1
GATE ME 2024
MCQ (Single Correct Answer)
+1
-0.33

Let $f(.)$ be a twice differentiable function from $ \mathbb{R}^{2} \rightarrow \mathbb{R}$. If $P, \mathbf{x}_{0} \in \mathbb{R}^{2}$ where $\vert \vert P\vert \vert$ is sufficiently small (here $\vert \vert . \vert \vert$ is the Euclidean norm or distance function), then $f (\mathbf{x}_{0} + p) = f(\mathbf{x}_{0}) + \nabla f(\mathbf{x}_{0})^{T}p + \dfrac{1}{2} p^{T} \nabla^{2}f(\psi)p$ where $\psi \in \mathbb{R}^{2}$ is a point on the line segment joining $\mathbf{x}_{0}$ and $\mathbf{x}_{0} + p$. If $\mathbf{x}_{0}$ is a strict local minimum of $f (\mathbf{x})$, then which one of the following statements is TRUE?

A

$\nabla f(x_{0})^{T}p > 0\ \ and\ \ p^{T} \nabla^{2} f( \psi)p = 0$

B

$\nabla f(x_{0})^{T}p = 0\ and\ p^{T} \nabla^{2} f( \psi)p > 0$

C

$\nabla f(x_{0})^{T}p = 0\ and\ p^{T} \nabla^{2} f( \psi)p = 0$

D

$\nabla f(x_{0})^{T}p = 0\ and\ p^{T} \nabla^{2} f( \psi)p < 0$

2
GATE ME 2023
MCQ (Single Correct Answer)
+1
-0.33

The figure shows the plot of a function over the interval [-4, 4]. Which one of the options given CORRECTLY identifies the function?

GATE ME 2023 Engineering Mathematics - Calculus Question 1 English
A
|2 − 𝑥|
B
|2 − |𝑥||
C
|2 + |𝑥||
D
2 − |𝑥|
3
GATE ME 2022 Set 2
MCQ (Single Correct Answer)
+1
-0.33

Given $\int^{\infty}_{-\infty}e^{-x^2}dx=\sqrt{\pi}$

If a and b are positive integers, the value of $\int^{\infty}_{-\infty}e^{-a(x+b)^2}dx$ is _________.

A
$\sqrt{\pi a}$
B
$\sqrt{\frac{\pi}{a}} $
C
$b\sqrt{\pi a}$
D
$b\sqrt{\frac{\pi}{a}}$
4
GATE ME 2022 Set 2
MCQ (Single Correct Answer)
+1
-0.33
A polynomial ψ(s) = ansn + an-1sn-1 + ......+ a1s + a0 of degree n > 3 with constant real coefficients an, an-1, ... a0 has triple roots at s = -σ. Which one of the following conditions must be satisfied?
A
ψ(s) = 0 at all the three values of s satisfying s3 + σ3 = 0
B
ψ(s) = 0, $\frac{d\psi(s)}{ds}=0$ and $\frac{d^2\psi(s)}{ds^2}=0$ at s = -σ
C
ψ(s) = 0, $\frac{d^2\psi(s)}{ds^2}=0$ and $\frac{d^4\psi(s)}{ds^4}=0$ at s = -σ
D
ψ(s) = 0, $\frac{d^3\psi(s)}{ds^3}=0$  at s = -σ
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