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:
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:
Only statements I, II and III are correct. | |
Only statements I, II and IV are correct. | |
Only statements I and IV are correct. | |
Only statements II and IV are correct. |
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.
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
\frac{1}{1+x}+constant | |
\frac{1}{1+x^2}+constant | |
-\frac{1}{1-x}+constant | |
-\frac{1}{1-x^2}+constant |
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.
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\} ?
It is continuous on its domain. | |
It has a local minimum at x = 2. | |
It has a local maximum at x = 2. | |
It is differentiable on its domain. |
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}
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
(1,0,\sqrt{2}) | |
(1,0,-\sqrt{2}) | |
(-1,0,\sqrt{2}) | |
(-1,0,-\sqrt{2}) |
Question 4 Explanation:
Using potential matrix about x-axis
\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}
\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 _______
0.5 | |
1 | |
1.5 | |
2.2 |
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 ________
1.2 | |
3.1 | |
2.4 | |
0 |
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
\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} | |
\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} | |
\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}} | |
\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}} |
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
0 | |
1 | |
0.5 | |
\infty |
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) _______
85 | |
35 | |
15 | |
28 |
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 _______
\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)
The limit (correct up to one decimal place) is _______
0.25 | |
0.5 | |
0.75 | |
0.95 |
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
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.
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