2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 28th June Morning Shift
Let f : R $\to$ R be defined as
$f(x) = \left[ {\matrix{
{[{e^x}],} & {x < 0} \cr
{a{e^x} + [x - 1],} & {0 \le x < 1} \cr
{b + [\sin (\pi x)],} & {1 \le x < 2} \cr
{[{e^{ - x}}] - c,} & {x \ge 2} \cr
} } \right.$
where a, b, c $\in$ R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?
A.
There exists a, b, c $\in$ R such that f is continuous on R.
B.
If f is discontinuous at exactly one point, then a + b + c = 1
C.
If f is discontinuous at exactly one point, then a + b + c $\ne$ 1
D.
f is discontinuous at at least two points, for any values of a, b and c
Show Answer
Correct Answer: C
Explanation:
$f(x) = \left\{ {\matrix{
0 & {x < 0} \cr
{a{e^x} - 1} & {0 \le x < 1} \cr
b & {x = 1} \cr
{b - 1} & {1 < x < 2} \cr
{ - c} & {x \ge 2} \cr
} } \right.$
To be continuous at x = 0
a $-$ 1 = 0
to be continuous at x = 1
ae $-$ 1 = b = b $-$ 1 $\Rightarrow$ not possible
to be continuous at x = 2
b $-$ 1 = $-$ c $\Rightarrow$ b + c = 1
If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 27th June Morning Shift
Let a be an integer such that $\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exists, where [t] is greatest integer $\le$ t. Then a is equal to :
Show Answer
Correct Answer: A
Explanation:
$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exist & $a \in I$.
$ = \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}$ exist
$RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}$ $\left[ {a \ne {7 \over 3}} \right]$
$LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}$ $\left[ {a \ne 2} \right]$
For limit to exist
$LHL = RHL$
${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$
$ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$
$\therefore$ $a = - 6$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 26th June Evening Shift
$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}}$ is equal to :
Show Answer
Correct Answer: C
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2\sin (x + \sin x)\,.\,\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$
$ = \mathop {\lim }\limits_{x \to 0} 2\,.\,\left( {{{\left( {{{x + \sin x} \over 2}} \right)\left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}} \right)$
$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {{{\left( {x + x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}}...} \right)\left( {x - x + {{{x^3}} \over {3!}}...} \right)} \over {{x^4}}}} \right)$
$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {2 - {{{x^2}} \over {3!}} + {{{x^4}} \over {5!}}...} \right)\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - 1} \right)$
$ = {1 \over 6}$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 26th June Evening Shift
Let f(x) = min {1, 1 + x sin x}, 0 $\le$ x $\le$ 2$\pi $. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair (m, n) is equal to
Show Answer
Correct Answer: B
Explanation:
$f(x) = \min \{ 1,\,1 + x\sin x\} $, $0 \le x \le x$
$f(x) = \left\{ {\matrix{
{1,} & {0 \le x < \pi } \cr
{1 + x\sin x,} & {\pi \le x \le 2\pi } \cr
} } \right.$
Now at $x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)$
$\therefore$ f(x) is continuous in [0, 2$\pi$]
Now, at x = $\pi$ $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0$
$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi $
$\therefore$ f(x) is not differentiable at x = $\pi$
$\therefore$ (m, n) = (1, 0)
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 26th June Morning Shift
$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$ is equal to :
C.
${1 \over {\sqrt 2 }}$
D.
