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CBSE Class 12-science Answered

do it pls
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Asked by sdmbotch1123 | 16 Jan, 2023, 20:18: PM
answered-by-expert Expert Answer
slope of tangent = begin mathsize 14px style fraction numerator d y over denominator d x end fraction space equals space fraction numerator 2 e to the power of 2 x end exponent minus space 6 space e to the power of negative x end exponent plus 9 over denominator 2 plus 9 space e to the power of negative 2 x end exponent end fraction end style
Let ex = α
 
begin mathsize 14px style fraction numerator d y over denominator d x end fraction space equals space fraction numerator 2 space alpha squared minus space begin display style 6 over alpha end style plus 9 over denominator 2 plus begin display style 9 over alpha squared end style end fraction space equals space fraction numerator 2 space alpha to the power of 4 space plus space 9 space alpha squared space minus 6 alpha over denominator 2 space alpha squared space plus space 9 end fraction space equals space alpha squared space minus space fraction numerator 6 alpha over denominator 2 space alpha squared plus 9 end fraction end style
 
begin mathsize 14px style fraction numerator d y over denominator d x end fraction space equals space alpha squared space minus space fraction numerator 3 space alpha over denominator alpha squared plus begin display style 9 over 2 end style end fraction end style
 
If we resubstitute α = ex in above expression , then we have
 
begin mathsize 14px style fraction numerator d y over denominator d x end fraction space equals space e to the power of 2 x end exponent space minus space fraction numerator 3 space e to the power of x over denominator e to the power of 2 x end exponent space plus space begin display style 9 over 2 end style end fraction end style
 
Let us integrate RHS to get y(x)
 
begin mathsize 14px style y left parenthesis x right parenthesis space equals space integral e to the power of 2 x end exponent d x space minus space 3 space integral fraction numerator e to the power of x space d x over denominator e to the power of 2 x end exponent plus space begin display style left parenthesis 9 divided by 2 right parenthesis end style end fraction end style
to perform the second integration in above expression,
 
let us use the substitution  u = ex  and du = ex dx
 
begin mathsize 14px style y left parenthesis x right parenthesis space equals space 1 half e to the power of 2 x space end exponent space minus space 3 space integral fraction numerator d u over denominator u squared plus left parenthesis 9 divided by 2 right parenthesis end fraction end style
 
begin mathsize 14px style y left parenthesis x right parenthesis space equals space 1 half e to the power of 2 x space end exponent space minus space square root of 2 space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 space e to the power of x over denominator 3 end fraction close parentheses space plus space C end style  ............................(1)
where C is constant of integration.
 
we are given that y(x) passes through begin mathsize 14px style open parentheses 0 comma space 1 half plus fraction numerator straight pi over denominator 2 square root of 2 end fraction close parentheses end style
 
hence if we substitute x and y values as above in eqn.(1) , then we have

begin mathsize 14px style 1 half plus fraction numerator pi over denominator 2 square root of 2 end fraction space equals space 1 half space minus space square root of 2 space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 space over denominator 3 end fraction close parentheses space plus space C end style
 
Hence we get ,    begin mathsize 14px style C space equals space fraction numerator pi over denominator 2 square root of 2 end fraction space plus space space square root of 2 space tan to the power of negative 1 end exponent stretchy left parenthesis fraction numerator square root of 2 space over denominator 3 end fraction stretchy right parenthesis space end style
 
Now if y(x) passes through begin mathsize 14px style open parentheses alpha space comma space 1 half space e to the power of 2 alpha end exponent close parentheses end style , then we have
begin mathsize 14px style 1 half space e to the power of 2 alpha end exponent space equals space 1 half e to the power of 2 alpha space end exponent space minus space square root of 2 space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 space e to the power of alpha over denominator 3 end fraction close parentheses space plus space fraction numerator pi over denominator 2 square root of 2 end fraction space plus space square root of 2 space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 3 end fraction close parentheses end style
 
begin mathsize 14px style space space space tan to the power of negative 1 end exponent stretchy left parenthesis fraction numerator square root of 2 space e to the power of alpha over denominator 3 end fraction stretchy right parenthesis space equals space space space pi over 4 space plus space space tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 3 end fraction close parentheses end style
 
begin mathsize 14px style space space space tan to the power of negative 1 end exponent stretchy left parenthesis fraction numerator square root of 2 space e to the power of alpha over denominator 3 end fraction stretchy right parenthesis space equals space space space pi over 4 space plus space 0.441 end style
begin mathsize 14px style space space e to the power of alpha space equals space space space fraction numerator 3 over denominator square root of 2 end fraction space space tan open parentheses pi over 4 space plus space 0.441 close parentheses space equals space 5.9 end style
 
 
 
 
Answered by Thiyagarajan K | 17 Jan, 2023, 00:16: AM

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