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Question

If $log subscript 10 end subscript invisible function application open square brackets fraction numerator 1 over denominator 2 to the power of x end exponent plus blank x minus 1 end fraction close square brackets equals x open square brackets log subscript 10 end subscript invisible function application 5 minus 1 close square brackets comma blank t h e n blank x equals$

4
3
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1

The correct answer is: 1

$R. H. S. equals x equals open square brackets log subscript 10 end subscript invisible function application 5 minus log subscript 10 end subscript invisible function application 10 close square brackets equals x log subscript 10 end subscript invisible function application fraction numerator 5 over denominator 10 end fraction equals log subscript 10 end subscript invisible function application fraction numerator 1 over denominator 2 to the power of x end exponent end fraction rightwards double arrow fraction numerator 1 over denominator 2 to the power of x end exponent plus x minus 1 end fraction equals fraction numerator 1 over denominator 2 to the power of x end exponent end fraction comma blank$ therefore $x minus 1 equals 0 blank o r blank x equals 1$

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r triangle $A B C comma blank R equals 5 divided by 2 blank$ and $blank r equals 1.$ Let $I$ be the incentre of the triangle and $D comma blank E$ and $blank F$ be the feet of the perpendiculars from $I blank$ to $blank B C comma blank C A blank$ and $blank A B comma$ respectively. The value of $fraction numerator I D cross times I E cross times I F over denominator I A cross times I B cross times I C end fraction blank$ is equal to

r triangle $A B C comma blank R equals 5 divided by 2 blank$ and $blank r equals 1.$ Let $I$ be the incentre of the triangle and $D comma blank E$ and $blank F$ be the feet of the perpendiculars from $I blank$ to $blank B C comma blank C A blank$ and $blank A B comma$ respectively. The value of $fraction numerator I D cross times I E cross times I F over denominator I A cross times I B cross times I C end fraction blank$ is equal to

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any $increment A B C comma$ the value of $a open parentheses b to the power of 2 end exponent plus c to the power of 2 end exponent close parentheses cos invisible function application A plus b left parenthesis c to the power of 2 end exponent plus a to the power of 2 end exponent right parenthesis cos invisible function application B plus c left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent right parenthesis cos invisible function application C equals$

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$fraction numerator sin invisible function application x over denominator sin invisible function application y end fraction equals fraction numerator 1 over denominator 2 end fraction comma fraction numerator cos invisible function application x over denominator cos invisible function application y end fraction equals fraction numerator 3 over denominator 2 end fraction blank w h e r e blank x comma blank y blank element of open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses comma blank t h e n blank t h e blank v a l u e blank o f blank left parenthesis x plus y right parenthesis blank i s blank e q u a l blank t o$

parallel

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$tan invisible function application 3 theta plus tan invisible function application theta equals 2 tan invisible function application 2 theta$ , then $blank theta$ is equal to $open parentheses n element of Z close parentheses$

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

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

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

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parallel

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$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ 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

parallel

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

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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

parallel

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

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

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

parallel

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