GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 22

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GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 22

GO Classes asked May 24, 2023 • retagged Jun 12 by GO Classes

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Sachin Mittal 1

commented Jul 4, 2023

@Anuj Negi

First, let's consider the function $a x^2+b x+c$. We can rewrite it as the sum of an even function and an odd function:
$a x^2+b x+c=\left(a x^2+c\right)+b x$
The first term $a x^2+c$ is an even function because it is a constant added to an even function. The second term $b x$ is an odd function because it is a linear function of $x$ which is odd.

Now, let's consider the integral of an odd function over a symmetric interval. For any odd function $f(x)$ and any symmetric interval $[-a, a]$, we have:
$$
\int_{-a}^a f(x) d x=0
$$
Applying this property to the second term $b x$ in our equation, we have:
$$
\int_{-5}^5 b x d x=0
$$
Since -5 and 5 are symmetric about the origin, the integral of an odd function $b x$ over the interval $[-5,5]$ is zero.

Now, let's focus on the first term $a x^2+c$. We know that it is an even function. For any even function $g(x)$ and any symmetric interval $[-a, a]$, we have:
$$
\int_{-a}^a g(x) d x=2 \int_0^a g(x) d x
$$
Applying this property to the first term $a x^2+c$ in our equation, we have:
$$
\int_{-5}^5\left(a x^2+c\right) d x=2 \int_0^5\left(a x^2+c\right) d x
$$
Thus, by utilizing the properties of even and odd functions, we have proven that:
$$
\int_{-5}^5\left(a x^2+b x+c\right) d x=2 \int_0^5\left(a x^2+c\right) d x
$$

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GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 20

Suppose $f(x)$ is continuous and diffrentiable for all $x$. And $-1 \leq f^{\prime}(x) \leq 3$ fo all $x$. Which of the following is/are ALWAYS true?$f(5) \leq f(3)+6$f(5) \geq f(3)-2$f(5) \leq f(3)+10$f(5) \geq f(3)-10$

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GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 18

If $g$ is continuous (but not differentiable) at $x=0, g(0)=8$, and $f(x)=x g(x),$ find $f^{\prime}(0)$.$0$8$1$f(x)$ is also not differentiable at $x=0$.

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GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 13

The function$f(x)= \begin{cases}e^x & \text { if } \quad x \leq 1 \\ m x+b & \text { if } \quad x>1\end{cases}$is continuous and differentiable at $x=1$.Find the value of $m-b?$e$-e$\mathrm{e}-1$1-e$

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GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 15

Find the values of $A$ and $B$ that make$f(x)=\left\{\begin{array}{lll}x^2+1 & \text { if } & x \geq 0 \\A \sin x+B \cos x & \text { if } & x<0\end{array}\right.$differentiable at $x=0$.$A=0, B=1$A=1, B=0$A=0, B=-1$A=-1, B=0$

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GO Classes asked May 24, 2023

in Option D :
counterexample let f(x) = x then $\sqrt{x}$ is not differentiable at x = 0 — ArturoGangwar, May 25, 2023
@Sachin Mittal 1 sir how option a is correct in LHS, RHS both are different function. — Anuj Negi, Jul 3, 2023
@Anuj Negi

First, let's consider the function $a x^2+b x+c$. We can rewrite it as the sum of an even function and an odd function:
$a x^2+b x+c=\left(a x^2+c\right)+b x$
The first term $a x^2+c$ is an even function because it is a constant added to an even function. The second term $b x$ is an odd function because it is a linear function of $x$ which is odd.

Now, let's consider the integral of an odd function over a symmetric interval. For any odd function $f(x)$ and any symmetric interval $[-a, a]$, we have:
$$
\int_{-a}^a f(x) d x=0
$$
Applying this property to the second term $b x$ in our equation, we have:
$$
\int_{-5}^5 b x d x=0
$$
Since -5 and 5 are symmetric about the origin, the integral of an odd function $b x$ over the interval $[-5,5]$ is zero.

Now, let's focus on the first term $a x^2+c$. We know that it is an even function. For any even function $g(x)$ and any symmetric interval $[-a, a]$, we have:
$$
\int_{-a}^a g(x) d x=2 \int_0^a g(x) d x
$$
Applying this property to the first term $a x^2+c$ in our equation, we have:
$$
\int_{-5}^5\left(a x^2+c\right) d x=2 \int_0^5\left(a x^2+c\right) d x
$$
Thus, by utilizing the properties of even and odd functions, we have proven that:
$$
\int_{-5}^5\left(a x^2+b x+c\right) d x=2 \int_0^5\left(a x^2+c\right) d x
$$ — Sachin Mittal 1, Jul 4, 2023
How can $x \geq 2x$ ?
I am assuming this counter example to be a silly mistake …

Because this property (B) is a well known property from integration…

Source: Tom M. Apostol – Calculus Vol-1 2^nd Edition

Note:
Saying, f and g are continuous for all $x \epsilon [a,b]$ is equivalent of saying both f and g are integrabele on [a,b] — RohanB14, May 24, 2023

RohanB14 · Answer 1 · 2023-05-24T09:12:20+0000

This property (B) is a well known property from integration… Please add it to the correct options

Source: Tom M. Apostol – Calculus Vol-1 2nd Edition

Note:
Saying, f and g are continuous for all $x \epsilon [a,b]$ is equivalent of saying both f and g are integrabele on [a,b]

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GO Classes 2025 | Weekly Quiz 21 | Calculus | Question: 22

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