Variables and Data Types
Learn how to store information using variables and understand the different data types available in JavaScript, such as strings, numbers, and booleans.
extract all questions and give solutions
Sure! Here are the extracted questions from all the images, along with their solutions:
1. Match the functions with the correct area under the curve and graph (Image 4)
Question:
Match the functions in Column A with the corresponding (signed) area between its graph and the interval $[-1, 1]$ on the X-axis in column B and the images of their graphs and the highlighted region corresponding to the area computed in Column C. Options (in terms of matching):
- i) → b) → 1), ii) → a) → 2), iii) → c) → 3)
- i) → b) → 3), ii) → c) → 1), iii) → a) → 2)
- i) → c) → 1), ii) → b) → 2), iii) → a) → 3)
- i) → b) → 1), ii) → c) → 3), iii) → a) → 2)
Column A (Functions): i) $f(x) = x$ ii) $f(x) = |x|$ iii) $f(x) = x^2$
Column B (Areas): a) $\frac{2}{3}$ b) 0 c) 1
Column C (Graphs): — 1) 2) 3) (See image for mapping)
Solution:
- $f(x) = x$ is an odd function, so the area under the curve from $[-1, 1]$ is 0 (b), and matches graph 1.
- $f(x) = |x|$ is symmetric and positive, area is the integral of $|x|$ over $[-1, 1]$, which is 1 (c), and matches graph 3.
- $f(x) = x^2$ is always positive, and $\int_{-1}^1 x^2 dx = 2/3$ (a), matches graph 2.
So, Correct Option: i) → b) → 1), ii) → c) → 3), iii) → a) → 2) (First option)
2. Maximum volume and surface area of an inscribed cylinder (Image 3)
Question:
A cylinder of radius $x$ and height $2h$ is to be inscribed in a sphere of radius $R$… Choose the correct set of options.
Given:
- Volume: $V = 2\pi x^2 h$
- Curved Surface Area: $S = 4\pi x h$
- Cylinder fits in sphere: $x^2 + h^2 = R^2$
Options:
- The cylinder has maximum volume amongst all cylinders when $h = R$.
- The cylinder has maximum volume amongst all cylinders when $h = \sqrt{3}R$.
- The cylinder has maximum volume amongst all cylinders when $h = \frac{R}{\sqrt{3}}$.
- The cylinder has maximum surface area of curved surface when $h = 2R$.
- The cylinder has maximum surface area when $h = \frac{R}{\sqrt{2}}$.
- The cylinder has maximum surface area when $h = \sqrt{2}$.
Solution:
Maximum Volume:
Let us maximize $V = 2\pi x^2 h$ under constraint $x^2 + h^2 = R^2$:
Substitute $x^2 = R^2 - h^2$: $V = 2\pi (R^2 - h^2) h = 2\pi h (R^2 - h^2)$
Let $V = 2\pi (R^2 h - h^3)$. Set derivative to 0 for critical points:
$$ \frac{dV}{dh} = 2\pi (R^2 - 3h^2) = 0 \implies R^2 = 3h^2 \implies h = \frac{R}{\sqrt{3}} $$
Maximum Surface Area:
$S = 4\pi x h$, substitute $x = \sqrt{R^2 - h^2}$
$$ S = 4\pi h \sqrt{R^2 - h^2} $$
Let $u = h^2$, so $S^2 \sim h^2 (R^2 - h^2) = h^2 R^2 - h^4$ Take derivative, set to 0:
$$ \frac{d}{dh}(h^2(R^2 - h^2)) = 2hR^2 - 4h^3 = 2h(R^2 - 2h^2) = 0 \implies h = 0 \text{ or } h^2 = \frac{R^2}{2} \implies h = \frac{R}{\sqrt{2}} $$
Correct Options:
- The cylinder has maximum volume amongst all cylinders when $h = \frac{R}{\sqrt{3}}$.
- The cylinder has maximum surface area of curved surface when $h = \frac{R}{\sqrt{2}}$.
