12th Class Mathematics Applications of Derivatives Question Bank Case Based (MCQs) - Derivatives

  • question_answer
    Directions: (21 - 25)
    Derivative of \[y\text{ }=\text{ }f\left( x \right)\]w.r.t. x (if exists) is denoted by \[\frac{dy}{dx}\] of \[f'\left( x \right)\]and is called the first order derivative of y.
    If we take derivative of again, then we get \[\frac{dy}{dx}\left( \frac{dy}{dx} \right)=\frac{{{d}^{2}}y}{d{{x}^{2}}}\] or \[f''\left( x \right)\] and is called the second order derivative of y. Similarly, \[\frac{d}{dx}\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] is denoted and defined as \[\frac{{{d}^{3}}y}{d{{x}^{3}}}\]or \[f'''\left( x \right)\] and is known as third order derivative of y and so on.
    Based on the above information, answer the following questions.
    If \[y={{\tan }^{-1}}\left( \frac{\log \left( e/{{x}^{2}} \right)}{\log \left( e{{x}^{2}} \right)} \right)+{{\tan }^{-1}}\left( \frac{3+2\log \,x}{1-6\log \,x} \right)\], then \[\frac{{{d}^{2}}y}{d{{s}^{2}}}\]is equal to

    A) 2

    B) 1

    C) 0

    D) -1

    Correct Answer: C

    Solution :

    Given, \[y={{\tan }^{-1}}\left( \frac{\log \left( \frac{e}{{{x}^{2}}} \right)}{\log e{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \frac{3+2\log \,x}{1-6\,\log \,x} \right)\] \[={{\tan }^{-1}}\left( \frac{1-\log \,{{x}^{2}}}{1+\log \,{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \frac{3+2\log \,x}{1-6\log \,x} \right)\] \[={{\tan }^{-1}}\left( 1 \right)-{{\tan }^{-1}}\left( 2\,\log \,x \right)+{{\tan }^{-1}}\left( 3\, \right)+{{\tan }^{-1}}\left( 2\,\log \,x \right)\]\[\Rightarrow \,y={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 3 \right)\] \[\Rightarrow \,\frac{dy}{dx}=0\,\,\,\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner