Commerce (English Medium) Class 12 - CBSE Question Bank Solutions for Mathematics

Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x)

If the function f(x) = x² − kx + 5 is increasing on [2, 4], then

The function f(x) = −x/2 + sin x defined on [−π/3, π/3] is

Find the equation of the line passing through the point (1, 2, −4) and parallel to the line \[\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3} .\] 

If the function f(x) = x³ − 9kx² + 27x + 30 is increasing on R, then

Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line  \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]

The function f(x) = x⁹ + 3x⁷ + 64 is increasing on

Find the equation of the line passing through the point (2, −1, 3) and parallel to the line  \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]

Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines  \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{  and  } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]

Find the equation of the line passing through the point  \[\hat{i}  + \hat{j}  - 3 \hat{k} \] and perpendicular to the lines  \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + \hat{j}  - 3 \hat{k}  \right) \text { and }  \overrightarrow{r} = \left( 2 \hat{i}  + \hat{j}  - \hat{ k}  \right) + \mu\left( \hat{i}  + \hat{j}  + \hat{k}  \right) .\]

Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).

Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{    and    } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]

Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.

Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. 

If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{     and     } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\]  are perpendicular, find the value of λ.

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD. 

Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]

Find the direction cosines of the line 

\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]  Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.  

Show that the lines  \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{          and         } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection. 

Show that the lines \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{           and                } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\]  do not intersect.