JEE MAIN Continuity Previous Year Questions (PYQs) – Page 3 of 4

The value of k for which the function

$f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr } } \right.$

is continuous at x = ${\pi \over 2},$ is :

Let \; f:\mathbb{R}\to\mathbb{R} \text{ be defined as} \[ f(x) = \begin{cases} \dfrac{\sin\!\big((a+1)x\big)+\sin 2x}{2x}, & x<0 \\[8pt] b, & x=0 \\[8pt] \dfrac{\sqrt{x+bx^{3}}-\sqrt{x}}{b\,x^{5/2}}, & x>0 \end{cases} \] If f is continuous at x = 0, then the value of a + b is equal to :

If the function $f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.$ is continuous at x = 0, then ${1 \over a} + {1 \over b} + {4 \over k}$ is equal to :

Let a function f : R $\to$ R be defined as $f(x) = \left\{ {\matrix{ {\sin x - {e^x}} & {if} & {x \le 0} \cr {a + [ - x]} & {if} & {0 < x < 1} \cr {2x - b} & {if} & {x \ge 1} \cr } } \right.$ where [ x ] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to:

Let f : R $ \to $ R be a function defined as
$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$ Then, f is

Let $\quad f(x)= \begin{cases}(1+a x)^{1 / x} & , x<0 \\ 1+b, & x=0 \\ \frac{(x+4)^{1 / 2}-2}{(x+c)^{1 / 3}-2}, & x>0\end{cases}$ be continuous at $x=0$. Then $e^a b c$ is equal to:

Let $f:\mathbf{R}\rightarrow\mathbf{R}$ be defined as $ f(x)= \begin{cases} \dfrac{a - b\cos 2x}{x^2}, & x < 0, \\[6pt] x^2 + cx + 2, & 0 \le x \le 1, \\[6pt] 2x + 1, & x > 1. \end{cases} $ If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is **not differentiable**, then $m + a + b + c$ equals :

If$f(x) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} & {,x > 0} \cr } } \right.$
is continuous at x = 0, then the ordered pair (p, q) is equal to

Let f : R $\to$ R be defined as $f(x) = \left\{ {\matrix{ {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$ If f is continuous at x = 0, then $\alpha$ is equal to :

Let f : R $\to$ R be defined as $f(x) = \left\{ {\matrix{ {{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr {\mu ,} & {x = 2} \cr } } \right.$ where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then $\lambda$ + $\mu$ is equal to :