Question 1 |
The maximum value of the function f(x)=ln(1+x)-x(\; where \; x>-1) occurs
at x = ____.
0 | |
1 | |
0.5 | |
-1 |
Question 1 Explanation:
\begin{aligned} f^{\prime}(x) &=\frac{1}{1+x}-1=0 \\ \frac{1-1-x}{1+x} &=0 \\ \frac{x}{1+x} &=0 \\ x &=0\\ f^{\prime \prime}(x)&=\frac{-1}{(1+x)^{2}} \\ f^{\prime}(0)&=-1 \lt 0 \end{aligned}
f(x) have maximum value at x=0
\begin{aligned} f(0)&=\ln (1+0)-0=0 \\ t_{\max }&=0 \end{aligned}
f(x) have maximum value at x=0
\begin{aligned} f(0)&=\ln (1+0)-0=0 \\ t_{\max }&=0 \end{aligned}
Question 2 |
Which ONE of the following is a linear non-homogeneous differential equation,
where x and y are the independent and dependent variables respectively ?
\frac{dy}{dx}+xy=e^{-x} | |
\frac{dy}{dx}+xy=0 | |
\frac{dy}{dx}+xy=e^{-y} | |
\frac{dy}{dx}+e^{-y}=0 |
Question 2 Explanation:
General form of linear differential equation
\frac{d y}{d x}+p y=\theta when P and \theta can be function of x
Only option (A) is in this form.
\frac{d y}{d x}+p y=\theta when P and \theta can be function of x
Only option (A) is in this form.
Question 3 |
Match the application to appropriate numerical method.
P1-M3, P2-M2, P3-M4, P4-M1 | |
P1-M3, P2-M1, P3-M4, P4-M2 | |
P1-M4, P2-M1, P3-M3, P4-M2 | |
P1-M2, P2-M1, P3-M3, P4-M4 |
Question 4 |
An unbiased coin is tossed an infinite number of times. The probability that the
fourth head appears at the tenth toss is
0.067 | |
0.073 | |
0.082 | |
0.091 |
Question 4 Explanation:
It means 3-head appears in 1^{\text {st }} 9 trials.
Probability of getting exactly 3 head in 1^{\text {st }} 9 trials
\begin{aligned} &=\text{ Coefficient } p^{3}\text{ in }(4+p) 9 \\ \text{When, }&[\overline{p+q=1}] \\ &={ }^{9} C_{3} q^{6} p^{3}\\ \text{when, }\quad \rho&= \text{ probability of occure of head}\\ q &=\text { probability of occure of tail } \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{3} \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{9} \end{aligned}
and in 10^{\text {th }} trial head must appears.
So required probability
\begin{aligned} &={ }^{9} C_{3}\left(\frac{1}{2}\right)^{9} \times \frac{1}{2} \\ &=\frac{9 \times 8 \times 7}{3 !} \times\left(\frac{1}{2}\right)^{10}=\frac{84}{1024}=0.082 \end{aligned}
Probability of getting exactly 3 head in 1^{\text {st }} 9 trials
\begin{aligned} &=\text{ Coefficient } p^{3}\text{ in }(4+p) 9 \\ \text{When, }&[\overline{p+q=1}] \\ &={ }^{9} C_{3} q^{6} p^{3}\\ \text{when, }\quad \rho&= \text{ probability of occure of head}\\ q &=\text { probability of occure of tail } \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{3} \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{9} \end{aligned}
and in 10^{\text {th }} trial head must appears.
So required probability
\begin{aligned} &={ }^{9} C_{3}\left(\frac{1}{2}\right)^{9} \times \frac{1}{2} \\ &=\frac{9 \times 8 \times 7}{3 !} \times\left(\frac{1}{2}\right)^{10}=\frac{84}{1024}=0.082 \end{aligned}
Question 5 |
If z = xy ln(xy), then
x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0 | |
y\frac{\partial z}{\partial x}=x\frac{\partial z}{\partial y} | |
x\frac{\partial z}{\partial x}=y\frac{\partial z}{\partial y} | |
y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0 |
Question 5 Explanation:
\begin{aligned} \frac{\partial z}{\partial x}&=yln(x y)+\frac{x y}{x y} y\\ \frac{\partial z}{\partial x}&=y[\ln (x y)+1] &\ldots(i)\\ \frac{\partial z}{\partial y}&=x \ln (x y)+\frac{x y}{x y} \times x\\ \frac{\partial z}{\partial y}&=x[\ln (x y)+1]\\ \text{Here}\quad x\frac{\partial z}{\partial x}&=y \frac{\partial z}{\partial y} \end{aligned}
There are 5 questions to complete.