If `int_-2^-1 (ax^2-5)dx=0 and 5 + int_1^2 (bx + c) dx = 0`, then

Options

A. `ax^(2)- bx+x=0` has aleast one root in` (1,2)`
B. `ax^(2)-bx+c =0` has aleast one root in `(-2,-1)`
C. `ax^(2)+bx+c=0` has atleast one root in`(-2,-1)`
D. None of these

Answer

To solve the given problem, we need to evaluate two definite integrals and set them equal to zero. Let's break it down step by step.

### Step 1: Evaluate the first integral
We are given the integral:
\[ \int_{-2}^{-1} (ax^2 - 5) \, dx = 0 \]
First, we compute the integral:
\[ \int (ax^2 - 5) \, dx = \frac{a}{3}x^3 - 5x + C \]
Now, we evaluate it from -2 to -1:
\[ \left[ \frac{a}{3}(-1)^3 - 5(-1) \right] - \left[ \frac{a}{3}(-2)^3 - 5(-2) \right] \]
Calculating the values:
\[ = \left[ \frac{a}{3}(-1) + 5 \right] - \left[ \frac{a}{3}(-8) + 10 \right] \]
\[ = \left[ -\frac{a}{3} + 5 \right] - \left[ -\frac{8a}{3} + 10 \right] \]
\[ = -\frac{a}{3} + 5 + \frac{8a}{3} - 10 \]
\[ = \frac{7a}{3} - 5 \]
Setting this equal to zero:
\[ \frac{7a}{3} - 5 = 0 \]
\[ \frac{7a}{3} = 5 \]
\[ 7a = 15 \quad \Rightarrow \quad a = \frac{15}{7} \]

### Step 2: Evaluate the second integral
Next, we have:
\[ 5 + \int_{1}^{2} (bx + c) \, dx = 0 \]
First, compute the integral:
\[ \int (bx + c) \, dx = \frac{b}{2}x^2 + cx + C \]
Now, evaluate it from 1 to 2:
\[ \left[ \frac{b}{2}(2^2) + c(2) \right] - \left[ \frac{b}{2}(1^2) + c(1) \right] \]
Calculating the values:
\[ = \left[ \frac{b}{2}(4) + 2c \right] - \left[ \frac{b}{2}(1) + c \right] \]
\[ = \left[ 2b + 2c \right] - \left[ \frac{b}{2} + c \right] \]
\[ = 2b + 2c - \frac{b}{2} - c \]
\[ = 2b - \frac{b}{2} + 2c - c \]
\[ = \frac{4b}{2} - \frac{b}{2} + c \]
\[ = \frac{3b}{2} + c \]
Setting this equal to -5:
\[ 5 + \left( \frac{3b}{2} + c \right) = 0 \]
\[ \frac{3b}{2} + c = -5 \]

### Step 3: Summary of results
We have two equations:
1. \( a = \frac{15}{7} \)
2. \( \frac{3b}{2} + c = -5 \)