If ' $a$ ' is a complex number such that $|a|=1$. Find the value of $a$, so that the equation $a z^2+z+1=0$ has one purely imaginary root.
$\cos \left\{\cos ^{-1}\left(\frac{-\sqrt{5}+1}{4}\right)\right\}$
$\cos \left\{\sin ^{-1}\left(\frac{\sqrt{5}+1}{4}\right)\right\}+i \sin \left\{\cos ^{-1}\left(\frac{\sqrt{5}+1}{4}\right)\right\}$
$\sin \left\{\cos ^{-1}\left(\frac{\sqrt{5}-1}{4}\right)\right\}+i \sin ^{-1}\left(\frac{-\sqrt{5}+1}{2}\right)$
None of the above
If $\mathbf{a , b , c}$ are vectors such that $|\mathbf{b}|=|\mathbf{c}|$ then $\{(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}+\mathbf{c})\} \times(\mathbf{b} \times \mathbf{c}) \cdot(\mathbf{b}+\mathbf{c})$ is equal to
1
4
2
0
For what value of $a, 6$ lies between the roots of the equation $x^2+2(a-3) x+9=0$.
$\left(\frac{-3}{4}, \infty\right)$
$\left(-\infty, \frac{-3}{4}\right) \cup(2, \infty)$
$\left(-\infty, \frac{-3}{4}\right)$
$\left(\frac{3}{4}, \infty\right)$
What is the probability of getting a sum of 9 in a single throw of three fair dice?
$\frac{6}{216}$
$\frac{36}{216}$
$\frac{9}{216}$
$\frac{25}{216}$
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