The maximum value of the function \(7 + 6 x - 9 x^{2}\) is:
1. \(8\)
2. \(-8\)
3. \(4\)
4. \(-4\)

If \(f \left(x\right) = x^{2} - 2 x + 4\), then \(f(x)\) has:

The resultant of the forces \(\overrightarrow {P}\) and \(\overrightarrow {Q}\) is \(\overrightarrow {R}\). If \(\overrightarrow {Q}\) is doubled, then the resultant also doubles in magnitude. Find the angle between \(\overrightarrow {P}\) and \(\overrightarrow {Q}\)?
1. \(\cos\theta = \frac{Q}{2P}\)
2. \(\cos\theta = \frac{-4Q}{3P}\)
3. \(\cos\theta = \frac{-2Q}{3P}\)
4. \(\cos\theta = \frac{-3P}{4Q}\)

If \(\overrightarrow{A}={2}\hat{i}+\hat{j}\;{\&}\;\overrightarrow{B}=\hat{i}{-}\hat{j},\)  then the components of \(\overrightarrow {A}\) along with \(\overrightarrow {B}\) & perpendicular to \(\overrightarrow {B}\) respectively will be:
1. \(\frac{\hat{i} - \hat{j}}{2} ,~\frac{3}{2}\left(\hat i+\hat j\right)\)
2. \(\frac{\hat{i} - \hat{j}}{2} ,~-\frac{2}{3}\left(\hat i+\hat j\right)\)
3. \(\frac{\hat{i} - \hat{j}}{2} ,~-\frac{3}{2}\left(\hat i-\hat j\right)\)
4. \(\frac{\hat{i} - \hat{j}}{2} ,~\frac{2}{3}\left(\hat i-\hat j\right)\)

If \(\overrightarrow{a}\) is a vector and \(x\) is a non-zero scalar, then which of the following is correct?

The vector \(\overrightarrow b\) which is collinear with the vector \(\overrightarrow a = \left(2, 1, -1\right)\) and satisfies the condition \(\overrightarrow a. \overrightarrow b=3\) is:
1. \(\left(1, \frac{1}{2}, \frac{-1}{2}\right)\)
2. \(\left(\frac{2}{3}, \frac{1}{3}, \frac{-1}{3}\right)\)
3. \(\left(\frac{1}{2}, \frac{1}{4}, \frac{-1}{4}\right)\)
4. \(\left(1, 1, 0\right)\)

If a, b and c are three non-zero vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$, then the value of $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}$ will be:

If $\stackrel{}{\mathrm{}}$\(\overrightarrow P +\overrightarrow Q\) = \(\overrightarrow P-\overrightarrow Q\) and \(\theta\) is the angle between \(\overrightarrow {P}\) and \(\overrightarrow {Q}\), then:
1. \(\theta = 0^{\circ}\)
2. \(\theta = 90^{\circ}\)
3. \(P=0\)
4. \(Q=0\)

What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }}<mspace\; width="1em"></mspace>=\overrightarrow{\mathrm{r}<mspace\; width="1em"></mspace>}\times \overrightarrow{\mathrm{F}}$)

1. $6\hat{\mathrm{i}}-6\hat{\mathrm{j}}+12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

2. $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

3. $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

4. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

The momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right)<mspace\; width="1em"></mspace>\mathrm{kg}<mspace\; width="1em"></mspace>\mathrm{m}/\mathrm{s}$. Force acting on the body at t=2 sec will be: \(\left(\text{Given:}~ F=\frac{dp}{dt}\right)\)
1. 6 N
2. 8 N
3. 4 N
4. 2 N