$ - {1 \over {\sqrt 2 }}$
Show Answer
Correct Answer: D
Explanation:
$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$
Let ${\cos ^{ - 1}}x = t$
$ \Rightarrow x = \cos t$
When $x \to {1 \over {\sqrt 2 }}$, then $t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}$
$\therefore$ $\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}$
$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}$
$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}$
$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t$
$ = - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t$
$ = - {1 \over {\sqrt 2 }}$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 26th June Morning Shift
Let f, g : R $\to$ R be two real valued functions defined as $f(x) = \left\{ {\matrix{
{ - |x + 3|} & , & {x < 0} \cr
{{e^x}} & , & {x \ge 0} \cr
} } \right.$ and $g(x) = \left\{ {\matrix{
{{x^2} + {k_1}x} & , & {x < 0} \cr
{4x + {k_2}} & , & {x \ge 0} \cr
} } \right.$, where k1 and k2 are real constants. If (gof) is differentiable at x = 0, then (gof) ($-$ 4) + (gof) (4) is equal to :
Show Answer
Correct Answer: D
Explanation:
$\because$ gof is differentiable at x = 0
So R.H.D = L.H.D
${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$
$ \Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$
Also $f(f({0^ + })) = g(f({0^ - }))$
$ \Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$
Now $g(f( - 4)) + g(f(4))$
$ = g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$
$ = 4{e^4} - 2$
$ = 2(2{e^4} - 1)$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 25th June Evening Shift
$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$ is equal to
Show Answer
Correct Answer: A
Explanation:
$\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }^2}x + 6\sin x + 2} }}$
$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(1 - \sin x)(2 - \sin x)} \over {{{\cos }^2}x}}\,.\,{\sin ^2}x$
$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(2 - \sin x){{\sin }^2}x} \over {1 + \sin x}}$
$ = {1 \over {12}}$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 25th June Morning Shift
Let f(x) be a polynomial function such that $f(x) + f'(x) + f''(x) = {x^5} + 64$. Then, the value of $\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$ is equal to:
Show Answer
Correct Answer: A
Explanation:
Given, $f(x) + f'(x) + f''(x) = {x^5} + 64$ .........(i)
$\Rightarrow f(x)$ is a polynomial in $x$ whose degree is 5.
Let $f(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e$
$f'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d$
$f''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c$
On substituting the value of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$ in Eq. (i), we get
${x^5}+(a + 5){x^4} + (b + 4a + 20){x^3} + (c + 3b + 12a){x^2} + (d + 2c + 6b)x + e + d + 2c = {x^5} + 64$
Now, equating the coefficient, we get
$ \Rightarrow a + 5 = 0$
$b + 4a + 20 = 0$
$c + 3b + 12a = 0$
$d + 2c + 6b = 0$
$e + d + 2c = 64$
$\therefore$ $a = - 5,\,b = 0,\,c = 60,\,d = - 120,\,e = 64$
$\therefore$ $f(x) = {x^5} - 5{x^4} + 60{x^2} - 120x + 64$
Now, $\mathop {\lim }\limits_{x \to 1} {{{x^5} - 5{x^4} + 60{x^2} - 120x + 64} \over {x - 1}}$ is (${0 \over 0}$ form)
By L' Hospital rule
$\mathop {\lim }\limits_{x \to 1} {{5{x^4} - 20{x^3} + 120x - 120} \over 1}$
$ = - 15$
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 24th June Evening Shift
Let $f(x) = \left\{ {\matrix{
{{{\sin (x - [x])} \over {x - [x]}}} & {,\,x \in ( - 2, - 1)} \cr
{\max \{ 2x,3[|x|]\} } & {,\,|x| < 1} \cr
1 & {,\,otherwise} \cr
} } \right.$
where [t] denotes greatest integer $\le$ t. If m is the number of points where $f$ is not continuous and n is the number of points where $f$ is not differentiable, then the ordered pair (m, n) is :
Show Answer
Correct Answer: C
Explanation:
$f(x) = \left\{ {\matrix{
{{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr
{\max \{ 2x,3[|x|]\} } & , & {|x| < 1} \cr
1 & , & {otherwise} \cr
} } \right.$
$f(x) = \left\{ {\matrix{
{{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr
0 & , & {x \in ( - 1,0]} \cr
{2x} & , & {x \in (0,1)} \cr
1 & , & {otherwise} \cr
} } \right.$
It clearly shows that f(x) is discontinuous
At x = $-$1, 1 also non differentiable
and at $x = 0$, $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0$
$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h} = 2$
$\therefore$ f(x) is not differentiable at x = 0
$\therefore$ m = 2, n = 3
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 29th July Evening Shift
If $[t]$ denotes the greatest integer $\leq t$, then the number of points, at which the function $f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$ is not differentiable in the open interval $(-20,20)$, is __________.
Show Answer
Correct Answer: 79
Explanation:
$f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 28th July Morning Shift
Let $f:[0,1] \rightarrow \mathbf{R}$ be a twice differentiable function in $(0,1)$ such that $f(0)=3$ and $f(1)=5$. If the line $y=2 x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$, then the least number of points $x \in(0,1)$, at which $f^{\prime \prime}(x)=0$, is ____________.
Show Answer
Correct Answer: 2
Explanation:
If a graph cuts $y = 2x + 5$ in (0, 1) twice then its concavity changes twice.
$\therefore$ $f'(x) = 0$ at at least two points.
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 28th July Morning Shift
$\lim\limits_{x \rightarrow 0}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)^{\frac{100}{x}}$ is equal to ___________.
Show Answer
Correct Answer: 1
Explanation:
Let $x + 2\cos x = a$
$x + 2 = b$
as $x \to 0$, $a \to 2$ and $b \to 2$
$\mathop {\lim }\limits_{x \to 0} {\left( {{{{a^3} + 2{a^2} + 3\sin a} \over {{b^3} + 2{b^2} + 3\sin b}}} \right)^{{{100} \over x}}}$
$ = {e^{\mathop {\lim }\limits_{x \to 0} \,.\,{{100} \over x}\,.\,{{({a^3} - {b^3}) + 2({a^2} - {b^2}) + 3(\sin a - \sin b)} \over {{b^3} + 2{b^2} + 3\sin b}}}}$
$\because$ $\mathop {\lim }\limits_{x \to 0} {{a - b} \over x} = \mathop {\lim }\limits_{x \to 0} {{2(\cos x - 1)} \over x} = 0$
$ = {e^0}$
$ = 1$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 25th July Morning Shift
Let $f(x)=\left\{\begin{array}{l}\left|4 x^{2}-8 x+5\right|, \text { if } 8 x^{2}-6 x+1 \geqslant 0 \\ {\left[4 x^{2}-8 x+5\right], \text { if } 8 x^{2}-6 x+1<0,}\end{array}\right.$ where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbf{R}$ where $f$ is not differentiable is ___________.
Show Answer
Correct Answer: 3
Explanation:
$f(x)= \begin{cases}\left|4 x^{2}-8 x+5\right|, & \text { if } 8 x^{2}-6 x+1 \geq 0 \\ {\left[4 x^{2}-8 x+5\right],} & \text { if } 8 x^{2}-6 x+1<0\end{cases}$
$
= \begin{cases}4 x^{2}-8 x+5, & \text { if } x \in\left[-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ {\left[4 x^{2}-8 x+5\right]} & \text { if } x \in\left(\frac{1}{4}, \frac{1}{2}\right)\end{cases}
$
$f(x)=\left\{\begin{array}{cc}
4 x^2-8 x+5 & \text { if } x \in\left(-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\
3 & x \in\left(\frac{1}{4}, \frac{2-\sqrt{2}}{2}\right) \\
2 & x \in\left[\frac{2-\sqrt{2}}{2}, \frac{1}{2}\right)
\end{array}\right.$
$\therefore \quad$ Non-diff at $x=\frac{1}{4}, \frac{2-\sqrt{2}}{2}, \frac{1}{2}$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 30th June Morning Shift
Suppose $\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$ exists and is equal to L, where
$F(x) = \left| {\matrix{
{a + \sin {x \over 2}} & { - b\cos x} & 0 \cr
{ - b\cos x} & 0 & {a + \sin {x \over 2}} \cr
0 & {a + \sin {x \over 2}} & { - b\cos x} \cr
} } \right|$.
Then, $-$112 L is equal to ___________.
Show Answer
Correct Answer: 14
Explanation:
Given,
$F(x) = \left| {\matrix{
{a + \sin {x \over 2}} & { - b\cos x} & 0 \cr
{ - b\cos x} & 0 & {a + \sin {x \over 2}} \cr
0 & {a + \sin {x \over 2}} & { - b\cos x} \cr
} } \right|$
$ = \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \right) + b\cos x \times {b^2}{\cos ^2}x$
$ = - {\left( {a + \sin {x \over 2}} \right)^3} - {b^3}{\cos ^3}x$
Now,
$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$
$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {a + \sin {x \over 2}} \right)}^3} - {b^3}{{\cos }^3}x} \over {{x^3}}}$
Given limit exists, it only possible when a = 0 and b = 0.
$ = \mathop {\lim }\limits_{x \to 0} {{ - {{\left( {\sin {x \over 2}} \right)}^3}} \over {{x^3}}}$
$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2} \times \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)} \right)^3}$
$ = \mathop {\lim }\limits_{x \to 0} - {\left( {{1 \over 2}} \right)^3} \times {\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^3}$
$ = - {1 \over 8} \times 1 = L$
$\therefore$ $ - 112L = - 112 \times - {1 \over 8} = 14$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 28th June Evening Shift
If $\mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2$, then the value of (a $-$ b) is equal to ___________.
