Q.17. Let be a singular matrix. Let , . If M and m are respectively the maximum and the minimum val

Q: What is the correct answer for this exam Mathematics Q17?

The correct answer is B. First, find $\alpha$. A matrix is singular if its determinant is 0.$\det(A) = 1(1(-1) - \alpha(1)) - 3(2(-1) - \alpha(0)) + (-1)(2(1) - 1(0)) = 0$.$1(-1 - \alpha) - 3(-2 - 0) - 1(2 - 0) = 0$.$-1 - \alpha + 6 - 2 = 0$.$3 - \alpha = 0 \Rightarrow \alpha = 3$.Now, $f(x) = \int_0^x (t^2 + 2t + 3)\,dt$ for $x \in [1, 3]$.First, evaluate the integral:$\int (t^2 + 2t + 3)\,dt = \frac{t^3}{3} + t^2 + 3t +

Q: Q.17. Let be a singular matrix. Let , . If M and m are respectively the maximum and the minimum values of f in then is equal to :

The correct answer is B. First, find $\alpha$. A matrix is singular if its determinant is 0.$\det(A) = 1(1(-1) - \alpha(1)) - 3(2(-1) - \alpha(0)) + (-1)(2(1) - 1(0)) = 0$.$1(-1 - \alpha) - 3(-2 - 0) - 1(2 - 0) = 0$.$-1 - \alpha + 6 - 2 = 0$.$3 - \alpha = 0 \Rightarrow \alpha = 3$.Now, $f(x) = \int_0^x (t^2 + 2t + 3)\,dt$ for $x \in [1, 3]$.First, evaluate the integral:$\int (t^2 + 2t + 3)\,dt = \frac{t^3}{3} + t^2 + 3t +

Q: What are the options for this this exam Mathematics question?

A. 64 · B. 68 · C. 72 · D. 76

Question

Q.17. Let  be a singular matrix. Let , . If M and m are respectively the maximum and the minimum values of f in  then  is equal to :

FirstInTest · Accepted Answer

First, find $\alpha$. A matrix is singular if its determinant is 0.$\det(A) = 1(1(-1) - \alpha(1)) - 3(2(-1) - \alpha(0)) + (-1)(2(1) - 1(0)) = 0$.$1(-1 - \alpha) - 3(-2 - 0) - 1(2 - 0) = 0$.$-1 - \alpha + 6 - 2 = 0$.$3 - \alpha = 0 \Rightarrow \alpha = 3$.Now, $f(x) = \int_0^x (t^2 + 2t + 3)\,dt$ for $x \in [1, 3]$.First, evaluate the integral:$\int (t^2 + 2t + 3)\,dt = \frac{t^3}{3} + t^2 + 3t + C$.So, $f(x) = [\frac{t^3}{3} + t^2 + 3t]_0^x = \frac{x^3}{3} + x^2 + 3x$.To find maximum and minimum values, we check critical points and endpoints.Find $f'(x)$: $f'(x) = \frac{d}{dx} (\frac{x^3}{3} + x^2 + 3x) = x^2 + 2x + 3$.The discriminant of $f'(x)$ is $D = 2^2 - 4(1)(3) = 4 - 12 = -8 < 0$.Since the leading coefficient is positive (1), $f'(x)$ is always positive.This means $f(x)$ is strictly increasing on its domain.Therefore, the minimum value $m$ occurs at $x=1$ and the maximum value $M$ occurs at $x=3$.$m = f(1) = \frac{1^3}{3} + 1^2 + 3(1) = \frac{1}{3} + 1 + 3 = \frac{1}{3} + 4 = \frac{13}{3}$.$M = f(3) = \frac{3^3}{3} + 3^2 + 3(3) = \frac{27}{3} + 9 + 9 = 9 + 9 + 9 = 27$.We need to calculate $3(M-m)$.$3(M-m) = 3(27 - \frac{13}{3}) = 3(\frac{81-13}{3}) = 3(\frac{68}{3}) = 68$.

Answer

64

Answer

68

Answer

72

Answer

76

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