Download App

Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
The function f(x) is befined as follows: $\text{f(x)}=\begin{cases}\text{x}^2+\text{ax}+\text{b},&0\leq\text{x}<2\\3\text{x}+2,&2\leq\text{x}\leq4\\2\text{ax}+5\text{b},&4<\text{x}\leq8\end{cases}$ if is continuous on [0, 8] find the value of a and b.
✓

Answer

It is given that the f(x) is continuous on [0, 8]
f(x) is continuous at x = 2 and x = 4
Now, At x = 2
LHL = RHL = f(2) ....(A)
f(2) = 3 × 2 + 2 = 8 ....(i)
$\text{LHL}=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2-\text{h}^2)+\text{a}(2-\text{h})+\text{b}=4+2\text{a}+\text{b}$
From (A)
4 + 2a + b = 8
2a + b = 4 ....(B)
Now, At x = 4
LHL = RHL = f(4) ....(C)
f(4) = 3 × 4 + 2 = 14 ....(ii)
$\text{RHL}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(4+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}2\text{a}(4+\text{h})+5\text{b}=8\text{a}+5\text{b}$
From (c) we get
8a + 5b = 14 ....(D)
Solving (B) and (D) we get
a = 3 and b = -2

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 "5 Marks Questions" 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

The mean and variance of the binomial distribution are 4 and $\frac{4}{3}$ respectively. Find the distribution and P(X>1).
→
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
→
Solve the following system of equations by matrix method:$\frac{2}{\text{x}}-\frac{3}{\text{y}}+\frac{3}{\text{z}}=10$
$\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=10$
$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\frac{2}{\text{z}}=13$
→
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\text{ at }(\text{a}\cos\theta,\text{b}\sin\theta)$
→
$\overrightarrow{A B}=3 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{C D}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6 \hat{i}+7 \hat{j}+4 \hat{k}$ and $-9 \hat{j}+2 \hat{k}$, respectively. Find the position vector of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{P Q}$ is perpendicular to $\overrightarrow{A B}$ and $\overrightarrow{C D}$ both.
→
Solve the following differential equation:
$\frac{\text{y}}{\text{x}}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}-\Big\{\frac{\text{x}}{\text{y}}\sin\Big(\frac{\text{y}}{\text{x}}\Big)+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
→
Consider the binary operations *: R × R → R and o: R × R → R defined as $\text{a}∗\text{b} = |\text{a } – \text{ b}|\text{ and}\text{ a} o \text{b} = \text{a},\forall\text{a},\text{b}\in\text{R}.$ Show that * is commutative but not associative, o is associative but not commutative. Further, show that $\forall\text{a},\text{b},\text{c}\in\text{R},\text{a} *(\text{b}o\text{c}) = (\text{a} *\text{b})o(\text{a}*\text{b}).$ [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
→
Let $\vec{\text{u}},\vec{\text{v}}$ and $\vec{\text{w}}$ be vectors such $\vec{\text{u}}+\vec{\text{v}}+\vec{\text{w}}=\vec{0}.$ If $|\vec{\text{u}}|=3,|\vec{\text{v}}|=4$ and $|\vec{\text{w}}|=5,$ then find $\vec{\text{u}}.\vec{\text{v}}+\vec{\text{v}}.\vec{\text{w}}+\vec{\text{w}}.\vec{\text{u}}.$
→
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 2&1&3 \\ 4&{ - 1}&0 \\ { - 7}&2&1 \end{array}} \right]$
→
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}}+\cos\text{x},&\text{if }\text{ x}\neq0\\5,&\text{if }\text{ x}=0\end{cases}$
→