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  • Help me answer: - Limit , continuity and differentiability - JEE Main-12

Let f(x) = \begin{vmatrix} \sin x & \cos x & 1 \\ \cos x& -\sin x & 2\\ \sin x& \cos x & 3 \end{vmatrix} then f'(x) = ?

  • Option 1)

    -1

  • Option 2)

    0

  • Option 3)

    1

  • Option 4)

    2

 

Answers (1)

best_answer

As we have learnt,

 

Differentiation of a determinant -

y=\begin{vmatrix} f(x) & g(x) & h(x) \\ p(x)& q(x) & r(x) \\ u(x)& v(x)& w(x)) \end{vmatrix} then 

\frac{dy}{dx}=\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ p(x)& q(x) & r(x) \\ u(x)& v(x)& w(x)) \end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x) \\ p'(x)& q'(x) & r'(x) \\ u(x)& v(x)& w(x)) \end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x) \\ p(x)& q(x) & r(x) \\ u'(x)& v'(x)& w'(x)) \end{vmatrix}

 


 

-

 

 Differenciating columnwise,

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

(1st and 2nd diterminant are 0 as columns are propotional)

\therefore f'(x) = 0 + 0 + 0 = 0


Option 1)

-1

Option 2)

0

Option 3)

1

Option 4)

2

Posted by

Himanshu

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