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UPSC NDA I 2016 Mathematics Solved Paper

For the next two (2) items that follow :

Consider the function

f(x)=\begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

Where p is constant.

[NDA I 2016]

Question : 35 of 120

Marks: +1, -0

f''(0)=0

Solution:

f''(x)=\begin{vmatrix} 6x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

f''(0)=\begin{vmatrix} 0 & 0 & -\cos 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}

=-1(6p^2+p)

f''(0)=0

-(6p^2+p)=0

p(6p+1)=0

p=0

p=-\frac{1}{6}

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