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Answer» Solution :`ararrq,brarrr,drarrp.` a. `(dy)/(dx)=(dy//dt)/(dx//dt)=(12t^(2)-6t-18)/(5t^(4)-15t^(2)-20)` `"or "(dy)/(dx):|_(t=1)=(12-6-18)/(5-15-20)=(2)/(5)` `"or "-5(dy)/(dx):|_(t=1)=-2" at "t=1` B. Let us take `P(x)=a(x-2)^(4)+b(x-2)^(3)+c(x-2)^(2)+d(x-2)-1` `therefore""P(2)=-1` `0=P'(2)=d` `2=P''(2)=2 c or c=1` `-12=P'''(2)=6 b or b=-2` `24=P^(iv)(2)=24a or a=1` Thus, `P''(x)=12 (x-2)^(2)-12(x-2)+2` `"or "P''(3)=12-12(1)+2=2` c. `"Here, "sqrt((1+y^(4)))=sqrt((1+(1)/(x^(4))))=(sqrt(1+x^(4)))/(x^(2))""(becausey=(1)/(x))` `"or "(sqrt(1+y^(4)))/(sqrt(1+x^(4)))=(1)/(x^(2))"(1)"` `"But "y=(1)/(x)` `therefore""(dy)/(dx)=-(1)/(x^(2))"(2)"` `"From (1) and (2), "(sqrt(1+y^(4)))/(sqrt(1+x^(4)))=-(dy)/(dx)` `"or "((dy)/sqrt(1+y^(4)))/((dx)/sqrt(1+x^(4)))=-1` d. Obviously, f(x) is a linear function. Also, from f'(0) = p and f(0) = Q, f(x) = px +q `"or "f''(0)=0` |
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