Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
Let $\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$ Then f(x) is continuous on the set:
  • A
    R
  • B
    R - {1}
  • C
    R - {2}
  • ✓
    R - {1, 2}
✓

Answer

Correct option: D.
R - {1, 2}
Given:

$\text{f(x)=}\begin{cases}\frac{\text{x}^4-5\text{x}^2+4}{|(\text{x}-1)(\text{x-2})|},\text{x}\neq1,2\\6, \text{x}=1\\12,\text{x}=2\end{cases}$

Now,

$\Rightarrow\text{x}^4-5\text{x}^2+4=\text{x}^4-\text{x}^2-4\text{x}^2+4\\=\text{x}^2(\text{x}^2-1)-4(\text{x}^2-1)$

$=(\text{x}^2-1)(\text{x}^2-4)=(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)$

$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-1)(\text{x}+1)(\text{x}-2)(\text{x}+2)}{|(\text{x}-2)(\text{x}-1)|},&\text{x}\neq1,2\\6,& \text{x}=1\\12,&\text{x}=2\end{cases}$

$\Rightarrow\text{f(x)}=\begin{cases}(\text{x}+1)(\text{x}+2),&\text{x}<1\\-(\text{x+1})(\text{x}+2),&1<\text{x}<2\$\text{x+1})(\text{x}+2),&\text{x}>2\\6,&\text{x=1}\\12,&\text{x}=2\end{cases}$

So,

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(1-\text{h}+1)(1-\text{h}+2)\\=2\times3=6$

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(1+\text{h}+1)(1+\text{h}+2)\\=-2\times3=-6$

Also,

$\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=-\lim\limits_{\text{h}\rightarrow0}(2-\text{h}+1)(2-\text{h}+2)=-12$

$\Rightarrow\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2+\text{h}+1)(2+\text{h}+2)=12$

Thus,

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}$

$\therefore$ The only point of discontinuities of the function f(x) are x = 1 and x = 2. Hence, the given function is continuous on the set R - {1, 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 "M.C.Q (1 Marks)" 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

If $x$ denotes the number of sixes in four consecutive throws of a dice, then $P\,(x = 4)$ is
→
If $\text{f(x)}=(\text{x+1})^{\cot\text{x}}$ be continuous at x = 0, then f(0) is equal to:
→
If $\begin{vmatrix}x&2\\18&x\end{vmatrix}=\begin{vmatrix}6&2\\18&6\end{vmatrix},$ then x is equal to:
→
If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $f ^{\prime}\left(\frac{4}{3}\right)=0,$ then ordered pair $( a , b )$ is equal to
→
If the function $f(x)=\frac{\cos (\sin x)-\cos x}{x^{4}}$ is continuous at each point in its domain and $f (0)=\frac{1}{ k },$ then $k$ is ........
→
If $P = \left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{ - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right],\,A = \,\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $Q=PAP^T,$ then $P^T$ $Q^{2015}$ $P$ is 
→
${d \over {dx}}(x{e^{{x^2}}}) = $
→
If $\text{f}(\text{x})=\tan^{-1}(\text{g}(\text{x})),$ where g(x) is monotonically increasing for $0<\text{x}<\frac{\pi}{2}.$ Then, f(x) is:
→
The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let z = px + qy where p, q > 0. Condition on p and q so that the maximum of z occurs at both the points (15, 15) and (0, 20) is __________.
→
If the system of linear equations

$2 x+y-z=3$

$x-y-z=\alpha$

$3 x+3 y+\beta z=3$

has infinitely many solution, then $\alpha+\beta-\alpha \beta$ is equal to .... .

→