# Calculus

 Question 1
Consider the polynomial $f(x)=x^3-6x^2+11x-6$ on the domain S given by $1 \leq x \leq 3$. The first and second derivatives are $f'(x)$ and $f''(x)$.

Consider the following statements:
I. The given polynomial is zero at the boundary points x = 1 and x = 3.
II. There exists one local maxima of $f(x)$ within the domain S.
III. The second derivative $f''(x) \gt 0$ throughout the domain S.
IV. There exists one local minima of $f(x)$ within the domain S.

The correct option is:
 A Only statements I, II and III are correct. B Only statements I, II and IV are correct. C Only statements I and IV are correct. D Only statements II and IV are correct.
GATE CE 2022 SET-2   Engineering Mathematics
Question 1 Explanation:
\begin{aligned} f(x)&=x^3-6x^2+11x-6 \\ f(1)&=1-6+11-6=0 \\ f(3)&=3^3-6 \times 3^2+11 \times 3-6=0 \\ &\text{Statement (I) is correct.} \\ f'(x)&=3x^2-12x+11 \\ f'(x)&=0\Rightarrow x=2\pm \frac{1}{\sqrt{3}} \end{aligned}

$f(x)$ has local maxima at $x=2-\frac{1}{\sqrt{3}}$
Statement (II) is also true.
Now,
\begin{aligned} f''(x)&=6x-12\\ f''(x) &\gt 0\\ 6x-12 &\gt 0\\ x & \gt 0 \end{aligned}
Statement (III) is incorrect statement (IV) is also correct.
$\because x=2+\frac{1}{\sqrt{3}}$ is point of minima.
 Question 2
$\int \left ( x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \right )dx$ is equil to
 A $\frac{1}{1+x}+constant$ B $\frac{1}{1+x^2}+constant$ C $-\frac{1}{1-x}+constant$ D $-\frac{1}{1-x^2}+constant$
GATE CE 2022 SET-2   Engineering Mathematics
Question 2 Explanation:
MTA- Marks to All
$I=\int \left ( x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\infty \right )dx$
$I=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+...$
Option (A)
$\frac{1}{1+x}=(1+x)^{-1}=1-x+x^2-x^3...\infty$
So, its incorrect.
Option (B)
$\frac{1}{1+x^2}=(1+x^2)^{-1}=1-x^2+x^4-x^6...\infty$
So, its incorrect.
Similarly option (C) and (D) both are incorrect.
No-correct choice given.
 Question 3
Let $max \{a, b\}$ denote the maximum of two real numbers a and b. Which of the following statement(s) is/are TRUE about the function $f(x) = max\{3 -x, x - 1\}$ ?
 A It is continuous on its domain. B It has a local minimum at x = 2. C It has a local maximum at x = 2. D It is differentiable on its domain.
GATE CE 2022 SET-1   Engineering Mathematics
Question 3 Explanation:
$f(x)=max{3-x,x-1}$

both intersecting at
\begin{aligned} 3-x&=x-1 \\ 2x&= 4\\ x&=2 \\ y &= max\{3-x,x-1\} \end{aligned}

 Question 4
The Cartesian coordinates of a point P in a right-handed coordinate system are (1, 1, 1). The transformed coordinates of P due to a $45 ^{\circ}$ clockwise rotation of the coordinate system about the positive x-axis are
 A $(1,0,\sqrt{2})$ B $(1,0,-\sqrt{2})$ C $(-1,0,\sqrt{2})$ D $(-1,0,-\sqrt{2})$
GATE CE 2022 SET-1   Engineering Mathematics
Question 4 Explanation:
$\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & \cos \theta & \sin \theta \\ 0& \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 0& \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}=\begin{bmatrix} 1\\ 0\\ \sqrt{2} \end{bmatrix}$
 Question 5
A function is defined in Cartesian coordinate system as $f(x, y)=x e^{y}$. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point $\left(\frac{1}{2}, 2\right)$ is _______
 A 0.5 B 1 C 1.5 D 2.2
GATE CE 2021 SET-2   Engineering Mathematics
Question 5 Explanation:
\begin{aligned} f(x, y)&=x e^{y}\\ \mathrm{P}(2,0) \text{ and } Q\left(\frac{1}{2}, 2\right)\\ \operatorname{grad} f &=\hat{i}\left(e^{y}\right)+\hat{j}\left(x e^{y}\right)+\hat{k}(0) \\ \Rightarrow \qquad \qquad \qquad (\operatorname{grad} f)_{p} &=\hat{i}+2 \hat{j} ]\\ \overline{P Q}&=\left(\frac{1}{2}-2\right) \hat{i}+(2-0) \hat{j}=-\frac{3}{2} \hat{i}+2 \hat{j}\\ \text{Required directional derivative} &=(\text { grad } f)_{P} \widehat{P Q}\\ &=(\hat{i}+2 \hat{j}) \times \frac{\left(-\frac{3}{2} \hat{i}+2 \hat{j}\right)}{\sqrt{\frac{9}{4}+4}}=\frac{\frac{-3}{2}+4}{\sqrt{\frac{25}{4}}} \\ &=\frac{\frac{-3+8}{2}}{\left(\frac{5}{2}\right)}=1 \end{aligned}
 Question 6
The value (round off to one decimal place) of $\int_{-1}^{1} x e^{|x|} d x$ is ________
 A 1.2 B 3.1 C 2.4 D 0
GATE CE 2021 SET-2   Engineering Mathematics
Question 6 Explanation:
\begin{aligned} \int_{-1}^{1} x e^{-|x|} d x&=0\\ \text { As } \qquad \qquad \qquad f(-x)&=x e^{x} \text { is an odd function. } \end{aligned}
 Question 7
The unit normal vector to the surface $X^{2}+Y^{2}+Z^{2}-48=0$ at the point (4,4,4) is
 A $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$ B $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ C $\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}$ D $\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}$
GATE CE 2021 SET-2   Engineering Mathematics
Question 7 Explanation:
\begin{aligned} \phi &=x^{2}+y^{2}+z^{2}-48, P(4,4,4) \\ \operatorname{grad} \phi &=\vec{\nabla} \phi=\hat{i} \frac{\partial \phi}{\partial x}+\hat{j} \frac{\partial \phi}{\partial y}+\hat{k} \frac{\partial \phi}{\partial z} \\ &=(2 x) \hat{i}+(2 y) \hat{j}+(2 z) \hat{k} \\ \vec{n} &=(\operatorname{grad} \phi)_{P}=8 \hat{i}+8 \hat{j}+8 \hat{k} \\ \hat{n} &=\frac{\vec{n}}{|\vec{n}|}=\frac{8 \hat{i}+8 \hat{j}+8 \hat{k}}{\sqrt{64+64+64}}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \\ & \simeq\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\right) \end{aligned}
 Question 8
The value of $\lim _{x \rightarrow \infty} \frac{x \ln (x)}{1+x^{2}}$ is
 A 0 B 1 C 0.5 D $\infty$
GATE CE 2021 SET-2   Engineering Mathematics
Question 8 Explanation:
\begin{aligned} &\lim _{x \rightarrow \infty}\left(\frac{x \ln x}{x^{2}+1}\right) \qquad \qquad \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ &=\lim _{x \rightarrow \infty}\left(\frac{x\left(\frac{1}{x}\right)+\ln x}{2 x}\right) \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ \lim _{x \rightarrow \infty}\left(\frac{0+\frac{1}{x}}{2}\right)&=\lim _{x \rightarrow \infty}\left(\frac{1}{2 x}\right)=\frac{1}{2 \times \infty}=0 \end{aligned}
 Question 9
The volume determined from $\iiint_{V} 8 x y z d V \text { for } V=[2,3] \times[1,2] \times[0,1]$ will be (in integer) _______
 A 85 B 35 C 15 D 28
GATE CE 2021 SET-1   Engineering Mathematics
Question 9 Explanation:
\begin{array}{l} \begin{aligned} \text{Given,}\qquad \qquad \qquad V&=[2,3,] \times[1,2] \times[0,1] \text { i.e. } \\ 2& \lt x \lt 3,1 \lt y \lt 2,0 \lt z \lt 1 \\ I&=\iiint_{V} 8 x y z d V=8 \iiint_{V} x y z d x d y d z\\ So,\qquad \qquad \qquad \qquad &=8 \int_{x=2}^{3} x d x \times \int_{y=1}^{2} y d y \times \int_{z=0}^{1} z d z \\ &=8\left[\frac{x^{2}}{2}\right]_{2}^{3}\left[\frac{y^{2}}{2}\right]_{1}^{2}\left[\frac{z^{2}}{2}\right]_{0}^{1} \\ &=(9-4)(4-1)(1-0) \\ &=5 \times 3 \times 1 \\ &=15 \end{aligned} \end{array}
 Question 10
Consider the limit:
$\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$
The limit (correct up to one decimal place) is _______
 A 0.25 B 0.5 C 0.75 D 0.95
GATE CE 2021 SET-1   Engineering Mathematics
Question 10 Explanation:
$\lim _{x \rightarrow 1}\left[\frac{1}{\ln x}-\frac{1}{x-1}\right]=\lim _{x \rightarrow 1}\left[\frac{(x-1)-\ln x}{\ln x(x-1)}\right]\left(\frac{0}{0} \text { form }\right)$
So using L-Hospital's rule twices
$=\lim _{x \rightarrow 1}\left[\frac{1-\frac{1}{x}}{\log x+(x-1)\left(\frac{1}{x}\right)}\right]=\lim _{x \rightarrow 1}\left(\frac{1}{2+\log x}\right)=0.5$
There are 10 questions to complete.

### 2 thoughts on “Calculus”

1. In que 8,The sum of XY +XY =2Xy but it is taken as xy #doubt