Let \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) be vectors of magnitude 2,3 and 4 respectively. If \(\bar{a}\) is perpendicular to \((\overline{\mathrm{b}}+\overline{\mathrm{c}}), \overline{\mathrm{b}}\) is perpendicular to ( \(\overline{\mathrm{c}}+\overline{\mathrm{a}}\) ) and \(\overline{\mathrm{c}}\) is perpendicular to \((\bar{a}+\bar{b})\), then the magnitude of \(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}\) is
If
\[ \sqrt{y-\sqrt{y-\sqrt{y-\ldots \ldots \ldots \infty}}}=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots \infty}}} \]
then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)
\(\int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} d x\) equals
With usual notations in \(\triangle \mathrm{ABC}\), if \(\angle \mathrm{B}=\frac{\pi}{2}\), and \(\tan \frac{\mathrm{A}}{2}, \tan \frac{\mathrm{C}}{2}\) are roots of equation \(\mathrm{p} x^2+\mathrm{qx}+\mathrm{r}=0\), \(\mathrm{p} \neq 0\), then
\(\int \frac{\mathrm{d} x}{x\left(x^3+1\right)}=\)
\[ \int \frac{\mathrm{d} x}{(x+\mathrm{a})^{\frac{9}{7}}(x-\mathrm{b})^{\frac{5}{7}}}= \]
If \(y=\tan ^{-1}\left(\frac{12 x-64 x^3}{1-48 x^2}\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)
If \(\mathrm{A}=\left[\begin{array}{cc}5 \mathrm{a} & -\mathrm{b} \\ 3 & 2\end{array}\right]\) and A .adj \(\mathrm{A}=\mathrm{AA}^{\mathrm{T}}\), then \(5 \mathrm{a}+\mathrm{b}=\)
Consider the probability distribution
\[ \begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{X}=x) & \mathrm{K} & 2 \mathrm{~K} & \mathrm{~K}^2 & 2 \mathrm{~K} & 5 \mathrm{~K}^2 \\ \hline \end{array} \]
Then the value of \(\mathrm{P}(\mathrm{X}>2)\) is
If \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) are unit vectors and \(|\overline{\mathrm{a}}|=7\), \(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}} \times(\overline{\mathrm{c}} \times \overline{\mathrm{a}})=\frac{1}{2} \overline{\mathrm{a}}\), then angle between the vectors \(\bar{a}\) and \(\bar{c}\) and angle between the vectors \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) are respectively
A doctor assumes that patient has one of three diseases \(\mathrm{d} 1, \mathrm{~d} 2\) or d 3 . Before any test he assumes an equal probability for each disease. He carries out a test that will be positive with probability 0.7 if the patient has disease \(\mathrm{d} 1,0.5\) if the patient has disease d 2 and 0.8 if the patient has disease d3. Given that the outcome of the test was positive then probability that patient has disease d2 is
If a circle with centre at \((-1,1)\) touches the line \(x+2 y+4=0\) then the co-ordinates of the point of contact are
The probability distribution of a random variable X is given by\[ \begin{array}{|l|c|c|c|c|c|} \hline \mathrm{X}=x_i & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}\left(\mathrm{X}=x_i\right) & 0.4 & 0.3 & 0.1 & 0.1 & 0.1 \\ \hline \end{array} \]
Then the variance of X is
\[ \int \sqrt{x^2+3 x} d x= \]
\[ \int \frac{x}{1+x^4} d x= \]
The foci of the conic \(25 x^2+16 y^2-150 x=175\) are
Let \(\mathrm{A} \equiv(0,0), \mathrm{B}(3,0), \mathrm{C}(0,-4)\) are vertices of \(\triangle A B C\), then the co-ordinates of incentre of \(\triangle \mathrm{ABC}\) is
\[ \int \frac{x^3}{x^4+5 x^2+4} d x= \]
The differential equation satisfied by \(y = X\sin(6t+5) + Y\cos(6t+5)\) (where \(X\) and \(Y\) are constants) is
Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is
If the point \((1, \alpha, \beta)\) lies on the line of the shortest distance between the lines \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}\) and \(\frac{x+2}{-1}=\frac{y+6}{2}, \mathrm{z}=1\), then \(\alpha+\beta=\)
\[ \int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x d x= \]
\[ \mathop {\lim }\limits_{x \to 1}\left(\log _3 3 x\right)^{\log _x 8}=\ldots \]
If \(x=\operatorname{acos}^3 \theta y=\operatorname{asin}^3 \theta\)
Then \(\sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}=\)
\[ \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} d x= \]