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Application of Derivatives question

2020 · 6 Sep · Shift 1 · Q29
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  5. /2020 · 6 Sep · Shift 1 · Q29

Application of Derivatives question

2020 · 6 Sep · Shift 1 · Q29

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
The position of a moving car at time t is given by f(t) = at2 + bt + c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t1 , t2 ] is attained at the point :
  1. A
    (t1+t2)2{{\left( {{t_1} + {t_2}} \right)} \over 2}2(t1​+t2​)​
  2. B
    (t2−t1)2{{\left( {{t_2} - {t_1}} \right)} \over 2}2(t2​−t1​)​
  3. C
    2a(t1 + t2) + b
  4. D
    a(t2 – t1) + b
View written solutionFree

Correct answer: A

  1. The position function is
\quad t>0$$ where $a,b,c>1$. 2. Since $a>0$ and $t>0$, the velocity is $$f'(t)=2at+b$$ which is positive for all $t>0$ because $a,b>1$. Hence the car is always moving forward, so $$\text{average speed} = \text{average rate of change of position}.$$ 3. Average speed over $[t_1,t_2]$ is $$\frac{f(t_2)-f(t_1)}{t_2-t_1}.$$ Compute it: $$f(t_2)-f(t_1)=a(t_2^2-t_1^2)+b(t_2-t_1)$$ $$=a(t_2-t_1)(t_2+t_1)+b(t_2-t_1).$$ So, $$\frac{f(t_2)-f(t_1)}{t_2-t_1}=a(t_1+t_2)+b.$$ 4. By the Mean Value Theorem, there exists some $\xi\in(t_1,t_2)$ such that $$f'(\xi)=\frac{f(t_2)-f(t_1)}{t_2-t_1}.$$ Substitute: $$2a\xi+b=a(t_1+t_2)+b.$$ Cancelling $b$, $$2a\xi=a(t_1+t_2).$$ Since $a\neq 0$, $$2\xi=t_1+t_2$$ $$\xi=\frac{t_1+t_2}{2}.$$ 5. Therefore, the average speed is attained at $$\boxed{\frac{t_1+t_2}{2}}.$$ So the correct option is **A**.
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