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General
Easy

Question

The coefficient of x to the power of n in fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction is

  1. fraction numerator 3 to the power of n plus 2 end exponent minus 2 to the power of n plus 3 end exponent plus 1 over denominator 2 end fraction
  2. fraction numerator 3 to the power of n plus 2 end exponent plus 2 to the power of n plus 3 end exponent minus 1 over denominator 2 end fraction
  3. fraction numerator 3 to the power of n plus 2 end exponent minus 2 to the power of n plus 3 end exponent minus 1 over denominator 2 end fraction
  4. fraction numerator 3 to the power of n plus 2 end exponent plus 2 to the power of n plus 3 end exponent plus 1 over denominator 2 end fraction

The correct answer is: fraction numerator 3 to the power of n plus 2 end exponent minus 2 to the power of n plus 3 end exponent plus 1 over denominator 2 end fraction

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The greatest possible number of points of intersections of 8 straight line and 4 circles is :

 Detailed Solution
The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

For selecting r objects from n objects can be done by using the formula as follows
C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fraction
 As mentioned in the question, we have to find the total number of intersection points.
For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = C presuperscript 8 subscript 2 space cross times 1 space equals space 28
 For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 2 space cross times 2 space equals space 12

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 Hence, the total number of points of intersection is = 28 + 64 + 12 = 104

The greatest possible number of points of intersections of 8 straight line and 4 circles is :

Maths-General
 Detailed Solution
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For selecting r objects from n objects can be done by using the formula as follows
C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fraction
 As mentioned in the question, we have to find the total number of intersection points.
For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = C presuperscript 8 subscript 2 space cross times 1 space equals space 28
 For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 2 space cross times 2 space equals space 12

 For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 1 space cross times C presuperscript 8 subscript 1 cross times 2 space equals space 64

 Hence, the total number of points of intersection is = 28 + 64 + 12 = 104
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Assertion (A):Iffraction numerator 1 over denominator left parenthesis x minus 2 right parenthesis open parentheses x squared plus 1 close parentheses end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator x squared plus 1 end fraction ,then Aequals 1 fifth comma B equals 1 fifth comma C equals negative 2 over 5
Reason (R) : fraction numerator 1 over denominator left parenthesis x minus a right parenthesis open parentheses x squared plus b close parentheses end fraction equals fraction numerator 1 over denominator a squared plus b end fraction open square brackets fraction numerator 1 over denominator x minus a end fraction minus fraction numerator x plus a over denominator x squared plus b end fraction close square brackets

Assertion (A):Iffraction numerator 1 over denominator left parenthesis x minus 2 right parenthesis open parentheses x squared plus 1 close parentheses end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator x squared plus 1 end fraction ,then Aequals 1 fifth comma B equals 1 fifth comma C equals negative 2 over 5
Reason (R) : fraction numerator 1 over denominator left parenthesis x minus a right parenthesis open parentheses x squared plus b close parentheses end fraction equals fraction numerator 1 over denominator a squared plus b end fraction open square brackets fraction numerator 1 over denominator x minus a end fraction minus fraction numerator x plus a over denominator x squared plus b end fraction close square brackets

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