JEE MATHEMATICS MCQ: CONTINUITY

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:R \to R

be defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\lambda \left| {{x^2} - 5x + 6} \right|}}{{\mu (5x - {x^2} - 6)}}}&,&{x < 2}\\ {{e^{\frac{{\tan (x - 2)}}{{x - \left[ x \right]}}\,}}}&,&{x > 2}\\ \mu &,&{x = 2}\\ \,&\,&\, \end{array}} \right.

Where,

\left[ x \right]

is the greatest integer less than or equal to

x

. If

f

is continuous at

x = 2,

then

\lambda + \mu

is equal to

View answer and explanation

Correct option: A

e\,( - e + 1)

Option A (

e(-e+1)

) is correct because, for continuity at x=2, the left-hand limit, right-hand limit, and the function's value at x=2 must be equal, leading to

\lambda = -e^2

and

\mu = e

More Questions from CONTINUITY

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f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\left\{ {1 + \left| {\sin x} \right|} \right\}}^{\frac{a}{{\left| {\sin x} \right|}}}}, - \frac{\pi }{6} < x < 0}\\ {b,x = 0}\\ {{e^{\frac{{\tan 2x}}{{\tan 3x}}}},0 < x < \frac{\pi }{6}} \end{array}} \right.

x = 0

f(x)=\left\{\begin{array}{rr}1+\cos ^{-1}\{\cot x\}, & x<\frac{\pi}{2} \\ \pi[x]+1, & x \geq \frac{\pi}{2}\end{array}\right.

\{ \}

f:R \to R

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\cos 3x - \cos x}}{{{x^2}}},\mathrm{for}\,\,x \ne 0}\\ {\lambda ,\mathrm{for}\,\,x = 0} \end{array}} \right.

f

x = 0

\lambda

f

f(x) f(y)=

f(x)+f(y)+f(x y)-2

x, y \in {R}

f(2)=5

\lim \limits_{{x \rightarrow 4}} f(x)

\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\frac{x^2+3 x-10}{x^2+2 x-15} \text {, when } x \neq-5 \\ a, \quad \text { when } x=-5\end{array}\right.

x=-5

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x},x \ne 0}\\ {k,x = 0} \end{array}} \right.

x = 0

k