If | a b |=4, | a . b |=2, then | a |^2 | b |^2= 1. 6 2. 2 3. 20 4. 8

If | a b |=4, | a . b |=2, then | a |^2 | b |^2= 1. 6 2. 2 3. 20 …

The function f(x) = e|x| is:

Continuous everywhere but not differentiable at x = 0
Continuous and differentiable everywhere
Not continuous at x = 0
None of these.

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The order and the degree of the differential equation $\left(1+3 \frac{d y}{d x}\right)^2=4 \frac{d^3 y}{d x^3}$ respectively are

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If a vector makes an angle of $\frac{\pi}{4}$ with the positive directions of both x-axis and y-axis, then the angle which it makes with positive z-axis is:

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lf a line makes angles $\frac{\pi}{12},\frac{5\pi}{12}$ with oy, oz respectively where 0 = (0, 0, 0), then the angle made by that line with ox is:

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Direction cosines of ray from P(1, −2, 4) to Q(−1, 1, −2) are:

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If $f(x)=\left\{\begin{array}{cl}\frac{1}{1+e^{1 x}} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$ then $f ( x )$ is

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Solve: $\int\frac{\text{x}^2+1}{\text{x}^2+1}\text{dx}=$

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$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$ is equal to

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The vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ satisfy the equation $2\vec{\text{a}}+\vec{\text{b}}=\vec{\text{p}}$ and $\vec{\text{a}}+2\vec{\text{b}}=\vec{\text{q}},$ where $\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{q}}=\hat{\text{i}}-\hat{\text{j}}.$ If $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}},$ then:

$\cos \theta = \frac{4}{5}$
$\sin \theta = \frac{1}{\sqrt{2}}$
$\cos \theta = -\frac{4}{5}$
$\cos \theta = -\frac{3}{5}$

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The least value of the function $f(x)=2 \cos x+x$ in the closed interval $\left[0, \frac{\pi}{2}\right]$ is

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