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Question

If $A equals open curly brackets n colon fraction numerator n to the power of 3 end exponent plus 5 n to the power of 2 end exponent plus 2 over denominator n end fraction blank i s blank a n blank i n t e g e r blank a n d blank i t s e l f blank i s blank a n blank i n t e g e r close curly brackets comma$ then the number of elements in the set $A comma$ is

1
2
3
4

The correct answer is: 4

We have,
$fraction numerator n to the power of 3 end exponent plus 5 n to the power of 2 end exponent plus 2 over denominator n end fraction equals n to the power of 2 end exponent plus 5 n plus fraction numerator 2 over denominator n end fraction$
$therefore fraction numerator n to the power of 3 end exponent plus 5 n to the power of 2 end exponent plus 2 over denominator n end fraction$ is an integer, if $fraction numerator 2 over denominator n end fraction$ is an integer
$rightwards double arrow n equals plus-or-minus 1 comma plus-or-minus 2$
$rightwards double arrow$ A consists of four elements $v i z. negative 1 comma blank 1 comma negative 2 comma blank 2$

Related Questions to study

If $A equals open curly brackets 1 comma blank 2 comma blank 3 close curly brackets comma blank B equals left curly bracket a comma blank b right curly bracket$ , then $A cross times B$ mapped $A$ to $B$ is

If $A equals open curly brackets 1 comma blank 2 comma blank 3 close curly brackets comma blank B equals left curly bracket a comma blank b right curly bracket$ , then $A cross times B$ mapped $A$ to $B$ is

Statement 1: $1 to the power of 99 end exponent plus 2 to the power of 99 end exponent plus horizontal ellipsis plus 100 to the power of 99 end exponent$ is divisible by 10100
Statement 2: $a to the power of n end exponent plus b to the power of n end exponent$ is divisible by $a plus b$ if $n$ is odd

Statement 1: $1 to the power of 99 end exponent plus 2 to the power of 99 end exponent plus horizontal ellipsis plus 100 to the power of 99 end exponent$ is divisible by 10100
Statement 2: $a to the power of n end exponent plus b to the power of n end exponent$ is divisible by $a plus b$ if $n$ is odd

Statement 1:Let $F subscript 1 end subscript open parentheses n close parentheses equals 1 plus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction plus... plus fraction numerator 1 over denominator n end fraction$ , then $not stretchy sum from r equals 1 to n of F subscript 1 end subscript open parentheses r close parentheses equals open parentheses n plus 1 close parentheses F subscript 1 end subscript open parentheses n close parentheses minus n.$
Statement 2: $fraction numerator 1 to the power of negative 1 end exponent plus 2 to the power of negative 1 end exponent plus 3 to the power of negative 1 end exponent plus... plus n to the power of negative 1 end exponent over denominator n end fraction greater than open parentheses fraction numerator 1 plus 2 plus 3 plus... plus n over denominator n end fraction close parentheses to the power of negative 1 end exponent$ or $open parentheses 1 plus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction plus plus... fraction numerator 1 over denominator n end fraction close parentheses greater than fraction numerator n to the power of 2 end exponent over denominator not stretchy sum n end fraction$ or $open parentheses 1 plus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction plus... plus fraction numerator 1 over denominator n end fraction close parentheses greater than fraction numerator 2 n over denominator left parenthesis n plus 1 right parenthesis end fraction$

Statement 1:Let $F subscript 1 end subscript open parentheses n close parentheses equals 1 plus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction plus... plus fraction numerator 1 over denominator n end fraction$ , then $not stretchy sum from r equals 1 to n of F subscript 1 end subscript open parentheses r close parentheses equals open parentheses n plus 1 close parentheses F subscript 1 end subscript open parentheses n close parentheses minus n.$
Statement 2: $fraction numerator 1 to the power of negative 1 end exponent plus 2 to the power of negative 1 end exponent plus 3 to the power of negative 1 end exponent plus... plus n to the power of negative 1 end exponent over denominator n end fraction greater than open parentheses fraction numerator 1 plus 2 plus 3 plus... plus n over denominator n end fraction close parentheses to the power of negative 1 end exponent$ or $open parentheses 1 plus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction plus plus... fraction numerator 1 over denominator n end fraction close parentheses greater than fraction numerator n to the power of 2 end exponent over denominator not stretchy sum n end fraction$ or $open parentheses 1 plus fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 3 end fraction plus... plus fraction numerator 1 over denominator n end fraction close parentheses greater than fraction numerator 2 n over denominator left parenthesis n plus 1 right parenthesis end fraction$

parallel

Statement 1:If an infinite G.P. has 2^nd term $x$ and its sum is 4, then $x$ belongs to $left parenthesis negative 8 comma blank 1 right parenthesis$
Statement 2:Sum of an infinite G.P. is finite if for its common ratio $r comma blank 0 less than open vertical bar r close vertical bar less than 1$

