Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 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 AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Minimise $\text{Z}=13\text{x}-15\text{y},$ subject to the constraints: $\text{x}+\text{y}\leq7,2\text{x}-3\text{y}+6\geq0,\text{x}\geq0,\text{y}\geq0.$
→
show that the differential equation of which $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^2}$ is a solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
→
In a triangle OAB, $\angle\text{AOB}=90^\circ.$ If P and Q are points of trisection of AB, prove that $\text{OP}^2+\text{OQ}^2=\frac{5}{9}\text{AB}^2.$
→
Evaluate the following definite integrals:
$\int\limits_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$
→
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
→
$\text{If}\ \text{y}=\sin(\sin\text{x}),\ \text{prove that}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0.$
→
Evaluate the following:
$\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
→
Evaluate the following integrals:
$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
→
If y(x) is a solution of the different equation $\Big(\frac{2+\sin\text{x}}{1+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=-\cos\text{x}$ and $\text{y}(0)=1,$ then find the value of $\text{y}\Big(\frac{\pi}{2}\Big).$
→
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
→