Download App

Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity4 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
✓

Answer

Let $\text{x}-1=\text{y}$
$\Rightarrow\text{x}=\text{y}+1$
Thus,
$\lim_\limits{\text{x}\rightarrow 1}(\text{x}-1)\tan\frac{\pi\text{x}}{2}=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\frac{\pi(\text{y}+1)}{2}$
$=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\Big(\frac{\pi\text{y}}{2}+\frac{\pi}{2}\Big)$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\cot\frac{\pi\text{y}}{2}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\sin\frac{\pi\text{y}}{2}}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{2}{\pi}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\frac{2}{\pi}\lim_\limits{\text{y}\rightarrow0}\cos\frac{\pi\text{y}}{2}$
$=-\frac{2}{\pi}$
Since the function is continuous, LHL = RHL.
Thus, $\text{k}=-\frac{2}{\pi}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from Continuity→Continuity — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Express the matrix $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
→
If $\text{y}=\sin\Big[2\tan^{-1}\Big\{\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}\Big],$ find $\frac{\text{dy}}{\text{dx}}.$
→
A manufacturer makes two types A and B of tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:
  Machines
I II III
A 12 18 6
B 6 0 9
Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.
→
Find a unit vector perpendicular to the plane A B C, where the coordinates of A, B and C are A (3, -1, 2), B (1, -1, -3) and C (4, -3, 1).
→
Solve the following systems of linear equations by cramer's rule:
2x + 3y = 10,
x + 6y = 4
→
Solve the following system of homogeneous linear equations:
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
→
If $=\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0.$
→
If $\text{x}=\text{e}^{\cos2\text{t}}$ and $\text{y}=\text{e}^{\sin2\text{t}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
→
A cooperative society of farmers has 50 hectares of land to grow two crops X and Y. The profits from crops X and Y per hectare are estimated as Rs. 10,500 and Rs. 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society? 
→
Determine whether the following pair of lines intersect or not:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\text{z}$ and $\frac{\text{x}+1}{5}=\frac{\text{y}-2}{1};\text{z}=2$
→