Show Answer
Correct Answer: 11
Explanation:
$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(\frac{\sin \left(3 x^{2}-4 x+1\right)}{3 x^{2}-4 x+1}\right)\left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{2(x-1)^{2}}{2 x^{3}-7 x^{2}+a x+b}=-2
\end{aligned}
$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 27th June Evening Shift
Let [t] denote the greatest integer $\le$ t and {t} denote the fractional part of t. The integral value of $\alpha$ for which the left hand limit of the function
$f(x) = [1 + x] + {{{\alpha ^{2[x] + {\{x\}}}} + [x] - 1} \over {2[x] + \{ x\} }}$ at x = 0 is equal to $\alpha - {4 \over 3}$, is _____________.
Show Answer
Correct Answer: 3
Explanation:
$f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}$
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}$
$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}$
$ \Rightarrow \mathop {\lim }\limits_{h \to {0^ - }} 1 - 1 + {{{\alpha ^{ - h - 1}} - 1 - 1} \over { - h - 1}} = \alpha - {4 \over 3}$
$\therefore$ ${{{\alpha ^{ - 1}} - 2} \over { - 1}} = \alpha - {4 \over 3}$
$ \Rightarrow 3{\alpha ^2} - 10\alpha + 3 = 0$
$\therefore$ $\alpha = 3$ or ${1 \over 3}$
$\because$ $\alpha$ in integer, hence $\alpha$ = 3
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 25th June Evening Shift
Let $f(x) = \left[ {2{x^2} + 1} \right]$ and $g(x) = \left\{ {\matrix{
{2x - 3,} & {x < 0} \cr
{2x + 3,} & {x \ge 0} \cr
} } \right.$, where [t] is the greatest integer $\le$ t. Then, in the open interval ($-$1, 1), the number of points where fog is discontinuous is equal to ______________.
Show Answer
Correct Answer: 62
Explanation:
$
\mathrm{f}(\mathrm{g}(\mathrm{x}))=\left[2 \mathrm{~g}^2(\mathrm{x})\right]+1
$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2022 (Online) 24th June Morning Shift
The number of points where the function
$f(x) = \left\{ {\matrix{
{|2{x^2} - 3x - 7|} & {if} & {x \le - 1} \cr
{[4{x^2} - 1]} & {if} & { - 1 < x < 1} \cr
{|x + 1| + |x - 2|} & {if} & {x \ge 1} \cr
} } \right.$
[t] denotes the greatest integer $\le$ t, is discontinuous is _____________.
Show Answer
Correct Answer: 7
Explanation:
$\because f(-1)=2$ and $f(1)=3$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 31st August Evening Shift
If $\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$ and $\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$ are the roots of the equation, ax2 + bx $-$ 4 = 0, then the ordered pair (a, b) is :
Show Answer
Correct Answer: D
Explanation:
$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}$ form2 + bx $-$ 4 = 0 are the roots then
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 31st August Evening Shift
Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then
A.
f''(x) = 0 for all x $\in$ (0, 2)
B.
f''(x) = 0 for some x $\in$ (0, 2)
C.
f'(x) = 0 for some x $\in$ [0, 2]
D.
f''(x) > 0 for all x $\in$ (0, 2)
Show Answer
Correct Answer: B
Explanation:
f(0) = 0, f(1) = 1 and f(2) = 2
Let h(x) = f(x) $-$ x
Clearly h(x) is continuous and twice differentiable on (0, 2)
Also, h(0) = h(1) = h(2) = 0
$\therefore$ h(x) satisfies all the condition of Rolle's theorem.
$\therefore$ there exist C1 $\in$(0, 1) such that h'(c1 ) = 0
$\Rightarrow$ f'(1 ) $-$ 1 = 0 $\Rightarrow$ f'(c1 ) = 1
also there exist c2 $\in$(1, 2) such that h'(c2 ) = 0
$\Rightarrow$ f'(c2 ) = 1
Now, using Rolle's theorem on [c1 , c2 ] for f'(x)
We have f''(c) = 0, c$\in$(c1 , c2 )
Hence, f''(x) = 0 for some x$\in$(0, 2).
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 31st August Morning Shift
The function
Show Answer
Correct Answer: C
Explanation:
$f(x) = \left| {(x - 3)(x + 1)} \right|\,.\,{e^{{{(3x - 2)}^2}}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 31st August Morning Shift
If the function
Show Answer
Correct Answer: A
Explanation:
If f(x) is continuous at x = 0, RHL = LHL = f(0)
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 31st August Morning Shift
$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$ is equal to :
Show Answer
Correct Answer: C
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 27th August Evening Shift
If $\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$, then the ordered pair (a, b) is :
A.