Statement 1:If an infinite G.P. has 2^nd term $x$ and its sum is 4, then $x$ belongs to $left parenthesis negative 8 comma blank 1 right parenthesis$
Statement 2:Sum of an infinite G.P. is finite if for its common ratio $r comma blank 0 less than open vertical bar r close vertical bar less than 1$

Statement 1:Coefficient of $x to the power of 14 end exponent$ in $open parentheses 1 plus 2 x plus 3 x to the power of 2 end exponent plus horizontal ellipsis plus 16 x to the power of 15 end exponent close parentheses to the power of 2 end exponent$ is 560
Statement 2: $not stretchy sum from r equals 1 to n of r left parenthesis n minus r right parenthesis equals fraction numerator n left parenthesis n to the power of 2 end exponent minus 1 right parenthesis over denominator 6 end fraction$

Statement 1:Coefficient of $x to the power of 14 end exponent$ in $open parentheses 1 plus 2 x plus 3 x to the power of 2 end exponent plus horizontal ellipsis plus 16 x to the power of 15 end exponent close parentheses to the power of 2 end exponent$ is 560
Statement 2: $not stretchy sum from r equals 1 to n of r left parenthesis n minus r right parenthesis equals fraction numerator n left parenthesis n to the power of 2 end exponent minus 1 right parenthesis over denominator 6 end fraction$

Statement 1: $x equals 1111 midline horizontal ellipsis 91$ times is composite number
Statement 2:91 is composite number

Statement 1: $x equals 1111 midline horizontal ellipsis 91$ times is composite number
Statement 2:91 is composite number

parallel

Statement 1:The numbers $square root of 2 comma blank square root of 3 comma blank square root of 5$ cannot be the terms of a single A.P. with non-zero common difference
Statement 2:If $p comma blank q comma blank r left parenthesis p not equal to q right parenthesis$ are terms (not necessarily consecutive) of an A.P., then there exists a rational number $k$ such that $left parenthesis r minus q right parenthesis divided by left parenthesis q minus p right parenthesis equals k$

Statement 1:The numbers $square root of 2 comma blank square root of 3 comma blank square root of 5$ cannot be the terms of a single A.P. with non-zero common difference
Statement 2:If $p comma blank q comma blank r left parenthesis p not equal to q right parenthesis$ are terms (not necessarily consecutive) of an A.P., then there exists a rational number $k$ such that $left parenthesis r minus q right parenthesis divided by left parenthesis q minus p right parenthesis equals k$

Statement 1:3, 6, 12 are in GP, then 9, 12, 18 are in HP.
Statement 2:If middle term is added in three consecutive terms of a GP, resultant will be in HP.

Statement 1:3, 6, 12 are in GP, then 9, 12, 18 are in HP.
Statement 2:If middle term is added in three consecutive terms of a GP, resultant will be in HP.

Statement 1:If sum of $n$ terms of a series is $6 n to the power of 2 end exponent plus 3 n plus 1$ then the series is in AP $.$
Statement 2:Sum of $n$ terms of an AP is always of the form $a n to the power of 2 end exponent plus b n.$

Statement 1:If sum of $n$ terms of a series is $6 n to the power of 2 end exponent plus 3 n plus 1$ then the series is in AP $.$
Statement 2:Sum of $n$ terms of an AP is always of the form $a n to the power of 2 end exponent plus b n.$

parallel

Statement 1:If the arithmetic mean of two numbers is 5/2, geometric mean of the numbers is 2, then the harmonic mean will be 8/5
Statement 2:For a group of positive numbers $open parentheses G. M. close parentheses to the power of 2 end exponent equals open parentheses A. M. close parentheses cross times left parenthesis H. M. right parenthesis$

Statement 1:If the arithmetic mean of two numbers is 5/2, geometric mean of the numbers is 2, then the harmonic mean will be 8/5
Statement 2:For a group of positive numbers $open parentheses G. M. close parentheses to the power of 2 end exponent equals open parentheses A. M. close parentheses cross times left parenthesis H. M. right parenthesis$

Statement 1:The sum of $n$ terms of two arithmetic progressions are in the ratio $open parentheses 7 n plus 1 close parentheses colon open parentheses 4 n plus 17 close parentheses comma$ then the ratio of their $n$ th terms is 7: 4.
Statement 2:If $S subscript n end subscript equals a x to the power of 2 end exponent plus b x plus c comma$ then $T subscript n end subscript equals S subscript n end subscript minus S subscript n minus 1 end subscript$

Statement 1:The sum of $n$ terms of two arithmetic progressions are in the ratio $open parentheses 7 n plus 1 close parentheses colon open parentheses 4 n plus 17 close parentheses comma$ then the ratio of their $n$ th terms is 7: 4.
Statement 2:If $S subscript n end subscript equals a x to the power of 2 end exponent plus b x plus c comma$ then $T subscript n end subscript equals S subscript n end subscript minus S subscript n minus 1 end subscript$

