Number of distinct solutions of $ x^{2} = y^{2} $ and $ (x - a)^{2} + y^{2} = 1 $ where $a$ is any real number:
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Lines $2x + 3y - 6 = 0$ and $9x + 6y - 18 = 0$ cut coordinate axes in concyclic points.
Center of circle is:
Topic: NIMCET 2011
$ \vec{a} = x\hat{i} - 3\hat{j} - \hat{k},\quad \vec{b} = 2x\hat{i} + x\hat{j} - \hat{k} $
Angle between $ \vec{a} $ an…
Topic: NIMCET 2011
If $ a,b,c $ are coplanar, evaluate
$ [,2a - b, 2b - c,2c - a,] $
Topic: NIMCET 2011
If area between $ y=x^{2} $ and $ y=x $ is $ A $, then area between $ y=x^{2} $ and $ y=1 $ is:
Topic: NIMCET 2011
$ \displaystyle \int_{0}^{1/2} \frac{dx}{\sqrt{x - x^{2}}} $
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$ \displaystyle \text{The value of } \frac{1 - \tan^{2} 15^\circ}{1 + \tan^{2} 15^\circ} \text{ is:} $
Topic: NIMCET 2011
The general solution of
$ \sqrt{3}\cos x + \sin x = 3 $
is:
Topic: NIMCET 2011
If $ \displaystyle \tan \theta = \frac{b}{a} $, then the value of
$ a\cos 2\theta + b\sin 2\theta $
is
Topic: NIMCET 2011
In $\triangle ABC$, $B = 45^\circ, C = 105^\circ, c=\sqrt{2}$.
Find side $a$ and $b$.
Topic: NIMCET 2011
The value of
$ \sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ $
Topic: NIMCET 2011
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Vertex of parabola $ y^{2} - 8y + 19 = 0 $
Topic: NIMCET 2011
Eccentricity of ellipse
$ 9x^{2} + 5y^{2} - 30y = 0 $
Topic: NIMCET 2011
$ \vec{v} = 2\hat{i} + \hat{j} - \hat{k},\quad \vec{w} = \hat{i} + 3\hat{k} $
If $ \vec{u} $ is a unit vector, maximum …
Topic: NIMCET 2011
If the function $f:[1,\infty)\to[1,\infty)$ is defined by
$f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is:
Topic: NIMCET 2011
A random variable $X$ has the probability distribution:
\[\begin{array}{c|ccccccccc}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 …
Topic: NIMCET 2011
$ \text{The sum of } 11^{2} + 12^{2} + \cdots + 30^{2} \text{ is} $
Topic: NIMCET 2011
$ \text{If } B = -A^{-1}BA,\ \text{then } (A+B)^{2} = $
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Roots of $x^{2} - 2x + 4 = 0$ are $\alpha, \beta$.
Compute $ \alpha^{6} + \beta^{6} $.
Topic: NIMCET 2011
If $ |\vec{a}\times \vec{b}| = |\vec{a}\cdot \vec{b}| $, then angle $\theta$ between $\vec{a},\vec{b}$ is:
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$ABCD$ is a parallelogram with diagonals $AC$ and $BD$.
Compute $ \overrightarrow{AC} - \overrightarrow{BD} $.
Topic: NIMCET 2011
