Find the real roots of the equation \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\] if ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $
A. $ \pm 1$
B. $ \pm \dfrac{1}{2}$
C. $ \pm \dfrac{1}{2}$
D. 0 and 1
Answer
Verified
504.9k+ views
Hint: In order to solve this problem use the concept that differentiation is the inverse of integration and vice-versa. Using this concept you can get the roots of the equation given.
Complete step-by-step answer:
The given equations are ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ and \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\]
On solving ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ we get f(x) in terms of x as in limit there is x so, t will be replaced by x.
So, f(x) can be written as ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $
And we know integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x).
If ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $
So, ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $
The other equation is \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\].
On putting the value of ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $ in the above equation we get the equation as:
\[{{\text{x}}^{\text{2}}} - \sqrt {2 - {x^2}} = 0\]
\[{{\text{x}}^{\text{2}}}{\text{ = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} \]
On squaring both sides we get,
${{\text{x}}^{\text{4}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 2 = 0}}$
${{\text{x}}^{\text{2}}}{\text{ = 1, - 2}}$
So, the real values of ${\text{x} = \pm 1}$
So, the correct option for this question is A.
Note: Whenever you face such types of problems you have to use the concept that integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x). Here in this question we have then found the roots of the equation obtained then eliminated the imaginary roots as only real roots have been asked in the question. Proceeding like this will take you to the right answer.
Complete step-by-step answer:
The given equations are ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ and \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\]
On solving ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ we get f(x) in terms of x as in limit there is x so, t will be replaced by x.
So, f(x) can be written as ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $
And we know integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x).
If ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $
So, ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $
The other equation is \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\].
On putting the value of ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $ in the above equation we get the equation as:
\[{{\text{x}}^{\text{2}}} - \sqrt {2 - {x^2}} = 0\]
\[{{\text{x}}^{\text{2}}}{\text{ = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} \]
On squaring both sides we get,
${{\text{x}}^{\text{4}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 2 = 0}}$
${{\text{x}}^{\text{2}}}{\text{ = 1, - 2}}$
So, the real values of ${\text{x} = \pm 1}$
So, the correct option for this question is A.
Note: Whenever you face such types of problems you have to use the concept that integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x). Here in this question we have then found the roots of the equation obtained then eliminated the imaginary roots as only real roots have been asked in the question. Proceeding like this will take you to the right answer.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Master Class 12 Biology: Engaging Questions & Answers for Success
Class 12 Question and Answer - Your Ultimate Solutions Guide
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Draw a labelled sketch of the human eye class 12 physics CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
What is a transformer Explain the principle construction class 12 physics CBSE
Explain sex determination in humans with the help of class 12 biology CBSE