Maths-
General
Easy

Question

where represents the integral part function, then:

  1. f is continuous but not differentiable at x=0    
  2. f is continuous & differentiable at x=0    
  3. the differentiability of 'f at x=0 depends on the value of a    
  4. f is continuous & differentiable at x=0 and for a=e only.    

The correct answer is: f is continuous & differentiable at x=0

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Which of the following function(s) has/have removable discontinuity at

Which of the following function(s) has/have removable discontinuity at

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Consider the function f(x)=open square brackets table row cell x left curly bracket x right curly bracket plus 1 end cell cell 0 less or equal than x less than 1 end cell row cell 2 minus left curly bracket x right curly bracket end cell cell 1 less or equal than x less or equal than 2 end cell end table close where {x} denotes the fractional part function. Which one of the following statements is NOT correct?

Consider the function f(x)=open square brackets table row cell x left curly bracket x right curly bracket plus 1 end cell cell 0 less or equal than x less than 1 end cell row cell 2 minus left curly bracket x right curly bracket end cell cell 1 less or equal than x less or equal than 2 end cell end table close where {x} denotes the fractional part function. Which one of the following statements is NOT correct?

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Statement-I : If a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis horizontal ellipsis a subscript n end subscript greater than 0, then l i m subscript x rightwards arrow infinity end subscript   open curly brackets fraction numerator a subscript 1 end subscript superscript 1 divided by x end superscript plus a subscript 2 end subscript superscript 1 divided by x end superscript plus a subscript 3 end subscript superscript 1 divided by x end superscript plus horizontal ellipsis.. plus a subscript n end subscript superscript 1 divided by x end superscript over denominator n end fraction close curly brackets to the power of n x end exponent equals product subscript i equals 1 end subscript superscript n end superscript   a subscript n end subscript

Statement-I : If a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis horizontal ellipsis a subscript n end subscript greater than 0, then l i m subscript x rightwards arrow infinity end subscript   open curly brackets fraction numerator a subscript 1 end subscript superscript 1 divided by x end superscript plus a subscript 2 end subscript superscript 1 divided by x end superscript plus a subscript 3 end subscript superscript 1 divided by x end superscript plus horizontal ellipsis.. plus a subscript n end subscript superscript 1 divided by x end superscript over denominator n end fraction close curly brackets to the power of n x end exponent equals product subscript i equals 1 end subscript superscript n end superscript   a subscript n end subscript

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L i m subscript x rightwards arrow 0 end subscript invisible function application fraction numerator left parenthesis 1 minus c o s invisible function application 2 x right parenthesis left parenthesis 3 plus c o s invisible function application x right parenthesis over denominator x t a n invisible function application 4 x end fraction is equal to

L i m subscript x rightwards arrow 0 end subscript invisible function application fraction numerator left parenthesis 1 minus c o s invisible function application 2 x right parenthesis left parenthesis 3 plus c o s invisible function application x right parenthesis over denominator x t a n invisible function application 4 x end fraction is equal to

maths-General
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l i m subscript s rightwards arrow pi divided by 2 end subscript   fraction numerator left square bracket 1 minus t a n invisible function application x divided by 2 right square bracket left square bracket 1 minus s i n invisible function application x right square bracket over denominator left square bracket 1 plus t a n invisible function application x divided by 2 right square bracket left square bracket pi minus 2 x right square bracket to the power of 3 end exponent end fraction is equal to-

l i m subscript s rightwards arrow pi divided by 2 end subscript   fraction numerator left square bracket 1 minus t a n invisible function application x divided by 2 right square bracket left square bracket 1 minus s i n invisible function application x right square bracket over denominator left square bracket 1 plus t a n invisible function application x divided by 2 right square bracket left square bracket pi minus 2 x right square bracket to the power of 3 end exponent end fraction is equal to-

maths-General
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Statement negative I colon The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400) is 8000.
Statement - II : sum subscript k equals 1 end subscript superscript n end superscript   open parentheses k to the power of 3 end exponent minus left parenthesis k minus 1 right parenthesis to the power of 3 end exponent close parentheses equals n to the power of 3 end exponent, for any natural number n.

Statement negative I colon The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400) is 8000.
Statement - II : sum subscript k equals 1 end subscript superscript n end superscript   open parentheses k to the power of 3 end exponent minus left parenthesis k minus 1 right parenthesis to the power of 3 end exponent close parentheses equals n to the power of 3 end exponent, for any natural number n.

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The value of the definite integral stretchy integral subscript 0 end subscript superscript 1 end superscript   open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x is

So here we used the concept of integrals and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the minimum and maximum method which makes problem to solve easily. So the answer is non of these.

The value of the definite integral stretchy integral subscript 0 end subscript superscript 1 end superscript   open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x is

Maths-General

So here we used the concept of integrals and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the minimum and maximum method which makes problem to solve easily. So the answer is non of these.

parallel
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Suppos ef, f' andf'' are continuous on[0, e] and that f times x f times 1 equals 1 text  and  end text stretchy integral subscript 1 end subscript superscript c end superscript   fraction numerator f left parenthesis x right parenthesis over denominator x to the power of 2 end exponent end fraction d x equals fraction numerator 1 over denominator 2 end fraction then the value of stretchy integral subscript 1 end subscript superscript e end superscript   f to the power of ´ ´ end exponent left parenthesis x right parenthesis l n invisible function application x d x text  equals - end text

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maths-General
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I I subscript n end subscript equals stretchy integral subscript 0 end subscript superscript pi divided by 4 end superscript   t a n to the power of n end exponent invisible function application x d x text end textther text end text stack l i m with n rightwards arrow infinity below   blank subscript n end subscript I subscript n end subscript plus l subscript m end subscript 2 equals

I I subscript n end subscript equals stretchy integral subscript 0 end subscript superscript pi divided by 4 end superscript   t a n to the power of n end exponent invisible function application x d x text end textther text end text stack l i m with n rightwards arrow infinity below   blank subscript n end subscript I subscript n end subscript plus l subscript m end subscript 2 equals

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Equation of the circle of radius square root of 2 text  ´ containing the point  end text left parenthesis 3 , 1 right parenthesis text  and touching the line  end text vertical line x minus 1 vertical line equals vertical line y minus 1 vertical line is

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maths-General
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In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If AG=2,GF=13,FC=1 and HJ=7, then DE=

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