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Let \mathrm{f(x) } be a polynomial of degree 4 with \mathrm{f(2)=-1, } \mathrm{f^{\prime}(2)=0, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(2)=-12, f^{\prime \prime \prime \prime}(2)=24 } then \mathrm{f^{\prime \prime}(1)= }

Option: 1

26


Option: 2

13


Option: 3

29


Option: 4

62


Answers (1)

best_answer

\mathrm{ Let f(x)=a x^4+b x^3+c x^2+d x+e }
\mathrm{\begin{aligned} & \because \quad f(2)=-1 \Rightarrow 16 a+8 b+4 c+2 d+e=-1...........(i) \\ & f^{\prime}(x)=4 a x^3+3 b x^2+2 c x+d \\ & \because \quad f^{\prime}(2)=0 \Rightarrow 32 a+12 b+4 c+d=0........(ii)\\ & f^{\prime \prime}(x)=12 a x^2+6 b x+2 c \\ & \because \quad f^{\prime \prime}(2)=2 \Rightarrow 48 a+12 b+2 c=2......(iii) \\ & f^{\prime \prime \prime}(x)=24 a x+6 b \end{aligned} }
\mathrm{\begin{aligned} & \because \quad f^{\prime \prime \prime}(2)=-12 \Rightarrow 48 a+6 b=-12.........(iv) \\ & f^{\prime \prime \prime \prime}(x)=24 a \\ & \because \quad f^{\prime \prime \prime \prime}(2)=24 \Rightarrow 24 a=24 \Rightarrow a=1 \\ & \text { (iv) } \Rightarrow 48+6 b=-12 \Rightarrow 6 b=-60 \Rightarrow b=-10 \\ & \text { (iii) } \Rightarrow 48-120+2 c=2 \Rightarrow 2 c=74 \Rightarrow c=37 \\ & \text { (ii) } \Rightarrow 32-120+148+d=0 \Rightarrow d=-60 \\ & \text { (i) } \Rightarrow 16-80+148-120+e=-1 \Rightarrow e=35 \\ & \therefore \quad f^{\prime \prime}(1)=12 \times 1 \times 1+6(-10) \times 1+2 \times 37 \\ & \quad=12-60+74=26 \end{aligned} }

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Deependra Verma

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