$\left( {1,{1 \over 2}} \right)$
B.
$\left( {1, - {1 \over 2}} \right)$
C.
$\left( { - 1,{1 \over 2}} \right)$
D.
$\left( { - 1, - {1 \over 2}} \right)$
Show Answer
Correct Answer: B
Explanation:
$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$ ($\infty$ $-$ $\infty$)
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 27th August Morning Shift
If $\alpha$, $\beta$ are the distinct roots of x2 + bx + c = 0, then
Show Answer
Correct Answer: C
Explanation:
$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 26th August Evening Shift
Let [t] denote the greatest integer less than or equal to t. Let
A.
continuous in [$-$2, 2] but not differentiable at more than
B.
not continuous at exactly three points in [$-$2, 2]
C.
continuous in [$-$2, 2] but not differentiable at exactly
D.
not continuous at exactly four points in [$-$2, 2]
Show Answer
Correct Answer: A
Explanation:
min{x $-$ [x], 1 $-$ x + [x]}
h(x) = min{x $-$ [x], 1 $-$ [x $-$ [x])}
$\Rightarrow$ always continuous in [$-$2, 2] but not differentiable at 7 points.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 26th August Evening Shift
$\mathop {\lim }\limits_{x \to 2} \left( {\sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} } \right)$ is equal to :
Show Answer
Correct Answer: A
Explanation:
$S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} $
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 27th July Evening Shift
The value of
Show Answer
Correct Answer: C
Explanation:
$\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 27th July Evening Shift
Let $f:[0,\infty ) \to [0,3]$ be a function defined by
A.
f is continuous everywhere but not differentiable exactly at one point in (0, $\infty$)
B.
f is differentiable everywhere in (0, $\infty$)
C.
f is not continuous exactly at two points in (0, $\infty$)
D.
f is continuous everywhere but not differentiable exactly at two points in (0, $\infty$)
Show Answer
Correct Answer: B
Explanation:
Graph of $\max \{ \sin t:0 \le t \le x\} $ in $x \in [0,\pi ]$
& graph of cos x for $x \in [\pi ,\infty )$
So graph of
$f(x) = \left\{ {\matrix{
{\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \cr
{2 + \cos x,} & {x > \pi} \cr
} } \right.$
f(x) is differentiable everywhere in (0, $\infty$)
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 27th July Morning Shift
Let $f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R$ be defined as $f(x) = \left\{ {\matrix{
{{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr
b & , & {x = 0} \cr
{{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr
} } \right.$2 is equal to :
Show Answer
Correct Answer: C
Explanation:
$\mathop {\lim }\limits_{x \to 0} f(x) = b$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 27th July Morning Shift
Let f : R $\to$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$ is equal to :
Show Answer
Correct Answer: D
Explanation:
This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 25th July Morning Shift
Let f : R $\to$ R be defined as
Show Answer
Correct Answer: A
Explanation:
$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {e^{{{\tan (x - 2)} \over {x - 2}}}} = {e^1}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 22th July Evening Shift
Let f : R $\to$ R be defined as $f(x) = \left\{ {\matrix{
{{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$
Show Answer
Correct Answer: A
Explanation:
For continuity
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 20th July Evening Shift
If $f:R \to R$ is given by $f(x) = x + 1$, then the value of $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$ is :
Show Answer
Correct Answer: D
Explanation:
$f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 20th July Morning Shift
Let a function f : R $\to$ R be defined as $f(x) = \left\{ {\matrix{
{\sin x - {e^x}} & {if} & {x \le 0} \cr
{a + [ - x]} & {if} & {0 < x < 1} \cr
{2x - b} & {if} & {x \ge 1} \cr
} } \right.$
Show Answer
Correct Answer: B
Explanation:
Continuous x = 0+ ) = f(0$-$ ) $\Rightarrow$ a $-$ 1 = 0 $-$ e0 + ) = f(1$-$ )
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 18th March Evening Shift
Let f : R $ \to $ R be a function defined as
Show Answer
Correct Answer: D
Explanation:
Given, $f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x}, & x<0 \\ b, & x=0 \\ \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}, & x>0\end{array}\right.$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 18th March Morning Shift
If $\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$ is equal to L, then the value of (6L + 1) is
Show Answer
Correct Answer: D
Explanation:
$L = \mathop {\lim }\limits_{x \to 0} {{\left( {x + {{{x^3}} \over 6} + .....} \right) - \left( {x - {{{x^3}} \over 3}.....} \right)} \over {3{x^3}}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 18th March Morning Shift
If $f(x) = \left\{ {\matrix{
{{1 \over {|x|}}} & {;\,|x|\, \ge 1} \cr
{a{x^2} + b} & {;\,|x|\, < 1} \cr
} } \right.$ is differentiable at every point of the domain, then the values of a and b are respectively :
A.
