Question:
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\left|4x^{2}+3x-10\right|+x^{2}-2x+5,\ x\in[-3,1]\) is ________.
The sum of absolute maximum and absolute minimum values of the function \(f(x)=\left|4x^{2}+3x-10\right|+x^{2}-2x+5,\ x\in[-3,1]\) is ________.
Show Hint
Find where \(4x^2+3x-10\) changes sign, write \(f\) piecewise on \([-3,1]\), and compare the values at the end points, break point and vertex.
Updated On: Nov 12, 2026
Show Solution
Verified By Collegedunia
Correct Answer: 44
Solution and Explanation
\(4x^2+3x-10=(4x-5)(x+2)\), so it is zero at \(x=-2\) and \(x=\frac54\). On \([-3,1]\) it is positive for \(-3\le x<-2\) and negative for \(-2<x\le 1\).
For \(-3\le x\le -2\): \(f(x)=5x^2+x-5\). Its vertex is at \(x=-\frac1{10}\), outside this interval, so \(f\) is decreasing here, from \(f(-3)=37\) to \(f(-2)=13\).
For \(-2\le x\le 1\): \(f(x)=-3x^2-5x+15\). Its vertex is at \(x=-\frac56\) with \(f\left(-\frac56\right)=\frac{205}{12}\approx 17.08\).
Values to compare: \(f(-3)=37\), \(f(-2)=13\), \(f\left(-\frac56\right)\approx17.08\), \(f(1)=7\).
Absolute maximum = 37 and absolute minimum = 7.
Sum = 37 + 7 = 44.
Answer: 44
For \(-3\le x\le -2\): \(f(x)=5x^2+x-5\). Its vertex is at \(x=-\frac1{10}\), outside this interval, so \(f\) is decreasing here, from \(f(-3)=37\) to \(f(-2)=13\).
For \(-2\le x\le 1\): \(f(x)=-3x^2-5x+15\). Its vertex is at \(x=-\frac56\) with \(f\left(-\frac56\right)=\frac{205}{12}\approx 17.08\).
Values to compare: \(f(-3)=37\), \(f(-2)=13\), \(f\left(-\frac56\right)\approx17.08\), \(f(1)=7\).
Absolute maximum = 37 and absolute minimum = 7.
Sum = 37 + 7 = 44.
Answer: 44
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Application of derivatives Questions
- Let the set of all values of \( p \), for which \[ f(x) = (p^2 - 6p + 8)(\sin^2 2x - \cos^2 2x) + 2(2 - p)x + 7 \] does not have any critical point, be the interval \( (a, b) \). Then \( 16ab \) is equal to _____ .
- For the function \(f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x), \quad x \in \left[0, \frac{\pi}{2}\right],\)consider the following two statements:
1. \( f \) is increasing in \( \left(0, \frac{\pi}{2}\right) \).
2. \( f' \) is decreasing in \( \left(0, \frac{\pi}{2}\right) \).
Between the above two statements - Let \(A\) be the region enclosed by the parabola \(y^2 = 2x\) and the line \(x = 24\). Then the maximum area of the rectangle inscribed in the region \(A\) is ________.
- f(x) = x\(^{2025}\) - x\(^{2000}\), x \(\in\) [0, 1], then minimum value of f(x) is :
- Let \[ f(x)=x^{2025}-x^{2000},\quad x\in[0,1] \] and the minimum value of the function $f(x)$ in the interval $[0,1]$ be \[ (80)^{80}(n)^{-81}. \] Then $n$ is equal to
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions