Limits, Continuity and Differentiability
278 Questions
Start JEE Mains Test
2021
Q151
JEE Mains
MCQ
14 Mar 2026
The value of
$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$, where [ x ] denotes the greatest integer $ \le $ x is :
$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$, where [ x ] denotes the greatest integer $ \le $ x is :
A.
$\pi$
B.
${\pi \over 4}$
C.
${\pi \over 2}$
D.
0
2021
Q152
JEE Mains
MCQ
14 Mar 2026
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 :
A.
1
B.
${{187} \over {144}}$
C.
${{197} \over {144}}$
D.
${{205} \over {144}}$
2021
Q153
JEE Mains
MCQ
14 Mar 2026
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
B.
$\alpha$ = 0
C.
$\alpha$ = ${\pi \over 4}$
D.
$\alpha$ = ${\pi \over {\sqrt 2 }}$
2021
Q154
JEE Mains
MCQ
14 Mar 2026
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.
${\cot ^{ - 1}}\left( {{3 \over 2}} \right)$
B.
${\pi \over 2}$
C.
tan$-$1 (3)
D.
${\tan ^{ - 1}}\left( {{3 \over 2}} \right)$
2021
Q155
JEE Mains
MCQ
14 Mar 2026
Let the functions f : R $ \to $ R and g : R $ \to $ R be defined as :
$f(x) = \left\{ {\matrix{ {x + 2,} & {x < 0} \cr {{x^2},} & {x \ge 0} \cr } } \right.$ and
$g(x) = \left\{ {\matrix{ {{x^3},} & {x < 1} \cr {3x - 2,} & {x \ge 1} \cr } } \right.$
Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :
$f(x) = \left\{ {\matrix{ {x + 2,} & {x < 0} \cr {{x^2},} & {x \ge 0} \cr } } \right.$ and
$g(x) = \left\{ {\matrix{ {{x^3},} & {x < 1} \cr {3x - 2,} & {x \ge 1} \cr } } \right.$
Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :
A.
0
B.
3
C.
1
D.
2
2021
Q156
JEE Mains
MCQ
14 Mar 2026
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4.
Then $\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$ equals :
Then $\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$ equals :
A.
4 $-$ 2a
B.
2a + 4
C.
a + 4
D.
2a $-$ 4
2021
Q157
JEE Mains
MCQ
14 Mar 2026
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.
$( - \infty , - 1] \cup [2,\infty )$
2021
Q158
JEE Mains
MCQ
14 Mar 2026
Let f : R $ \to $ R be defined as
$f(x) = \left\{ \matrix{ 2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr |a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr \sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$ If f(x) is continuous on R, then a + b equals :
$f(x) = \left\{ \matrix{ 2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr |a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr \sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$ If f(x) is continuous on R, then a + b equals :
A.
$-$3
B.
3
C.
$-$1
D.
1
2021
Q159
JEE Mains
MCQ
14 Mar 2026
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 :
A.
${4 \over 3}$
B.
${2 \over 3}$
C.
${3 \over 4}$
D.
${2 \over {\sqrt 3 }}$
2021
Q160
JEE Mains
MCQ
14 Mar 2026
$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}$ is equal to :
A.
${{1 \over 2}}$
B.
1
C.
0
D.
${{1 \over e}}$
2021
Q161
JEE Mains
MCQ
14 Mar 2026
If f : R $ \to $ R is a function defined by f(x)= [x - 1] $\cos \left( {{{2x - 1} \over 2}} \right)\pi $, where [.] denotes the greatest
integer function, then f is :
A.
continuous for every real x
B.
discontinuous at all integral values of x except at x = 1
C.
discontinuous only at x = 1
D.
continuous only at x = 1
2021
Q162
JEE Mains
Numerical
14 Mar 2026
Let $f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$, x $\in$ R. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$ is __________.
Correct Answer: 7
Explanation:
$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$
$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$
$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$
$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^2} + 2)} \over 1} = 44$
$\Rightarrow$ 9n $-$ (19) = 44
$\Rightarrow$ 9n = 63
$\Rightarrow$ n = 7
$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$
$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$
$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^2} + 2)} \over 1} = 44$
$\Rightarrow$ 9n $-$ (19) = 44
$\Rightarrow$ 9n = 63
$\Rightarrow$ n = 7
2021
Q163
JEE Mains
Numerical
14 Mar 2026
Let [t] denote the greatest integer $\le$ t. The number of points where the function $f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1],x \in ( - 2,2)$ is not continuous is _____________.
Correct Answer: 2
Explanation:
$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$
$f\left( x \right) = \left\{ {\matrix{ { - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.$
$ \therefore $ at x = -1, $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$ and $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$
Hence continuous at x = –1
Similarly check at x = 0,
$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$
So, f(x) discontinuous and at x = 0
$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$ and $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$
So, f(x) discontinuous and at x = 1
Hence 2 points of discontinuity.
$f\left( x \right) = \left\{ {\matrix{ { - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.$
$ \therefore $ at x = -1, $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$ and $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$
Hence continuous at x = –1
Similarly check at x = 0,
$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$
So, f(x) discontinuous and at x = 0
$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$ and $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$
So, f(x) discontinuous and at x = 1
Hence 2 points of discontinuity.
2021
Q164
JEE Mains
Numerical
14 Mar 2026
Let a, b $\in$ R, b $\in$ 0, Define a function
$f(x) = \left\{ {\matrix{ {a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr {{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr } } \right.$.
If f is continuous at x = 0, then 10 $-$ ab is equal to ________________.
$f(x) = \left\{ {\matrix{ {a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr {{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr } } \right.$.
If f is continuous at x = 0, then 10 $-$ ab is equal to ________________.
Correct Answer: 14
Explanation:
$f(x) = \left\{ {\matrix{
{a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr
{{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr
} } \right.$
For continuity at '0'
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$
$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$
$ \Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$
$ \Rightarrow 4 = - ab$
$ \Rightarrow 10 - ab = 14$
For continuity at '0'
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{\tan 2x - \sin 2x} \over {b{x^3}}} = - a$
$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {{{{8{x^3}} \over 3} + {{8{x^3}} \over {3!}}} \over {b{x^3}}} = - a$
$ \Rightarrow 8\left( {{1 \over 3} + {1 \over {3!}}} \right) = - ab$
$ \Rightarrow 4 = - ab$
$ \Rightarrow 10 - ab = 14$
2021
Q165
JEE Mains
Numerical
14 Mar 2026
Let $f:[0,3] \to R$ be defined by $f(x) = \min \{ x - [x],1 + [x] - x\} $ where [x] is the greatest integer less than or equal to x. Let P denote the set containing all x $\in$ [0, 3] where f i discontinuous, and Q denote the set containing all x $\in$ (0, 3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to ______________.
Correct Answer: 5
Explanation:
1 $-$ {x} = 1 $-$ x; 0 $\le$ x < 1
Non differentiable at
$x = {1 \over 2},1,{3 \over 2},2,{5 \over 2}$
2021
Q166
JEE Mains
Numerical
14 Mar 2026
Consider the function
where P(x) is a polynomial such that P'' (x) is always a constant and P(3) = 9. If f(x) is continuous at x = 2, then P(5) is equal to _____________.
where P(x) is a polynomial such that P'' (x) is always a constant and P(3) = 9. If f(x) is continuous at x = 2, then P(5) is equal to _____________.
Correct Answer: 39
Explanation:
$f(x) = \left\{ {\matrix{
{{{P(x)} \over {\sin (x - 2)}},} & {x \ne 2} \cr
{7,} & {x = 2} \cr
} } \right.$
P''(x) = const. $\Rightarrow$ P(x) is a 2 degree polynomial
f(x) is cont. at x = 2
f(2+) = f(2$-$)
$\mathop {\lim }\limits_{x \to {2^ + }} {{P(x)} \over {\sin (x - 2)}} = 7$
$\mathop {\lim }\limits_{x \to {2^ + }} {{(x - 2)(ax + b)} \over {\sin (x - 2)}} = 7 \Rightarrow 2a + b = 7$
P(x) = (x $-$ 2)(ax + b)
P(3) = (3 $-$ 2)(3a + b) = 9 $\Rightarrow$ 3a + b = 9
a = 2, b = 3
P(5) = (5 $-$ 2)(2.5 + 3) = 3.13 = 39
P''(x) = const. $\Rightarrow$ P(x) is a 2 degree polynomial
f(x) is cont. at x = 2
f(2+) = f(2$-$)
$\mathop {\lim }\limits_{x \to {2^ + }} {{P(x)} \over {\sin (x - 2)}} = 7$
$\mathop {\lim }\limits_{x \to {2^ + }} {{(x - 2)(ax + b)} \over {\sin (x - 2)}} = 7 \Rightarrow 2a + b = 7$
P(x) = (x $-$ 2)(ax + b)
P(3) = (3 $-$ 2)(3a + b) = 9 $\Rightarrow$ 3a + b = 9
a = 2, b = 3
P(5) = (5 $-$ 2)(2.5 + 3) = 3.13 = 39
2021
Q167
JEE Mains
Numerical
14 Mar 2026
Let f : R $\to$ R be a function defined as $f(x) = \left\{ {\matrix{
{3\left( {1 - {{|x|} \over 2}} \right)} & {if} & {|x|\, \le 2} \cr
0 & {if} & {|x|\, > 2} \cr
} } \right.$
Let g : R $\to$ R be given by $g(x) = f(x + 2) - f(x - 2)$. If n and m denote the number of points in R where g is not continuous and not differentiable, respectively, then n + m is equal to ______________.
Let g : R $\to$ R be given by $g(x) = f(x + 2) - f(x - 2)$. If n and m denote the number of points in R where g is not continuous and not differentiable, respectively, then n + m is equal to ______________.