${1 \over 2},{1 \over 2}$
B.
${1 \over 2}, - {3 \over 2}$
C.
${5 \over 2}, - {3 \over 2}$
D.
$ - {1 \over 2},{3 \over 2}$
Show Answer
Correct Answer: D
Explanation:
$f(x) = \left\{ {\matrix{
{{1 \over {|x|}},} & {|x| \ge 1} \cr
{a{x^2} + b,} & {|x| < 1} \cr
} } \right.$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 17th March Evening Shift
The value of the limit
Show Answer
Correct Answer: B
Explanation:
Given,
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 17th March Evening Shift
The value of $\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}}$, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to :
Show Answer
Correct Answer: B
Explanation:
We know,
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 17th March Morning Shift
The value of
Show Answer
Correct Answer: C
Explanation:
$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 16th March Evening Shift
Let f : S $ \to $ S where S = (0, $\infty $) be a twice differentiable function such that f(x + 1) = xf(x). If g : S $ \to $ R be defined as g(x) = loge f(x), then the value of |g''(5) $-$ g''(1)| is equal to :
Show Answer
Correct Answer: D
Explanation:
$f(x + 1) = xf(x)$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 16th March Evening Shift
Let $\alpha$ $\in$ R be such that the function $f(x) = \left\{ {\matrix{
{{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$ is continuous at x = 0, where {x} = x $-$ [ x ] is the greatest integer less than or equal to x. Then :
A.
no such $\alpha$ exists
C.
$\alpha$ = ${\pi \over 4}$
D.
$\alpha$ = ${\pi \over {\sqrt 2 }}$
Show Answer
Correct Answer: A
Explanation:
$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}} $
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 16th March Morning Shift
Let ${S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)} $. Then $\mathop {\lim }\limits_{k \to \infty } {S_k}$ is equal to :
A.
${\cot ^{ - 1}}\left( {{3 \over 2}} \right)$
D.
${\tan ^{ - 1}}\left( {{3 \over 2}} \right)$
Show Answer
Correct Answer: A
Explanation:
Sk = $\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 16th March Morning Shift
Let the functions f : R $ \to $ R and g : R $ \to $ R be defined as :
Show Answer
Correct Answer: C
Explanation:
$fog(x) = \left\{ {\matrix{
{{x^3} + 2,} & {x \le 0} \cr
{{x^6},} & {0 \le x \le 1} \cr
{{{(3x - 2)}^2},} & {x \ge 1} \cr
} } \right.$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 26th February Evening Shift
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4.
Show Answer
Correct Answer: A
Explanation:
$L = \mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$ [${0 \over 0}$ form]
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 26th February Evening Shift
Let $f(x) = {\sin ^{ - 1}}x$ and $g(x) = {{{x^2} - x - 2} \over {2{x^2} - x - 6}}$. If $g(2) = \mathop {\lim }\limits_{x \to 2} g(x)$, then the domain of the function fog is :
A.
$( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)$
B.
$( - \infty , - 2] \cup [ - 1,\infty )$
C.
$( - \infty , - 2] \cup \left[ { - {3 \over 2},\infty } \right)$
D.
$( - \infty , - 1] \cup [2,\infty )$
Show Answer
Correct Answer: A
Explanation:
$g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 26th February Evening Shift
Let f : R $ \to $ R be defined as
Show Answer
Correct Answer: C
Explanation:
$f( - {1^ - }) = 2$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 26th February Morning Shift
The value of $\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$ is :
D.
${2 \over {\sqrt 3 }}$
Show Answer
Correct Answer: A
Explanation:
Let L = $\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
JEE Main 2021 (Online) 25th February Morning Shift
$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}$ is equal to :
Show Answer
Correct Answer: B
Explanation:
It is ${1^\infty }$ form
