Arts (English Medium) कक्षा ११ - CBSE Question Bank Solutions for Mathematics

The mid-points of the sides of a triangle ABC are given by (–2, 3, 5), (4, –1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

A(1, 2, 3), B(0, 4, 1), C(–1, –1, –3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

Find the coordinates of the points which tisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6). 

Given that  P(3, 2, –4), Q(5, 4, –6) and R(9, 8, –10) are collinear. Find the ratio in which Qdivides PR. 

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, –8) is divided by the yz-plane. 

Find the coordinates of a point equidistant from the origin and points A (a, 0, 0), B (0, b, 0) andC(0, 0, c). 

Write the coordinates of the point P which is five-sixth of the way from A(−2, 0, 6) to B(10, −6, −12). 

If a parallelopiped is formed by the planes drawn through the points (2,3,5) and (5, 9, 7) parallel to the coordinate planes, then write the lengths of edges of the parallelopiped and length of the diagonal. 

Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3,2) and C(0, 0,1).

If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(−2, b −5) and C (4, 7, c), find the values of a, b, c.

Find k so that \[\lim_{x \to 2} f\left( x \right)\] \[f\left( x \right) = \begin{cases}2x + 3, & x \leq 2 \\ x + k, & x > 2\end{cases} .\] 

Show that \[\lim_{x \to 0} \frac{1}{x}\] does not exist. 

Let f(x) be a function defined by \[f\left( x \right) = \begin{cases}\frac{3x}{\left| x \right| + 2x}, & x \neq 0 \\ 0, & x = 0\end{cases} .\] Show that \[\lim_{x \to 0} f\left( x \right)\] does not exist.

Let \[f\left( x \right) = \left\{ \begin{array}{l}x + 1, & if x \geq 0 \\ x - 1, & if x < 0\end{array} . \right.\]Prove that \[\lim_{x \to 0} f\left( x \right)\] does not exist.

Let \[f\left( x \right) = \begin{cases}x + 5, & if x > 0 \\ x - 4, & if x < 0\end{cases}\] \[\lim_{x \to 0} f\left( x \right)\]  does not exist. 

Find \[\lim_{x \to 3} f\left( x \right)\] where \[f\left( x \right) = \begin{cases}4, & if x > 3 \\ x + 1, & if x < 3\end{cases}\] 

If \[f\left( x \right) = \left\{ \begin{array}{l}2x + 3, & x \leq 0 \\ 3 \left( x + 1 \right), & x > 0\end{array} . \right.\] find \[\lim_{x \to 0} f\left( x \right)\] 

If \[f\left(  x \right) = \left\{ \begin{array}{l}2x + 3, & x \leq 0 \\ 3 \left( x + 1 \right), & x > 0\end{array} . \right.\] find \[\lim_{x \to 1} f\left( x \right)\]

Find \[\lim_{x \to 1} f\left( x \right)\] if \[f\left( x \right) = \begin{cases}x^2 - 1, & x \leq 1 \\ - x^2 - 1, & x > 1\end{cases}\]