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Maths-

General

Easy

Question

The smallest possible value of S $equals a subscript 1 end subscript times a subscript 2 end subscript times a subscript 3 end subscript plus b subscript 1 end subscript times b subscript 2 end subscript times b subscript 3 end subscript plus c subscript 1 end subscript times c subscript 2 end subscript times c subscript 3 end subscript text end text$ where $text end text a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma b subscript 1 end subscript comma b subscript 2 end subscript comma b subscript 3 end subscript comma c subscript 1 end subscript comma c subscript 2 end subscript comma c subscript 3 end subscript$ is a permutation of the number 1, 2, 3, 4, 5, 6, 7, 8, and 9 is

213
216
324
214

The correct answer is: 214

The idea is to get 3 terms as close as possible. We have 214 = 70 + 72 + 72 = 2⋅5⋅7 + 1⋅8⋅9 + 3⋅4⋅6 by AM ³ GM
$S greater or equal than 3 times left parenthesis 9 factorial right parenthesis to the power of 1 divided by 3 end exponent open parentheses fraction numerator S over denominator 3 end fraction greater or equal than open parentheses a subscript 1 end subscript a subscript 2 end subscript a subscript 3 end subscript times b subscript 1 end subscript b subscript 2 end subscript b subscript 3 end subscript times c subscript 1 end subscript c subscript 2 end subscript c subscript 3 end subscript close parentheses to the power of 1 divided by 3 end exponent close parentheses$
$text Since, end text 9 factorial equals 70 times 72 times 72 greater than 71 to the power of 3 end exponent$ = 214

Related Questions to study

Statement 1 : The second degree equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 24 x plus 36 y minus 72 equals 0$ represents an ellipse Because
Statement 2 : $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0$ represents an ellipse if $capital delta equals a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent not equal to 0$ and $h to the power of 2 end exponent greater than a b$

Statement 1 : The second degree equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 24 x plus 36 y minus 72 equals 0$ represents an ellipse Because
Statement 2 : $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0$ represents an ellipse if $capital delta equals a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent not equal to 0$ and $h to the power of 2 end exponent greater than a b$

Statement-1: P is any point such that the chord of contact of tangents from P to the ellipse $x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6$ touches $x to the power of 2 end exponent plus 4 y to the power of 2 end exponent equals 4$ The the tangents from P of $x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6$ are at right angles and
Statement-2: The tangent from any point on the director circle of an ellipse are at right angles

Statement-1: P is any point such that the chord of contact of tangents from P to the ellipse $x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6$ touches $x to the power of 2 end exponent plus 4 y to the power of 2 end exponent equals 4$ The the tangents from P of $x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 6$ are at right angles and
Statement-2: The tangent from any point on the director circle of an ellipse are at right angles

Statement-1: The distance of the tangent through (0, (d) from a focus of the ellipse $fraction numerator x to the power of 2 end exponent over denominator 16 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 7 end fraction equals 1$ is 5
and
Statement-2: The locus of the foot of the perpendicular from a focus on any tangent to $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ is $x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent$

Statement-1: The distance of the tangent through (0, (d) from a focus of the ellipse $fraction numerator x to the power of 2 end exponent over denominator 16 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 7 end fraction equals 1$ is 5
and
Statement-2: The locus of the foot of the perpendicular from a focus on any tangent to $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ is $x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent$

parallel

There are ‘n’ numbered seats around a round table. Total number of ways in which n n1 1 () < n persons can sit around the round table is equal to

There are ‘n’ numbered seats around a round table. Total number of ways in which n n1 1 () < n persons can sit around the round table is equal to

The IUPAC name of is

The IUPAC name of is

Chemistry-General

If the normal at any given point P on the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ meets its
auxiliary circle at Q and R such that QOR = 90, where O is the centre
of ellipse, then

If the normal at any given point P on the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ meets its
auxiliary circle at Q and R such that QOR = 90, where O is the centre
of ellipse, then

parallel

8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places is

8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places is

Given an in equation $open vertical bar 1 plus fraction numerator 2 over denominator x end fraction close vertical bar greater than 3$ whose solution set is given by $left parenthesis a comma 0 right parenthesis union left parenthesis 0 comma b right parenthesis$ then answer the following questions If solution set for $left parenthesis x plus 1 right parenthesis to the power of 2 end exponent less than left parenthesis 7 x minus 3 right parenthesis blank$ is (c,d), then a+b+c+d=

Given an in equation $open vertical bar 1 plus fraction numerator 2 over denominator x end fraction close vertical bar greater than 3$ whose solution set is given by $left parenthesis a comma 0 right parenthesis union left parenthesis 0 comma b right parenthesis$ then answer the following questions If solution set for $left parenthesis x plus 1 right parenthesis to the power of 2 end exponent less than left parenthesis 7 x minus 3 right parenthesis blank$ is (c,d), then a+b+c+d=

