Maths-
General
Easy

Question

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

  1. Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I.    
  2. Statement-I is True, Statement-II is True; Statement-II is NOT a correct explanation for Statement-I    
  3. Statement-I is True, Statement-II is False    
  4. Statement-I is False, Statement-II is True    

The correct answer is: Statement-I is True, Statement-II is False

Related Questions to study

General
maths-

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
General
maths-

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
General
Maths-

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.

Maths-General
parallel
General
Maths-

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately.

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately.

Maths-General
General
maths-

The negation of the statement "If I become a teacher, then I will open a school", is :

The negation of the statement "If I become a teacher, then I will open a school", is :

maths-General
General
Maths-

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
General
maths-

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

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

maths-General
General
maths-

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

maths-General
General
maths-

If left parenthesis alpha beta comma right parenthesisis a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2), then the greatest value of a is

If left parenthesis alpha beta comma right parenthesisis a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2), then the greatest value of a is

maths-General
parallel
General
maths-

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

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

maths-General
General
maths-

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=

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=

maths-General
General
maths-

Two circles of unequal radii intersect in two distinct points P and Q. A line through P cuts 1st circle at A and 2nd circle at B. Let M be the mid point of AB. If the line MQ meets the 1st circle at X and the second circle at Z then M divides XZ internally in the ratio.

Two circles of unequal radii intersect in two distinct points P and Q. A line through P cuts 1st circle at A and 2nd circle at B. Let M be the mid point of AB. If the line MQ meets the 1st circle at X and the second circle at Z then M divides XZ internally in the ratio.

maths-General
parallel
General
maths-

A circle touches the lines y equals fraction numerator x over denominator square root of 3 end fraction comma y equals x square root of 3 and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

A circle touches the lines y equals fraction numerator x over denominator square root of 3 end fraction comma y equals x square root of 3 and has unit radius. If the centre of this circle lies in the first quadrant, then one possible equation of this circle is

maths-General
General
maths-

From a point R(5, 8) two tangents RP and RQ are drawn to a given circle S = 0 whose radius is 5. If circumcentre of the triangle PQR is (2, 3), then the equation of circle S = 0 is

From a point R(5, 8) two tangents RP and RQ are drawn to a given circle S = 0 whose radius is 5. If circumcentre of the triangle PQR is (2, 3), then the equation of circle S = 0 is

maths-General
General
maths-

PA and PB are tangents to a circle S touching S at points A and B.C is a point on S in between A and B as shown in the figure. LCM is a tangent to S intersecting PA and PB in points L and M respectively.Then the perimeter of the triangle PLM depends on

PA and PB are tangents to a circle S touching S at points A and B.C is a point on S in between A and B as shown in the figure. LCM is a tangent to S intersecting PA and PB in points L and M respectively.Then the perimeter of the triangle PLM depends on

maths-General
parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.