x)` |
| Answer» Correct Answer - `3/5` | |
| 49. |
Evaluate the following limits: `lim_(xrarr0)(sin(x//4))/(x)` |
| Answer» Correct Answer - `1/4` | |
| 50. |
Evaluate the following limits: `lim_(xrarr0)(sin4x)/(6x)` |
| Answer» Correct Answer - `4/7` | |