Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Application of Derivatives question

2022 · 25 Jul · Shift 2 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Application of Derivatives
  5. /2022 · 25 Jul · Shift 2 · Q40

Application of Derivatives question

2022 · 25 Jul · Shift 2 · Q40

JEE MainMathematicsApplication of DerivativesNumerical+4 / −1
The sum of the maximum and minimum values of the function f(x)=∣5x−7∣+[x2+2x]f(x)=|5 x-7|+\left[x^{2}+2 x\right]f(x)=∣5x−7∣+[x2+2x] in the interval [54,2]\left[\frac{5}{4}, 2\right][45​,2], where [t][t][t] is the greatest integer ≤t\leq t≤t, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 15

  1. We need the maximum and minimum of f(x)=∣5x−7∣+[x2+2x],  x∈[54,2].f(x)=|5x-7|+[x^2+2x], \,\, x\in\left[\frac54,2\right].f(x)=∣5x−7∣+[x2+2x],x∈[45​,2]. Here [t][t][t] denotes the greatest integer function.

  2. First, analyze the absolute value part:

\begin{cases} 7-5x, & x<\frac75,\\ 5x-7, & x\ge \frac75. \end{cases}$$ Since $\frac75=1.4$ lies in the interval $\left[\frac54,2\right]$, this split matters. 3. Now analyze the floor part: $$g(x)=x^2+2x.$$ On the interval $\left[\frac54,2\right]$, this is increasing because $$g'(x)=2x+2>0.$$ So its minimum and maximum on the interval are: $$g\left(\frac54\right)=\frac{25}{16}+\frac{10}{4}=\frac{25}{16}+\frac{40}{16}=\frac{65}{16}=4.0625,$$ $$g(2)=4+4=8.$$ Hence $$x^2+2x\in\left[\frac{65}{16},8\right].$$ Therefore $$[x^2+2x]$$ can only take values $4,5,6,7,8$. 4. Find where the floor changes value. We solve: - $x^2+2x=5 \Rightarrow x^2+2x-5=0 \Rightarrow x=-1+\sqrt6$. - $x^2+2x=6 \Rightarrow x^2+2x-6=0 \Rightarrow x=-1+\sqrt7$. - $x^2+2x=7 \Rightarrow x^2+2x-7=0 \Rightarrow x=-1+2\sqrt2$. - $x^2+2x=8 \Rightarrow x^2+2x-8=0 \Rightarrow x=2$ (relevant root). So on the interval: $$[x^2+2x]= \begin{cases} 4, & \frac54\le x< -1+\sqrt6,\\ 5, & -1+\sqrt6\le x< -1+\sqrt7,\\ 6, & -1+\sqrt7\le x< -1+2\sqrt2,\\ 7, & -1+2\sqrt2\le x<2,\\ 8, & x=2. \end{cases}$$ 5. Now combine with $|5x-7|$. Since $|5x-7|$ is piecewise linear, on each subinterval $f(x)$ is linear plus a constant. We compare the critical breakpoint $\frac75=1.4$ with the floor-change points: $$-1+\sqrt6\approx 1.449,$$ $$-1+\sqrt7\approx 1.646,$$ $$-1+2\sqrt2\approx 1.828.$$ Thus the interval splits as: - $\left[\frac54,\frac75\right)$: here $[x^2+2x]=4$ and $|5x-7|=7-5x$, $$f(x)=7-5x+4=11-5x.$$ This decreases from $$f\left(\frac54\right)=11-\frac{25}{4}=\frac{19}{4}=4.75$$ to approaching $$11-5\cdot\frac75=4.$$ - $\left[\frac75,-1+\sqrt6\right)$: here $[x^2+2x]=4$ and $|5x-7|=5x-7$, $$f(x)=5x-7+4=5x-3.$$ This increases from $$f\left(\frac75\right)=4$$ upward. - $\left[-1+\sqrt6,-1+\sqrt7\right)$: here $[x^2+2x]=5$, $$f(x)=5x-7+5=5x-2,$$ increasing. - $\left[-1+\sqrt7,-1+2\sqrt2\right)$: here $[x^2+2x]=6$, $$f(x)=5x-1,$$ increasing. - $\left[-1+2\sqrt2,2\right)$: here $[x^2+2x]=7$, $$f(x)=5x,$$ increasing. - At $x=2$: $$f(2)=|10-7|+[8]=3+8=11.$$ 6. Minimum value: From the first two pieces, the smallest value occurs at $$x=\frac75,$$ where $$f\left(\frac75\right)=|7-7|+[\tfrac{49}{25}+\tfrac{14}{5}]=0+[4.76]=4.$$ So the minimum value is $$\boxed{4}.$$ 7. Maximum value: For $x\ge \frac75$, each piece is increasing, and at the endpoint $x=2$ we get $$f(2)=11.$$ So the maximum value is $$\boxed{11}.$$ 8. Therefore, the required sum is $$11+4=\boxed{15}.$$
PreviousNext

More from Application of Derivatives

  • Water is being filled at the rate of 1 cm3 / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm2 / sec) at which the wet conical…2022 · MCQ
  • Let f(x)=∣(x−1)(x2−2x−3)∣+x−3,x∈R. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ​.2022 · Numerical
  • Let the function f(x)=2x2−loge​x,x>0, be decreasing in (0,a) and increasing in (a,4). A tangent to the parabola y2=4ax at a point P on it passes through the point (8a,8a−1)…2022 · Numerical
  • If the maximum value of a, for which the function fa​(x)=tan−12x−3ax+7 is non-decreasing in (−6π​,6π​), is aˉ, then faˉ​(8π​) is equal to :2022 · MCQ
  • The sum of the absolute minimum and the absolute maximum values of the function f(x) = |3x − x2 + 2|− x in the interval [− 1, 2] is :2022 · MCQ
  • Let f(x)=2cos−1x+4cot−1x−3x2−2x+10, x∈[−1,1]. If [a, b] is the range of the function f, then 4a − b is equal to :2022 · MCQ
  • Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :2022 · MCQ
  • A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is tan−143​. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter…2022 · Numerical