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If y is the solution of the differential equation<br>$$\eqalign{ & {{\text{y}}^3}\frac{{{\text{dy}}}}{{{\text{dx}}}} + {{\text{x}}^3} = 0, \cr & {\text{y}}\left( 0 \right) = 1 \cr} $$<br>the value of y(-1) is
A
-2
B
-1
C
0
D
1
Correct Answer:
0
Consider the 5 × 5 matrix \[{\text{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&1&2&3&4 \\ 4&5&1&2&3 \\ 3&4&5&1&2 \\ 2&3&4&5&1 \end{array}} \right
A
\
B
<br>It is given that A has only one real eigen value.<br>Then the real eigen value of A is
C
<p><span>A.</span> -2.5
D
</span> 0
Eigen values of the matrix \[\left[ {\begin{array}{*{20}{c}} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&{ - 2i} \\ 0&0&{2i}&0 \end{array}} \right
A
\
B
are
C
<p><span>A.</span> -2, -1, 1, 2
D
</span> -1, 1, 0, 2
Let A be an m × n matrix and Ban n × m matrix. It is given that determinant ($$I$$
m
+ AB) = determinant ($$I$$
n
+ BA), where $$I$$
k
is the k × k identity matrix. Using the above property, the determinant of the matrix given below is
\[\left[ {\begin{array}{*{20}{c}} 2&1&1&1 \\ 1&2&1&1 \\ 1&1&2&1 \\ 1&1&1&2 \end{array}} \right
A
\
B
<p><span>A.</span> 2
C
</span> 5
Let the function
\[{\text{f}}\left( \theta \right) = \left| {\begin{array}{*{20}{c}} {\sin \theta }&{\cos \theta }&{\tan \theta } \\ {\sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}&{\tan \left( {\frac{\pi }{6}} \right)} \\ {\sin \left( {\frac{\pi }{3}} \right)}&{\cos \left( {\frac{\pi }{3}} \right)}&{\tan \left( {\frac{\pi }{3}} \right)} \end{array}} \right|\
A
<br>where \\
B
and \ denote the derivative of f with respect to \. Which of the following statements is/are TRUE?<br>I. There exists \ such that \<br>II. There exists \ such that\
C
<p><span>A.</span> l only
D
</span> ll only
If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?
A
0
B
5
C
1
D
4
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?
A
0
B
1
C
4
D
1 + abcd
Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$ with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$ and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$ is
A
-2
B
-1
C
0
D
1
The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$ $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$ = ?
A
0
B
3
C
$$\frac{1}{3}$$
D
2
The figure shows the plot of y as a function of x
The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is
A
$$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 1$$
B
$$\frac{{{\text{dy}}}}{{{\text{dx}}}} = {\text{x}}$$
C
$$\frac{{{\text{dy}}}}{{{\text{dx}}}} = - {\text{x}}$$
D
$$\frac{{{\text{dy}}}}{{{\text{dx}}}} = \left| {\text{x}} \right|$$
The value of x for which all the eigen-values of the matrix given below are real is \[\left[ {\begin{array}{*{20}{c}} {10}&{5 + {\text{j}}}&4 \\ {\text{x}}&{20}&2 \\ 4&2&{ - 10} \end{array}} \right
A
\
B
<p><span>A.</span> 5 + j
C
</span> 5 - j