MCQ
Function $f(x)=\left\{\begin{array}{c}x+\lambda, \text { if } x<3 \\ 4, \text { if } x=3 \\ 3 x-5, \text { if } x>3\end{array}\right.$, is continuous at $x=3$, then value of $\lambda$ is :
  • 1
  • B
    2
  • C
    3
  • D
    4

Answer

Correct option: A.
1
(A)value of function at $x=3$
$
f(3)=4
$
value of L.H.L.$
\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}[3-h+\lambda]=3+\lambda
$because function is continuous.$
\therefore f(3)=\lim _{h \rightarrow 0} f(3-h) \quad \therefore 4=3+\lambda
$
$
\lambda=1
$
Hence correct option is (A).

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

Similar questions

If $\int {\frac{{dx}}{{{{\left( {{x^2} - 2x + 10} \right)}^2}}} = A\left( {{{\tan }^{ - 1}}\left( {\frac{{x - 1}}{3}} \right) + \frac{{f\left( x \right)}}{{{x^2} - 2x + 10}}} \right)}  + C$ Where $C$ is a constant of integration, then
If $x = – 4$ is a root of $\triangle=\begin{bmatrix}\text{x}&2&3\\1&\text{x}&1\\3&2&\text{x}\end{bmatrix}=0,$ then the other roots are:
The area enclosed by $y ^{2}=8 x$ and $y=\sqrt{2} x$ that lies outside the triangle formed by $y=\sqrt{2} x, x=$ $1, y=2 \sqrt{2}$, is equal to
A box open from top is made from a rectangular sheet of dimension $\mathrm{a} \times \mathrm{b}$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $\mathrm{x}$ is equal to :
If ${a_{ij}} = \frac{1}{2}(3i - 2j)$ and $A = {[{a_{ij}}]_{2 \times 2}}$, then $A$ is equal to
Let $A B C$ be an acute scalene triangle, and $O$ and $H$ be its circumcentre and orthocentre respectively. Further, let $N$ be the mid-point of $O$. The value of the vector sum $\overrightarrow{N A}+\overrightarrow{N B}+\overrightarrow{N C}$ is
What is the value of $\int \limits_0^1 \cos (\pi x) \cos ([2 x] \pi) d x$ ? (Here [t] denotes the integral part of the real numbert.)
$\int_{}^{} {\frac{{(x + 1){{(x + \log x)}^2}}}{x}dx = } $
If $y =\frac{{\cos 6x\, + \,6\cos 4x + 15\cos 2x\, + \,10}}{{\cos 5x + 5\cos 3x + 10\cos x}}\,\,\,\,\,\,\,$ , then $\frac{{dy}}{{dx}}=$
Find the cofactor of element -3 in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is: