1
GATE CE 2017 Set 2
MCQ (Single Correct Answer)
+2
-0.6
If $$A = \left[ {\matrix{ 1 & 5 \cr 6 & 2 \cr } } \right]\,\,and\,\,B = \left[ {\matrix{ 3 & 7 \cr 8 & 4 \cr } } \right]A{B^T}$$ is equal to
A
$$\left[ {\matrix{ {38} & {28} \cr {32} & {56} \cr } } \right]$$
B
$$\left[ {\matrix{ 3 & {40} \cr {42} & 8 \cr } } \right]$$
C
$$\left[ {\matrix{ {43} & {27} \cr {34} & {50} \cr } } \right]$$
D
$$\left[ {\matrix{ {38} & {32} \cr {28} & {56} \cr } } \right]$$
2
GATE CE 2017 Set 2
MCQ (Single Correct Answer)
+1
-0.3
Let $$\,\,W = f\left( {x,y} \right),\,\,$$ where $$x$$ and $$y$$ are functions of $$t.$$ Then, according to the chain rule, $${{dw} \over {dt}}$$ is equal to
A
$${{dw} \over {dx}}{{dx} \over {dt}} + {{dw} \over {dy}}{{dt} \over {dt}}$$
B
$${{\partial w} \over {\partial x}}{{\partial x} \over {\partial t}} + {{\partial w} \over {\partial y}}{{\partial y} \over {\partial t}}$$
C
$${{\partial w} \over {\partial x}}{{dx} \over {dt}} + {{\partial w} \over {\partial y}}{{dy} \over {dt}}$$
D
$${{dw} \over {dx}}{{\partial x} \over {\partial t}} + {{dw} \over {dy}}{{\partial y} \over {\partial t}}$$
3
GATE CE 2017 Set 2
MCQ (Single Correct Answer)
+2
-0.6
The tangent to the curve represented by $$y=x$$ $$ln$$ $$x$$ is required to have $${45^ \circ }$$ inclination with the $$x-$$axis. The coordinates of the tangent point would be
A
$$(1, 0)$$
B
$$(0,1)$$
C
$$(1,1)$$
D
$$\left( {\sqrt {2,} \sqrt 2 } \right)$$
4
GATE CE 2017 Set 2
MCQ (Single Correct Answer)
+2
-0.6
Consider the following definite integral $$${\rm I} = \int\limits_0^1 {{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} \over {\sqrt {1 - {x^2}} }}dx} $$$
The value of the integral is
A
$${{{\pi ^3}} \over {24}}$$
B
$${{{\pi ^3}} \over {12}}$$
C
$${{{\pi ^3}} \over {48}}$$
D
$${{{\pi ^3}} \over {64}}$$
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