3. Negative area enclosed by a curve (Image 1)
Question:
Which of the curves enclose a negative area on the $X$ axis in interval $$?1
Solution (By visual inspection):
- Curve 1: Always above x-axis ⇒ Positive area.
- Curve 2: Mostly below x-axis ⇒ Negative area.
- Curve 3: Always above x-axis ⇒ Positive area.
- Curve 4: Entire shaded region is below x-axis ⇒ Negative area.
Correct Curves: Curve 2 and Curve 4.
4. Profits, Errors, and minimum values (Image 2)
Given:
- $f_1(x) = x^3$, $f_2(x) = x$
- Predicted: $g_1(x) = \sqrt{x}$, $g_2(x) = e^x$
- Error in prediction = area bounded by the curves in $$.1
Q4:
What will the absolute difference between the minimum values of $f_2$ and $g_2$ in the interval $$ be?1
Solution:
- $f_2(x) = x$, minimum in $$ is at $x = 0$, so minimum is $0$1
- $g_2(x) = e^{x}$, minimum in $$ is at $x = 0$, so minimum is $e^0 = 1$1
Absolute difference: $|0 - 1| = 1$
Q5:
Choose the correct statements about prediction errors:
- The error in prediction for company A is $\frac{5}{12}$.
- The error in prediction for company A is $\frac{11}{12}$.
- The error for A is more than that for B.
- The error for B is more than that for A.
- Error cannot be compared using given data.
Solution:
- Company A: $f_1(x) = x^3$, $g_1(x) = \sqrt{x}$, shaded area in right graph in Figure M2W3G2.
- Company B: $f_2(x) = x$, $g_2(x) = e^x$, left graph in Figure M2W3G3.
- Error = area between the curves over $$.1
Calculate error for Company A:
$$ E_A = \int_0^1 |\sqrt{x} - x^3| dx = \int_0^1 (\sqrt{x} - x^3) dx = \left[ \frac{2}{3} x^{3/2} - \frac{1}{4} x^4 \right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{8-3}{12} = \frac{5}{12} $$
So, error for company A is $\frac{5}{12}$.
Company B:
$$ E_B = \int_0^1 |e^{x} - x| dx = \int_0^1 (e^{x} - x) dx = [e^x - \frac{1}{2} x^2]_0^1 = (e - \frac{1}{2}) - (1-0) = (e - \frac{1}{2} - 1) = (e - 1 - 0.5) $$
But the option “The error in prediction for Company A is $\frac{11}{12}$” is not correct.
Between Company A and B, the area for B (since $e^x$ grows rapidly) will be larger, so the error for Company B is more than A.
Correct Options:
- The error in prediction for company A is $\frac{5}{12}$.
- The error in prediction for company B is more than that for company A.
5. Cubic function local minimum (Image 2)
Question:
Let $f(x) = x^3 - 3x^2 + 37$. What is the local minimum value of $f$ attained at a critical point?
Solution:
Set $f’(x) = 3x^2 - 6x = 3x(x-2) = 0$ So, critical points at $x = 0$ and $x = 2$.
- $f’’(x) = 6x - 6$
- At $x = 0$: $f’’(0) = -6 < 0$ (local maximum)
- At $x = 2$: $f’’(2) = 12 - 6 = 6 > 0$ (local minimum)
Plug $x = 2$ into $f(x)$:
$$ f(2) = 8 - 12 + 37 = 33 $$
Local minimum value is $\boxed{33}$
Summary Table:
| Q. No. | Solution |
|---|---|
| 1 | i) → b) → 1), ii) → c) → 3), iii) → a) → 2) |
| 2 | Max volume: $h = \frac{R}{\sqrt{3}}$; Max curved surface area: $h = \frac{R}{\sqrt{2}}$ |
| 3 | Curves 2 and 4 enclose negative area |
| 4 | Absolute difference = 1 |
| 5 | Error for A = $\frac{5}{12}$; Error for B > A |
| 6 | Local minimum value is 33 |
Let me know if you want any particular solution in more detail!