Let $a comma blank r element of R minus left curly bracket 0 comma blank 1 comma blank minus 1 right curly bracket$ and $n$ be an even number
Statement 1: $a cross times a r cross times a r to the power of 2 end exponent midline horizontal ellipsis a r to the power of n minus 1 end exponent equals open parentheses a to the power of 2 end exponent r to the power of n minus 1 end exponent close parentheses to the power of n divided by 2 end exponent$
Statement 2:Product of $i to the power of t h end exponent$ term from the beginning and from the end in a G.P. is independent of $i$

Let $a comma blank r element of R minus left curly bracket 0 comma blank 1 comma blank minus 1 right curly bracket$ and $n$ be an even number
Statement 1: $a cross times a r cross times a r to the power of 2 end exponent midline horizontal ellipsis a r to the power of n minus 1 end exponent equals open parentheses a to the power of 2 end exponent r to the power of n minus 1 end exponent close parentheses to the power of n divided by 2 end exponent$
Statement 2:Product of $i to the power of t h end exponent$ term from the beginning and from the end in a G.P. is independent of $i$

parallel

Statement 1:Let $p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript$ and $x$ be distinct real number such that $open parentheses not stretchy sum from r equals 1 to n minus 1 of p subscript r end subscript superscript 2 end superscript close parentheses x to the power of 2 end exponent plus 2 open parentheses not stretchy sum from r equals 1 to n minus 1 of p subscript r end subscript blank p subscript r plus 1 end subscript close parentheses x plus not stretchy sum from r equals 2 to n of p subscript r end subscript superscript 2 end superscript less or equal than 0$ , then $p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript$ are in G.P. and when $a subscript 1 end subscript superscript 2 end superscript plus a subscript 2 end subscript superscript 2 end superscript plus a subscript 3 end subscript superscript 2 end superscript plus horizontal ellipsis plus a subscript n end subscript superscript 2 end superscript equals 0 comma blank a subscript 1 end subscript equals a subscript 2 end subscript equals a subscript 3 end subscript equals horizontal ellipsis equals a subscript n end subscript equals 0$
Statement 2:If $fraction numerator p subscript 2 end subscript over denominator p subscript 1 end subscript end fraction equals fraction numerator p subscript 3 end subscript over denominator p subscript 2 end subscript end fraction equals horizontal ellipsis equals fraction numerator p subscript n end subscript over denominator p subscript n minus 1 end subscript end fraction$ , then $p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript$ are in G.P.

Statement 1:Let $p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript$ and $x$ be distinct real number such that $open parentheses not stretchy sum from r equals 1 to n minus 1 of p subscript r end subscript superscript 2 end superscript close parentheses x to the power of 2 end exponent plus 2 open parentheses not stretchy sum from r equals 1 to n minus 1 of p subscript r end subscript blank p subscript r plus 1 end subscript close parentheses x plus not stretchy sum from r equals 2 to n of p subscript r end subscript superscript 2 end superscript less or equal than 0$ , then $p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript$ are in G.P. and when $a subscript 1 end subscript superscript 2 end superscript plus a subscript 2 end subscript superscript 2 end superscript plus a subscript 3 end subscript superscript 2 end superscript plus horizontal ellipsis plus a subscript n end subscript superscript 2 end superscript equals 0 comma blank a subscript 1 end subscript equals a subscript 2 end subscript equals a subscript 3 end subscript equals horizontal ellipsis equals a subscript n end subscript equals 0$
Statement 2:If $fraction numerator p subscript 2 end subscript over denominator p subscript 1 end subscript end fraction equals fraction numerator p subscript 3 end subscript over denominator p subscript 2 end subscript end fraction equals horizontal ellipsis equals fraction numerator p subscript n end subscript over denominator p subscript n minus 1 end subscript end fraction$ , then $p subscript 1 end subscript comma p subscript 2 end subscript comma horizontal ellipsis comma p subscript n end subscript$ are in G.P.

Statement 1:If $3 x plus 4 y equals 5 comma$ then the greatest value of $x to the power of 2 end exponent y to the power of 3 end exponent$ is $fraction numerator 3 over denominator 16 end fraction.$
Statement 2:Greatest value occurs when $9 x equals 8 y.$

Statement 1:If $3 x plus 4 y equals 5 comma$ then the greatest value of $x to the power of 2 end exponent y to the power of 3 end exponent$ is $fraction numerator 3 over denominator 16 end fraction.$
Statement 2:Greatest value occurs when $9 x equals 8 y.$

Statement 1:There are infinite geometric progressions for which 27, 8 and 12 are three of its terms (not necessarily consecutive)
Statement 2:Given terms are integers

Statement 1:There are infinite geometric progressions for which 27, 8 and 12 are three of its terms (not necessarily consecutive)
Statement 2:Given terms are integers

parallel

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