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Question

# Answer the following by appropriately matching the lists based on the information in Column I and Column II​​​​​​ Column IColumn IIa.y=f(x) is given by x=t5−5t3−20t+7 and y=4t3−3t2−18t+3. Then −5×dydx at t=1p. 0b. Let P(x) be a polynomial of degree 4, with P(2)=−1,P′(2)=0,P′′(2)=2,P′′′(2)=−12 and Piv(2)=24, then P′′(3) is q. −2c.y=1x, then dy√1+y4dx√1+x4r. 2d.f(2x+3y5)=2f(x)+3f(y)5 and f′(0)=p and f(0)=q. Then ,f′′(0) is s. −1

A
ap,br,cs,dq
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B
aq,br,cs,dp
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C
aq,bs,cr,dp
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D
aq,br,cp,ds
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Solution

## The correct option is B a→q,b→r,c→s,d→pa. dydx=dy/dtdx/dt=12t2−6t−185t4−15t2−20 Putting t=1 ⇒dydx=12−6−185−15−20=25⇒−5×dydx=−2 a→q b. P(x)=a(x−2)4+b(x−2)3+c(x−2)2+d(x−2)+e Given P(2)=−1⇒e=−1 P′(2)=0⇒d=0P′′(2)=2⇒2c=2⇒c=1P′′′(2)=−12⇒6b=−12⇒b=−2Piv(2)=24⇒24a=24⇒a=1 So, P′′(x)=12a(x−2)2+6b(x−2)+2c⇒P′′(x)=12(x−2)2−12(x−2)+2⇒P′′(3)=12−12+2=2 b→r c. √1+y4=√1+1x4[∵y=1x] ⇒√1+y4=√1+x4x2⇒√1+y4√1+x4=1x2⋯(1) Now, y=1x⇒dydx=−1x2 Using equation (1), ⇒dydx=−√1+y4√1+x4⇒dy√1+y4dx√1+x4=−1 c→s d. Obviously, f(x) is a linear function, Also f′(0)=p,f(0)=q f(x)=px+q⇒f′′(x)=0 d→p

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