Download App

Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity2 Marks
Question
Identify discontinuities for the following functions as either a jump or a removable discontinuity :
$f(x)=x^2-3 x-2$, for $x<-3=3+8 x$, for $x>-3$.
✓

Answer

$f(x)=x^2-3 x-2, x<-3=3+8 x, x>-3$
$f(x)$ is a polynomial function for both the intervals.
$\therefore f ( x )$ is continuous for both the given intervals.
Let us test the continuity at $x =-3$.
$\lim _{x \rightarrow-3^{-}} f (x)=\lim _{x \rightarrow-3^{-}}\left(x^2-3 x-2\right)$
$=(-3)^2-3(-3)-2$
$=9+9-2$
$=16$
$\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3^{+}}(3+8 x)$
$=3+8(-3)$
$=-21$
$\therefore \quad \lim _{x \rightarrow-3^{-}} f (x) \neq \lim _{x \rightarrow-3^{+}} f (x)$
$\therefore \quad \lim _{x \rightarrow-3} f (x) \text { does not exist. }$
$\therefore f ( x )$ is discontinuous at $x =-3$.
$\therefore f ( x )$ has a jump discontinuity at $x =-3$

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 "Solve the Following Question.(2 Marks)" from Continuity→Continuity — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

Find n, if

$\frac{(15-n) !}{(13-n) !}=12$

→
Examine the collinearity of the following set of points:A (3, – 1), B (0, – 3), C (12, 5)
→
Find the equation of the line which passes through the point (-4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
→
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{(25)^x-2(5)^x+1}{x \cdot \sin x}\right]$
→
Find x in the following.
$\frac{1}{4!}+\frac{1}{5!}=\frac{\text{x}}{6!}$
→
Show thet the sum of all odd integers between $1$ and $1000$ wich are divisibleby $3$ is $83667$.
→
Solve the following equations.

$\left|\begin{array}{lll}x+2 & x+6 & x-1 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6\end{array}\right|=0$

Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get

$\left|\begin{array}{ccc}x+2 & x+6 & x-1 \\ 4 & -7 & 3 \\ -3 & -4 & 7\end{array}\right|=0$

∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0 ∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0 ∴ -37(x + 2+ x + 6 + x – 1) = 0 ∴ 3x + 7 = 0

$\therefore x=\frac{-7}{3}$

→
Solve the following system of equations in R.
x + 3 > 0, 2x < 14.
→
A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j for j = 1, 2,….., 6. What is the probability, in one roil of the die, that an odd number of dots will turn up?
→
Find the angle between hour-hand and minute-hand in a clock at : 10:10
→