IUPAC name of ethers is

IUPAC name of ethers is

Chemistry-General

parallel

The functional group present in acylchlorides is

The functional group present in acylchlorides is

Chemistry-General

If $Z subscript 1 end subscript comma Z subscript 2 end subscript comma Z subscript 3 end subscript$ are complex numbers such that $A open parentheses Z subscript 1 end subscript close parentheses comma B open parentheses Z subscript 2 end subscript close parentheses comma C open parentheses Z subscript 3 end subscript close parentheses$ are vertices of a triangle ABC,
$angle A equals theta comma fraction numerator A C over denominator A B end fraction equals lambda$ and $open parentheses 1 plus lambda to the power of 2 end exponent minus 2 lambda c o s invisible function application theta close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 3 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses open parentheses lambda to the power of 2 end exponent minus lambda c o s invisible function application theta close parentheses Z subscript 2 end subscript equals left parenthesis 1 minus lambda c o s invisible function application theta right parenthesis Z subscript 3 end subscript close parentheses plus 2 lambda c o s invisible function application theta Z subscript 2 end subscript Z subscript 3 end subscript$ then
If $open parentheses 1 plus lambda to the power of 2 end exponent close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 2 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses lambda to the power of 2 end exponent Z subscript 2 end subscript plus Z subscript 3 end subscript close parentheses$ then the triangle is

If $Z subscript 1 end subscript comma Z subscript 2 end subscript comma Z subscript 3 end subscript$ are complex numbers such that $A open parentheses Z subscript 1 end subscript close parentheses comma B open parentheses Z subscript 2 end subscript close parentheses comma C open parentheses Z subscript 3 end subscript close parentheses$ are vertices of a triangle ABC,
$angle A equals theta comma fraction numerator A C over denominator A B end fraction equals lambda$ and $open parentheses 1 plus lambda to the power of 2 end exponent minus 2 lambda c o s invisible function application theta close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 3 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses open parentheses lambda to the power of 2 end exponent minus lambda c o s invisible function application theta close parentheses Z subscript 2 end subscript equals left parenthesis 1 minus lambda c o s invisible function application theta right parenthesis Z subscript 3 end subscript close parentheses plus 2 lambda c o s invisible function application theta Z subscript 2 end subscript Z subscript 3 end subscript$ then
If $open parentheses 1 plus lambda to the power of 2 end exponent close parentheses Z subscript 1 end subscript superscript 2 end superscript plus lambda to the power of 2 end exponent Z subscript 2 end subscript superscript 2 end superscript plus Z subscript 3 end subscript superscript 2 end superscript equals 2 Z subscript 1 end subscript open parentheses lambda to the power of 2 end exponent Z subscript 2 end subscript plus Z subscript 3 end subscript close parentheses$ then the triangle is

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation $f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater than 0$ has no solution.
ii) Solving the inequaiton $f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater than 0$ is equivalent to solving the equations $0 less than f left parenthesis x right parenthesis less than infinity$
iii) Solving the inequation $vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater than 0$ is equivalent to solving the equations $negative infinity less than f left parenthesis x right parenthesis less than 0$
iv) Solving the inequation $f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0$ is equivalent to solving the equations $0 less or equal than f left parenthesis x right parenthesis less than infinity$
v) The inequation $f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0$ is true for all $x element of$ domain of $f left parenthesis x right parenthesis$
vi) The inequation $vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater or equal than 0$ is true for all $x element of$ domain of $f left parenthesis x right parenthesis$
Set of all real values of x for which $fraction numerator 1 over denominator square root of left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis ∣ negative left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis end root end fraction$ is defined, is

The general procedure for solving inequation containing modulus functions is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach.
For example:
i) The inequation $f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater than 0$ has no solution.
ii) Solving the inequaiton $f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater than 0$ is equivalent to solving the equations $0 less than f left parenthesis x right parenthesis less than infinity$
iii) Solving the inequation $vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater than 0$ is equivalent to solving the equations $negative infinity less than f left parenthesis x right parenthesis less than 0$
iv) Solving the inequation $f left parenthesis x right parenthesis minus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0$ is equivalent to solving the equations $0 less or equal than f left parenthesis x right parenthesis less than infinity$
v) The inequation $f left parenthesis x right parenthesis plus vertical line f left parenthesis x right parenthesis vertical line greater or equal than 0$ is true for all $x element of$ domain of $f left parenthesis x right parenthesis$
vi) The inequation $vertical line f left parenthesis x right parenthesis vertical line minus f left parenthesis x right parenthesis greater or equal than 0$ is true for all $x element of$ domain of $f left parenthesis x right parenthesis$
Set of all real values of x for which $fraction numerator 1 over denominator square root of left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis ∣ negative left parenthesis 3 vertical line x minus 7 vertical line minus 6 right parenthesis end root end fraction$ is defined, is

parallel

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

Ten different letters of English alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated is

If two real valued functions f(x) & g(x) are given, for fog(x), g(x) is substituted in place of x in f(x) taking care of the domain of fog(x) which is subset of Domain of g(x), only those values of x such that g(x) lies in the Domain of f. Now, If $f left parenthesis x right parenthesis equals s i n to the power of 2 end exponent invisible function application x plus s i n to the power of 2 end exponent invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses plus c o s invisible function application x times c o s invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses text and end text g open parentheses fraction numerator 5 over denominator 4 end fraction close parentheses equals 1$ then g of (x) is

If two real valued functions f(x) & g(x) are given, for fog(x), g(x) is substituted in place of x in f(x) taking care of the domain of fog(x) which is subset of Domain of g(x), only those values of x such that g(x) lies in the Domain of f. Now, If $f left parenthesis x right parenthesis equals s i n to the power of 2 end exponent invisible function application x plus s i n to the power of 2 end exponent invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses plus c o s invisible function application x times c o s invisible function application open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses text and end text g open parentheses fraction numerator 5 over denominator 4 end fraction close parentheses equals 1$ then g of (x) is

where x, y, t R If t (5, k], then greatest value of k for which x is one-one function of t is

where x, y, t R If t (5, k], then greatest value of k for which x is one-one function of t is

parallel

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