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Question

e number of solutions of $blank not stretchy sum from r equals 1 to 5 of cos invisible function application r x equals 5$ in the interval $blank open square brackets 0 comma blank 2 pi close square brackets blank$ is

0
2
5
10

The correct answer is: 2

$not stretchy sum from r equals 1 to 5 of cos invisible function application r x equals 5$
$rightwards double arrow cos invisible function application x plus cos invisible function application 2 x plus cos invisible function application 3 x plus cos invisible function application 4 x plus cos invisible function application 5 x equals 5$
Which is possible only, when $cos invisible function application x equals cos invisible function application 2 x equals cos invisible function application 3 x equals cos invisible function application 4 x equals cos invisible function application 5 x equals 1$ and is satisfied by $x equals 0$ and $x equals 2 pi$

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Member of roots of $open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus sin invisible function application 2 theta close parentheses equals 1 plus tan invisible function application theta$ for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets$ is

Member of roots of $open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus sin invisible function application 2 theta close parentheses equals 1 plus tan invisible function application theta$ for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets$ is

r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

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parallel

$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ and $sin invisible function application x plus sin invisible function application y equals 2$ , then the value of $blank x plus y$ is

$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ and $sin invisible function application x plus sin invisible function application y equals 2$ , then the value of $blank x plus y$ is

e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

parallel

e total number of solutions of $ln invisible function application open vertical bar sin invisible function application x close vertical bar equals negative x to the power of 2 end exponent plus 2 x blank$ in $open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$ is equal to

e total number of solutions of $ln invisible function application open vertical bar sin invisible function application x close vertical bar equals negative x to the power of 2 end exponent plus 2 x blank$ in $open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$ is equal to

e general solution of the equation $8 cos invisible function application x cos invisible function application 2 x cos invisible function application 4 x equals sin invisible function application 6 x divided by sin invisible function application x$ is

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$alpha$ is a root of $25 cos to the power of 2 end exponent invisible function application theta plus 5 cos invisible function application theta minus 12 equals 0 comma fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma$ then $blank sin invisible function application 2 alpha blank$ is equal to

$alpha$ is a root of $25 cos to the power of 2 end exponent invisible function application theta plus 5 cos invisible function application theta minus 12 equals 0 comma fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma$ then $blank sin invisible function application 2 alpha blank$ is equal to

parallel

$x y to the power of 2 end exponent equals 4 blank a n d log subscript 3 end subscript invisible function application open parentheses log subscript 2 end subscript invisible function application x close parentheses plus log subscript 1 divided by 3 end subscript invisible function application open parentheses log subscript 1 divided by 2 end subscript invisible function application y close parentheses equals 1 comma$ then $x$ equals

$x y to the power of 2 end exponent equals 4 blank a n d log subscript 3 end subscript invisible function application open parentheses log subscript 2 end subscript invisible function application x close parentheses plus log subscript 1 divided by 3 end subscript invisible function application open parentheses log subscript 1 divided by 2 end subscript invisible function application y close parentheses equals 1 comma$ then $x$ equals

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

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parallel

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

e number of solutions of the equation $tan invisible function application x plus sec invisible function application x minus 2 cos invisible function application x$ lying in the interval $left square bracket 0 comma blank 2 pi right square bracket$ is

e number of solutions of the equation $tan invisible function application x plus sec invisible function application x minus 2 cos invisible function application x$ lying in the interval $left square bracket 0 comma blank 2 pi right square bracket$ is

a convex quadrilateral $A B C D comma blank A B equals a comma blank B C equals b comma blank C D equals c blank$ and $D A equals d.$ This quadrilateral is such that a circle can be inscribed in it and a circle can be also circumscribed about it, then $tan to the power of 2 end exponent invisible function application open parentheses A divided by 2 close parentheses$ is equal to

a convex quadrilateral $A B C D comma blank A B equals a comma blank B C equals b comma blank C D equals c blank$ and $D A equals d.$ This quadrilateral is such that a circle can be inscribed in it and a circle can be also circumscribed about it, then $tan to the power of 2 end exponent invisible function application open parentheses A divided by 2 close parentheses$ is equal to

parallel

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