Correct Answer: 4
Explanation:
$f(x) = \left\{ {\matrix{ {3\left( {{{1 - \left| x \right|} \over 2}} \right)} & {if\,\left| x \right| \le 2} \cr 0 & {if\,\left| x \right| > 2} \cr } } \right.$
$g(x) = f(x + 2) - f(x - 2)$
$f(x) = \left\{ {\matrix{ {0,} & {x < - 2} \cr {{3 \over 2}(1 + x),} & { - 2 \le x < 0} \cr {{3 \over 2}(1 - x),} & {0 \le x < 2} \cr {0,} & {x > 2} \cr } } \right.$
$f(x + 2) = \left\{ {\matrix{ {0,} & {x < - 4} \cr {{3 \over 2}( 3 + x),} & { - 4 \le x < - 2} \cr {{3 \over 2}( - 1 - x),} & { - 2 \le x < 0} \cr {0,} & {x > 4} \cr } } \right.$
$f(x - 2) = \left\{ {\matrix{ {0,} & {x < 0} \cr {{3 \over 2}(x - 1),} & {0 \le x < 2} \cr {{3 \over 2}( - 1 - x),} & {2 \le x < 4} \cr {0,} & {x > 4} \cr } } \right.$
$g(x) = f(x + 2) + f(x - 2)$
$ = \left\{ {\matrix{ {{{3x} \over 2} + 6,} & { - 4 \le x \le 2} \cr { - {{3x} \over 2},} & { - 2 < x < 2} \cr {{{3x} \over 2} - 6,} & {2 \le x \le 4} \cr {0,} & {\left| x \right| > 4} \cr } } \right.$
So, n = 0 and m = 4
$\therefore$ m + n = 4
2021
Q168
JEE Mains
Numerical
14 Mar 2026
Let a function g : [ 0, 4 ] $\to$ R be defined as
$g(x) = \left\{ {\matrix{ {\mathop {\max }\limits_{0 \le t \le x} \{ {t^3} - 6{t^2} + 9t - 3),} & {0 \le x \le 3} \cr {4 - x,} & {3 < x \le 4} \cr } } \right.$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.
$g(x) = \left\{ {\matrix{ {\mathop {\max }\limits_{0 \le t \le x} \{ {t^3} - 6{t^2} + 9t - 3),} & {0 \le x \le 3} \cr {4 - x,} & {3 < x \le 4} \cr } } \right.$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.
Correct Answer: 1
Explanation:
$f(x) = {x^3} - 6{x^2} + 9x - 3$
$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$
$f(1) = 1$, $f(3) = 3$
$g(x) = \left[ {\matrix{ {f(9x)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$
g(x) is continuous
$g'(x) = \left[ {\matrix{ {3(x - 1)(x - 3)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$
g(x) is non-differentiable at x = 3
$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$
$f(1) = 1$, $f(3) = 3$
$g(x) = \left[ {\matrix{ {f(9x)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$
g(x) is continuous
$g'(x) = \left[ {\matrix{ {3(x - 1)(x - 3)} & {0 \le x \le 1} \cr 0 & {1 \le x \le 3} \cr { - 1} & {3 < x \le 4} \cr } } \right.$
g(x) is non-differentiable at x = 3
2021
Q169
JEE Mains
Numerical
14 Mar 2026
If $\mathop {\lim }\limits_{x \to 0} {{\alpha x{e^x} - \beta {{\log }_e}(1 + x) + \gamma {x^2}{e^{ - x}}} \over {x{{\sin }^2}x}} = 10,\alpha ,\beta ,\gamma \in R$, then the value of $\alpha$ + $\beta$ + $\gamma$ is _____________.
Correct Answer: 3
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{\alpha x\left( {1 + x + {{{x^2}} \over x}} \right) - \beta \left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3}} \right) + \gamma {x^2}(1 - x)} \over {{x^3}}}$
$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$
For limit to exist
$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$ ..... (i)
$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$
Put in (i)
${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$
${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$
$ \Rightarrow \alpha = 6$
$\alpha$ = 6, $\beta$ = 6, $\gamma$ = $-$9
$\alpha$ + $\beta$ + $\gamma$ = 3
$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$
For limit to exist
$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$ ..... (i)
$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$
Put in (i)
${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$
${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$
$ \Rightarrow \alpha = 6$
$\alpha$ = 6, $\beta$ = 6, $\gamma$ = $-$9
$\alpha$ + $\beta$ + $\gamma$ = 3
2021
Q170
JEE Mains
Numerical
14 Mar 2026
If the value of $\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{\left( {{{x + 2} \over {{x^2}}}} \right)}}$ is equal to ea, then a is equal to __________.
Correct Answer: 3
Explanation:
$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{{{x + 2} \over {{x^2}}}}}$
form : 1$\infty$ $ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$
Now, $\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} {{\sin x\sqrt {\cos 2x} - \cos x \times {1 \over {2\sqrt {\cos 2x} }} \times ( - 2sin2x)} \over {2x}}$ (by L' Hospital Rule)
$\mathop {\lim }\limits_{x \to 0} {{\sin x\cos 2x + \sin 2x.\cos x} \over {2x}} = {1 \over 2} + 1 = {3 \over 2}$
So, ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right)(x + 2)}}$
$ = {e^{{3 \over 2} \times 2}} = {e^3}$
$\Rightarrow$ a = 3
form : 1$\infty$ $ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$
Now, $\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} {{\sin x\sqrt {\cos 2x} - \cos x \times {1 \over {2\sqrt {\cos 2x} }} \times ( - 2sin2x)} \over {2x}}$ (by L' Hospital Rule)
$\mathop {\lim }\limits_{x \to 0} {{\sin x\cos 2x + \sin 2x.\cos x} \over {2x}} = {1 \over 2} + 1 = {3 \over 2}$
So, ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right)(x + 2)}}$
$ = {e^{{3 \over 2} \times 2}} = {e^3}$
$\Rightarrow$ a = 3
2021
Q171
JEE Mains
Numerical
14 Mar 2026
Let f : R $ \to $ R satisfy the equation f(x + y) = f(x) . f(y) for all x, y $\in$R and f(x) $\ne$ 0 for any x$\in$R. If the function f is differentiable at x = 0 and f'(0) = 3, then
$\mathop {\lim }\limits_{h \to 0} {1 \over h}(f(h) - 1)$ is equal to ____________.
$\mathop {\lim }\limits_{h \to 0} {1 \over h}(f(h) - 1)$ is equal to ____________.
Correct Answer: 3
Explanation:
Given, $f(x + y) = f(x)\,.\,f(y)\,\forall x,y \in R$
$\therefore$ $f(x) = {a^x} \Rightarrow f'(x) = {a^x}\,.\,\log (a)$
Now, $f'(0) = \log (a) \Rightarrow 3 = \log (a) \Rightarrow a = {e^3}$
$\therefore$ $f(x) = {({e^3})^x} = {e^{3x}}$
$\therefore$ $f(h) = {e^{3h}}$
Now, $\mathop {\lim }\limits_{h \to 0} \left( {{{f(h) - 1} \over h}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over h}} \right)$
$ = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over {3h}} \times 3} \right) = 3 \times 1 = 3$
2021
Q172
JEE Mains
Numerical
14 Mar 2026
If the function $f(x) = {{\cos (\sin x) - \cos x} \over {{x^4}}}$ is continuous at each point in its domain and $f(0) = {1 \over k}$, then k is ____________.
Correct Answer: 6
Explanation:
$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {{\cos \left( {\sin x} \right) - \cos x} \over {{x^4}}}$
$ \Rightarrow $ ${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$
= $\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$
= $2 \times 2 \times {1 \over 6} \times {1 \over 4}$ = ${1 \over 6}$
$ \Rightarrow $ ${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$
= $\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$
= $2 \times 2 \times {1 \over 6} \times {1 \over 4}$ = ${1 \over 6}$
2021
Q173
JEE Mains
Numerical
14 Mar 2026
Let f : R $ \to $ R and g : R $ \to $ R be defined as
$f(x) = \left\{ {\matrix{ {x + a,} & {x < 0} \cr {|x - 1|,} & {x \ge 0} \cr } } \right.$ and
$g(x) = \left\{ {\matrix{ {x + 1,} & {x < 0} \cr {{{(x - 1)}^2} + b,} & {x \ge 0} \cr } } \right.$,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $\in$ R, then a + b is equal to ____________.
$f(x) = \left\{ {\matrix{ {x + a,} & {x < 0} \cr {|x - 1|,} & {x \ge 0} \cr } } \right.$ and
$g(x) = \left\{ {\matrix{ {x + 1,} & {x < 0} \cr {{{(x - 1)}^2} + b,} & {x \ge 0} \cr } } \right.$,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $\in$ R, then a + b is equal to ____________.
Correct Answer: 1
Explanation:
$g[f(x)] = \left[ {\matrix{
{f(x) + 1} & {f(x) < 0} \cr
{{{(f(x) - 1)}^2} + b} & {f(x) \ge 0} \cr
} } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x + a < 0\& x < 0} \cr {|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr {{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr } } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr {|x - 1| + 1} & {x \in \phi } \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr } } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)} \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr } } \right.$
g(f(x)) is continuous.
At x = $-$a
-a + a + 1 = (-a + a - 1)2 + b
$ \Rightarrow $ 1 = b + 1
$ \Rightarrow $ b = 0
at x = 0
(a $-$1)2 + b = (|0 - 1| - 1)2 + b
$ \Rightarrow $ (a $-$1)2 + b = b
$ \Rightarrow $ a = 1
$ \Rightarrow $ a + b = 1
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x + a < 0\& x < 0} \cr {|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr {{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr } } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr {|x - 1| + 1} & {x \in \phi } \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr } } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)} \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr } } \right.$
g(f(x)) is continuous.
At x = $-$a
-a + a + 1 = (-a + a - 1)2 + b
$ \Rightarrow $ 1 = b + 1
$ \Rightarrow $ b = 0
at x = 0
(a $-$1)2 + b = (|0 - 1| - 1)2 + b
$ \Rightarrow $ (a $-$1)2 + b = b
$ \Rightarrow $ a = 1
$ \Rightarrow $ a + b = 1
2021
Q174
JEE Mains
Numerical
14 Mar 2026
If $\mathop {\lim }\limits_{x \to 0} {{a{e^x} - b\cos x + c{e^{ - x}}} \over {x\sin x}} = 2$, then a + b + c is equal to ____________.
Correct Answer: 4
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{\left\{ {a\left( {1 + x + {{{x^2}} \over {2!}} + .....} \right) - b\left( {1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}}......} \right) + c\left( {1 - x + {{{x^2}} \over {2!}}......} \right)} \right\}} \over {x\left( {x - {{{x^3}} \over {3!}} + .....} \right)}} = 2$
$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{(a - b + c) + x(a - c) + {x^2}\left( {{a \over 2} + {b \over 2} + {c \over 2}} \right) + ....} \over {{x^2}\left( {1 - {{{x^2}} \over 6}....} \right)}} = 2$
For this limit to exist
a $-$ b + c = 0 & a $-$ c = 0
& ${a \over 2} + {b \over 2} + {c \over 2} = 2$
$ \Rightarrow $ a + b + c = 4
$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{(a - b + c) + x(a - c) + {x^2}\left( {{a \over 2} + {b \over 2} + {c \over 2}} \right) + ....} \over {{x^2}\left( {1 - {{{x^2}} \over 6}....} \right)}} = 2$
For this limit to exist
a $-$ b + c = 0 & a $-$ c = 0
& ${a \over 2} + {b \over 2} + {c \over 2} = 2$
$ \Rightarrow $ a + b + c = 4
2021
Q175
JEE Mains
Numerical
14 Mar 2026
A function f is defined on [$-$3, 3] as
$f(x) = \left\{ {\matrix{ {\min \{ |x|,2 - {x^2}\} ,} & { - 2 \le x \le 2} \cr {[|x|],} & {2 < |x| \le 3} \cr } } \right.$ where [x] denotes the greatest integer $ \le $ x. The number of points, where f is not differentiable in ($-$3, 3) is ___________.
$f(x) = \left\{ {\matrix{ {\min \{ |x|,2 - {x^2}\} ,} & { - 2 \le x \le 2} \cr {[|x|],} & {2 < |x| \le 3} \cr } } \right.$ where [x] denotes the greatest integer $ \le $ x. The number of points, where f is not differentiable in ($-$3, 3) is ___________.
Correct Answer: 5
Explanation:
Points of non-differentiability in ($-$3, 3) are at x = $-$2, $-$1, 0, 1, 2.
i.e. 5 points.
2021
Q176
JEE Mains
Numerical
14 Mar 2026
If $\mathop {\lim }\limits_{x \to 0} {{ax - ({e^{4x}} - 1)} \over {ax({e^{4x}} - 1)}}$ exists and is equal to b, then the value of a $-$ 2b is __________.
Correct Answer: 5
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$
Applying L' Hospital Rule
$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$
This is ${{a - 4} \over 0}$.
limit exist only when $a - 4 = 0$ $ \Rightarrow $ a = 4
Applying L' Hospital Rule
$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$
= ${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$
$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$
Applying L' Hospital Rule
$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$
This is ${{a - 4} \over 0}$.
limit exist only when $a - 4 = 0$ $ \Rightarrow $ a = 4
Applying L' Hospital Rule
$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$
= ${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$
$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$
2021
Q177
JEE Mains
Numerical
14 Mar 2026
The number of points, at which the function
f(x) = | 2x + 1 | $-$ 3| x + 2 | + | x2 + x $-$ 2 |, x$\in$R is not differentiable, is __________.
f(x) = | 2x + 1 | $-$ 3| x + 2 | + | x2 + x $-$ 2 |, x$\in$R is not differentiable, is __________.
Correct Answer: 2
Explanation:
$f(x) = |2x + 1| - 3|x + 2| + |{x^2} + x - 2|$
$f(x) = \left\{ {\matrix{ {{x^2} - 7;} & {x > 1} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr {{x^2} + 2x + 3;} & {x < - 2} \cr } } \right.$
$ \therefore $ $f'(x) = \left\{ {\matrix{ {2x;} & {x > 1} \cr {2x - 3;} & { - {1 \over 2} < x < 1} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr {2x + 2;} & {x < - 2} \cr } } \right.$
Check at 1, $-$2 and ${{ - 1} \over 2}$
Non. differentiable at x = 1 and ${{ - 1} \over 2}$
$f(x) = \left\{ {\matrix{ {{x^2} - 7;} & {x > 1} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr {{x^2} + 2x + 3;} & {x < - 2} \cr } } \right.$
$ \therefore $ $f'(x) = \left\{ {\matrix{ {2x;} & {x > 1} \cr {2x - 3;} & { - {1 \over 2} < x < 1} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr {2x + 2;} & {x < - 2} \cr } } \right.$
Check at 1, $-$2 and ${{ - 1} \over 2}$
Non. differentiable at x = 1 and ${{ - 1} \over 2}$
2021
Q178
JEE Mains
Numerical
14 Mar 2026
$\mathop {\lim }\limits_{n \to \infty } \tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right\}$ is equal to ______.
Correct Answer: 1
Explanation:
${\tan ^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)$
$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$
$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$
$ \therefore $ $\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $
$ = {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)$
$ = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)$
$ = {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)$
$ = {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)$
$\mathop {\lim }\limits_{n \to \infty } \tan \left( {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right)$
$ = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}\left( {{n \over {n + 2}}} \right)} \right)$
$ = \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}$
$ = 1$
$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$
$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$
$ \therefore $ $\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $
$ = {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)$
$ = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)$
$ = {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)$
$ = {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)$
$\mathop {\lim }\limits_{n \to \infty } \tan \left( {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right)$
$ = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}\left( {{n \over {n + 2}}} \right)} \right)$
$ = \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}$
$ = 1$
2020
Q179
JEE Mains
MCQ
14 Mar 2026
Let f : R $ \to $ R be a function defined by
f(x) = max {x, x2}. Let S denote the set of all points in R, where f is not differentiable. Then :
f(x) = max {x, x2}. Let S denote the set of all points in R, where f is not differentiable. Then :
A.
{0, 1}
B.
{0}
C.
$\phi $(an empty set)
D.
{1}
2020
Q180
JEE Mains
MCQ
14 Mar 2026
For all twice differentiable functions f : R $ \to $ R,
with f(0) = f(1) = f'(0) = 0
with f(0) = f(1) = f'(0) = 0
A.
f''(x) $ \ne $ 0, at every point x $ \in $ (0, 1)
B.
f''(x) = 0, for some x $ \in $ (0, 1)
C.
f''(0) = 0
D.
f''(x) = 0, at every point x $ \in $ (0, 1)
2020
Q181
JEE Mains
MCQ
14 Mar 2026
$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$
A.
is equal to 0.
B.
is equal to $\sqrt e $.
C.
is equal to 1.
D.
does not exist.
2020
Q182
JEE Mains
MCQ
14 Mar 2026
If the function
$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$ is
twice differentiable, then the ordered pair (k1, k2) is equal to :
$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$ is
twice differentiable, then the ordered pair (k1, k2) is equal to :
A.
$\left( {{1 \over 2},-1} \right)$
B.
(1, 1)
C.
(1, 0)
D.
$\left( {{1 \over 2},1} \right)$
2020
Q183
JEE Mains
MCQ
14 Mar 2026
If $\alpha $ is positive root of the equation, p(x) = x2 - x - 2 = 0, then
$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$ is equal to :
$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$ is equal to :
A.
${1 \over \sqrt2}$
B.
${1 \over 2}$
C.
${3 \over \sqrt2}$
D.
${3 \over 2}$
2020
Q184
JEE Mains
MCQ
14 Mar 2026
Let $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ be a differentiable function such that f(1) = e and
$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$. If f(x) = 1, then x is equal to :
$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$. If f(x) = 1, then x is equal to :
A.
${1 \over e}$
B.
e
C.
${1 \over 2e}$
D.
2e
2020
Q185
JEE Mains
MCQ
14 Mar 2026
The function
$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$ is :
$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$ is :
A.
continuous on R–{–1} and differentiable on R–{–1, 1}
B.
both continuous and differentiable on R–{1}
C.
both continuous and differentiable on R–{–1}
D.
continuous on R–{1} and differentiable on R–{–1, 1}
2020
Q186
JEE Mains
MCQ
14 Mar 2026
$\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}$ ($a$ $ \ne $ 0) is equal to :
A.
$\left( {{2 \over 9}} \right){\left( {{2 \over 3}} \right)^{{1 \over 3}}}$
B.
$\left( {{2 \over 3}} \right){\left( {{2 \over 9}} \right)^{{1 \over 3}}}$
C.
${\left( {{2 \over 3}} \right)^{{4 \over 3}}}$
D.
${\left( {{2 \over 9}} \right)^{{4 \over 3}}}$
2020
Q187
JEE Mains
MCQ
14 Mar 2026
Let [t] denote the greatest integer
$ \le $ t. If for some
$\lambda $ $ \in $ R - {1, 0}, $\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$ = L, then L is equal to :
$\lambda $ $ \in $ R - {1, 0}, $\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$ = L, then L is equal to :
A.
1
B.
2
C.
0
D.
${1 \over 2}$
2020
Q188
JEE Mains
MCQ
14 Mar 2026
$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$ is equal to :
A.
2
B.
1
C.
$e$
D.
$e$2
2020
Q189
JEE Mains
MCQ
14 Mar 2026
If a function f(x) defined by
$f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right.$
be continuous for some $a$, b, c $ \in $ R and f'(0) + f'(2) = e, then the value of of $a$ is :
$f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right.$
be continuous for some $a$, b, c $ \in $ R and f'(0) + f'(2) = e, then the value of of $a$ is :
A.
${e \over {{e^2} - 3e - 13}}$
B.
${1 \over {{e^2} - 3e + 13}}$
C.
${e \over {{e^2} - 3e + 13}}$
D.
${e \over {{e^2} + 3e + 13}}$
2020
Q190
JEE Mains
MCQ
14 Mar 2026
Let [t] denote the greatest integer $ \le $ t
and $\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right] = A$.
Then the function, f(x) = [x2]sin($\pi $x) is discontinuous, when x is equal to :
Then the function, f(x) = [x2]sin($\pi $x) is discontinuous, when x is equal to :
A.
$\sqrt {A + 1} $
B.
$\sqrt {A + 5} $
C.
$\sqrt {A + 21} $
D.
$\sqrt {A} $
2020
Q191
JEE Mains
MCQ
14 Mar 2026
If $f(x) = \left\{ {\matrix{
{{{\sin (a + 2)x + \sin x} \over x};} & {x < 0} \cr
{b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;} & {x = 0} \cr
{{{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{ {1 \over 3}}}} \over {{x^{{4 \over 3}}}}};} & {x > 0} \cr
} } \right.$
is continuous at x = 0, then a + 2b is equal to :
is continuous at x = 0, then a + 2b is equal to :
A.
0
B.
-1
C.
-2
D.
1
2020
Q192
JEE Mains
MCQ
14 Mar 2026
Let ƒ be any function continuous on [a, b] and
twice differentiable on (a, b). If for all x $ \in $ (a, b),
ƒ'(x) > 0 and ƒ''(x) < 0, then for any c $ \in $ (a, b),
${{f(c) - f(a)} \over {f(b) - f(c)}}$ is greater than :
A.
1
B.
${{b - c} \over {c - a}}$
C.
${{b + a} \over {b - a}}$
D.
${{c - a} \over {b - c}}$
2020
Q193
JEE Mains
MCQ
14 Mar 2026
Let S be the set of all functions ƒ : [0,1] $ \to $ R,
which are continuous on [0,1] and differentiable
on (0,1). Then for every ƒ in S, there exists a
c $ \in $ (0,1), depending on ƒ, such that
A.
$\left| {f(c) - f(1)} \right| < \left| {f'(c)} \right|$
B.
$\left| {f(c) + f(1)} \right| < \left( {1 + c} \right)\left| {f'(c)} \right|$
C.
$\left| {f(c) - f(1)} \right| < \left( {1 - c} \right)\left| {f'(c)} \right|$
D.
None
2020
Q194
JEE Mains
MCQ
14 Mar 2026
$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$ is equal to
A.
e
B.
e2
C.
${1 \over {{e^2}}}$
D.
${1 \over e}$
2020
Q195
JEE Mains
Numerical
14 Mar 2026
Let f : R $ \to $ R be defined as
$f\left( x \right) = \left\{ {\matrix{ {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr {0,} & {x = 0} \cr {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr } } \right.$
The value of $\lambda $ for which f ''(0) exists, is _______.
$f\left( x \right) = \left\{ {\matrix{ {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr {0,} & {x = 0} \cr {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr } } \right.$
The value of $\lambda $ for which f ''(0) exists, is _______.
Correct Answer: 5
Explanation:
If g(x) = x5sin$\left( {{1 \over x}} \right)$
and h(x) = x5cos$\left( {{1 \over x}} \right)$
then g''(0) = 0 and h''(0) = 0
So, f''(0+ ) = g''(0+ ) + 10 = 10
and f''(0–) = h''(0–) + 2$\lambda $ = f''(0+)
$ \Rightarrow $ 2$\lambda $ = 10
$ \Rightarrow $ $\lambda $ = 5
and h(x) = x5cos$\left( {{1 \over x}} \right)$
then g''(0) = 0 and h''(0) = 0
So, f''(0+ ) = g''(0+ ) + 10 = 10
and f''(0–) = h''(0–) + 2$\lambda $ = f''(0+)
$ \Rightarrow $ 2$\lambda $ = 10
$ \Rightarrow $ $\lambda $ = 5
2020
Q196
JEE Mains
Numerical
14 Mar 2026
Let $f(x) = x.\left[ {{x \over 2}} \right]$, for -10< x < 10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to _____.
Correct Answer: 8
Explanation:
$x \in ( - 10,10)$
$ \Rightarrow $ ${x \over 2} \in ( - 5,5) \to 9$ integers
check continuity at x = 0
$\left. {\matrix{ f & {(0) = } & 0 \cr f & {({0^ + }) = } & 0 \cr f & {({0^ - }) = } & 0 \cr } } \right\}continuous\,at\,x = 0$
function will be discontinuous when
${x \over 2} = \pm 4, \pm 3, \pm 2, \pm 1$
For example checking continuity at x = 4
$\left. {\matrix{ f & {(4) = } & 4 \cr f & {({4^ + }) = } & 4 \cr f & {({4^ - }) = } & 3 \cr } } \right\}discontinuous\,at\,x = 4$
$ \therefore $ 8 points of discontinuity.
$ \Rightarrow $ ${x \over 2} \in ( - 5,5) \to 9$ integers
check continuity at x = 0
$\left. {\matrix{ f & {(0) = } & 0 \cr f & {({0^ + }) = } & 0 \cr f & {({0^ - }) = } & 0 \cr } } \right\}continuous\,at\,x = 0$
function will be discontinuous when
${x \over 2} = \pm 4, \pm 3, \pm 2, \pm 1$
For example checking continuity at x = 4
$\left. {\matrix{ f & {(4) = } & 4 \cr f & {({4^ + }) = } & 4 \cr f & {({4^ - }) = } & 3 \cr } } \right\}discontinuous\,at\,x = 4$
$ \therefore $ 8 points of discontinuity.
2020
Q197
JEE Mains
Numerical
14 Mar 2026
Suppose a differentiable function f(x) satisfies the identity
f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y.
$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$, then f'(3) is equal to ______.
f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y.
$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$, then f'(3) is equal to ______.
Correct Answer: 10
Explanation:
Given, f(x + y) = f(x) + f(y) + xy2 + x2y ...(1)
differentiating partially with respect to x,
f'(x+y) = f'(x) + 0 + y2 + y(2x) [y = constant]
Put x = 0 and y = x
$ \therefore $ f'(x) = f'(0) + x2 ....(2)
putting x = y = 0 at equation (1),
f(0) = 2f(0)
$ \Rightarrow $ f(0) = 0
Given, $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = 1$
This is in $ \frac{0}{0} $ form, so we can apply L' hospital rule.
$\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over 1} = 1$
$ \Rightarrow f'(0) = 1$
Putting value of f'(0) at equation (2), we get
f'(x) = 1 + x2
$ \therefore $ f'(3) = 1 + 32 = 10
differentiating partially with respect to x,
f'(x+y) = f'(x) + 0 + y2 + y(2x) [y = constant]
Put x = 0 and y = x
$ \therefore $ f'(x) = f'(0) + x2 ....(2)
putting x = y = 0 at equation (1),
f(0) = 2f(0)
$ \Rightarrow $ f(0) = 0
Given, $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = 1$
This is in $ \frac{0}{0} $ form, so we can apply L' hospital rule.
$\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over 1} = 1$
$ \Rightarrow f'(0) = 1$
Putting value of f'(0) at equation (2), we get
f'(x) = 1 + x2
$ \therefore $ f'(3) = 1 + 32 = 10
2020
Q198
JEE Mains
Numerical
14 Mar 2026
If $\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\}$ = 2-k
then the value of k is _______ .
then the value of k is _______ .
Correct Answer: 8
Explanation:
$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\} = {2^{ - k}}$
$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $ = ${2^{ - k}}$
$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$ = 2-k
$ \Rightarrow $ $ {1 \over 8} \times {1 \over {32}} = {2^{ - k}}$
$ \Rightarrow $ 2-8 = 2-k
$ \Rightarrow $ k = 8
$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $ = ${2^{ - k}}$
$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$ = 2-k
$ \Rightarrow $ $ {1 \over 8} \times {1 \over {32}} = {2^{ - k}}$
$ \Rightarrow $ 2-8 = 2-k
$ \Rightarrow $ k = 8
2020
Q199
JEE Mains
Numerical
14 Mar 2026
If $\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$ = 820,
(n $ \in $ N) then the value of n is equal to _______.
(n $ \in $ N) then the value of n is equal to _______.
Correct Answer: 40
Explanation:
$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$ = 820
As it is $\left( {{0 \over 0}} \right)$ form, Apply L'Hospital's Rule.
$\mathop {\lim }\limits_{x \to 1} \left( {{{1 + 2x + 3{x^2} + ... + n{x^{n - 1}}} \over 1}} \right)$ = 820
$ \Rightarrow $ 1 + 2 + 3 + .....+ n = 820
$ \Rightarrow $ ${{n\left( {n + 1} \right)} \over 2}$ = 820
$ \Rightarrow $ n2 + n – 1640 = 0
$ \Rightarrow $ (n – 40)(n + 41) = 0
Since n $ \in $ N, so n = 40.
As it is $\left( {{0 \over 0}} \right)$ form, Apply L'Hospital's Rule.
$\mathop {\lim }\limits_{x \to 1} \left( {{{1 + 2x + 3{x^2} + ... + n{x^{n - 1}}} \over 1}} \right)$ = 820
$ \Rightarrow $ 1 + 2 + 3 + .....+ n = 820
$ \Rightarrow $ ${{n\left( {n + 1} \right)} \over 2}$ = 820
$ \Rightarrow $ n2 + n – 1640 = 0
$ \Rightarrow $ (n – 40)(n + 41) = 0
Since n $ \in $ N, so n = 40.
2020
Q200
JEE Mains
Numerical
14 Mar 2026
If the function ƒ defined on $\left( { - {1 \over 3},{1 \over 3}} \right)$ by
f(x) = $\left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr {k,} & {when\,x = 0} \cr } } \right.$
is continuous, then k is equal to_______.
f(x) = $\left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr {k,} & {when\,x = 0} \cr } } \right.$
is continuous, then k is equal to_______.
Correct Answer: 5
Explanation:
$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$
= $\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$
= $\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$
= 3 + 2 = 5
f(x) is continuous
$ \therefore $ $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ = f(0)
So f(0) = 5 = k
= $\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$
= $\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$
= 3 + 2 = 5
f(x) is continuous
$ \therefore $ $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ = f(0)
So f(0) = 5 = k