datavorous/entrance-exam-dataset huggingface dataset:
If $\int f(x) d x=F(x)+C$, then $\frac{d}{d t} \int_{g(t)}^{h(t)} f(x) d x=$

Options:
A $f(h(t))-f((t))$
B $F(h(t))-F(g(t))$
C $F(h(t)) h^{\prime}(t)-F(g(t)) g^{\prime}(t)$
D $f(h(t)) h^{\prime}(t)-f(g(t)) g^{\prime}(t)$

The correct answer is:
$f(h(t)) h^{\prime}(t)-f(g(t)) g^{\prime}(t)$

Solution:
$\frac{d}{d t} \int_{g(t)}^{h(t)} f(x) d x=f(h(t)) \cdot h^{\prime}(t)-f\left(g(t) \cdot g^{\prime}(t)\right.$
(by Newton - Leibnitz theorem)

Topic: Mathematics, Definite Integration, JEE Main
If ∫f(x)dx=ψ(x), then ∫x5f(x3)dx, is equal to

Options:
A 13x3ψ(x3)−∫x2ψ(x3)dx+c
B 13[x3ψ(x3)−∫x3ψ(x3)dx]+c
C 13[x3ψ(x3)−∫x2ψ(x3)dx]+c
D 13x3ψ(x3)−3∫x3ψ(x3)dx+c

The correct answer is:
13x3ψ(x3)−∫x2ψ(x3)dx+c

Solution:
∫x5fx3dx =∫x3.x2fx3dx

Applying integration by parts, we get
=x3∫x2fx3dx-∫ddxx3∫x2fx3dx
=13x3ψx3-∫x2ψx3dx+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2013 (07 Apr)
If fx=e1+1x-ae1x+1:xâ 0 b:x=0 (where a and b are arbitrary constants) is continuous at x=0, then the value of a2 is equal to

(use e=2.7)

Options:

The correct answer is:
7.29

Solution:
At x=0,
LHL=limh→0+f0-h=limh→0+e1-1h-ae-1h+1=0-a0+1=-a
RHL=limh→0+f0+h=limh→0+e1+1h-ae1h+1=limh→0e-e-1h×a1+e-1h=e
and f0=b
∵fx is continuous at x=0
∴-a=e=b⇒a=-e
a2=e2

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{array}{r}\frac{x}{e^{1 / x}+1}, \text { when } x \neq 0 \\ 0, \text { when } x=0\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 0^+} f(x)=1$
B $\lim _{x \rightarrow 0^-} f(x)=1$
C $f(x)$ is continuous at $x=0$
D None of these

The correct answer is:
$f(x)$ is continuous at $x=0$

Solution:
$f(0)=0$
$f(0-)=\lim _{h \rightarrow 0} \frac{-h}{e^{-1 / h}+1}=\lim _{h \rightarrow 0} \frac{-h}{1+\frac{1}{e^{1 / h}}}=0$
$f(0+)=\lim _{h \rightarrow 0} \frac{h}{e^{1 / h}+1}=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=e-1x2 ∀x≠0 and f0=0, then f ′ 0 is

Options:
A not defined
B 1
C 0
D 2

The correct answer is:
0

Solution:
Rf'0=limh→0e-10+h2−0h=limh→01he−1h2
=limh→0⁡1he1h2
= lim h→0 − 1 h 2 e 1 h 2 ⋅ − 2 h 3
=limh→0⁡h2e1h2=0∞=0
Lf'0=limh→0e−10−h2−0−h=limh→0−1he−1h2
=limh→0⁡-1he1h2
=limh→0⁡1h2e1h2-2h3
=limh→0⁡h-2e1h2=0∞=0

Hence, f'0=0

Topic: Mathematics, Application of Derivatives, JEE Main
If fx=e2x-1e2x+1:xâ 0 0:x=0, then fx is

Options:
A Continuous as well as differentiable at x=0
B Continuous but not differentiable at x=0
C Differentiable but not continuous at x=0
D None of these

The correct answer is:
None of these

Solution:
limhâ0+â¡f0-h=limhâ0+â¡e-2h-1e-2h+1=0-10+1=-1
and limhâ0+â¡f0+h=limhâ0+â¡e2h-1e2h+1=limhâ0+â¡1-e-2h1+e-2h=1-01+0=1
Since, limhâ0+â¡f0-hâ limhâ0+â¡f0+h, therefore, fx is not continuous and not differentiable at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=e2x+2x+1-12x+2x+1:xâ 01:x=0, then (where, . represents the greatest integer function)

Options:
A limx→0+⁡⁡fx=1
B limx→0-⁡⁡fx=e-1
C fx is continuous at x=0
D fx is discontinuous at x=0

The correct answer is:
fx is discontinuous at x=0

Solution:
limxâ0+â¡fxâ¡=limhâ0+â¡f0+h
=limhâ0+â¡e2h+2h+1-12h+2h+1
=limhâ0+â¡e2h+1-12h+1=e-1
And, limxâ0-â¡fx=limhâ0+â¡f0-h
=limhâ0+â¡e-2h-2h+1-1-2h-2h+1
=limhâ0+â¡e-2h-1-2h=1
Since, limxâ0+â¡fxâ¡â limxâ0-â¡fx, therefore, fx is not continuous at x=0

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=e2x+2x-12x+2x:x≠01:x=0, then (where, [.] represents the greatest integer function)

Options:
A limx→0+fx=-1
B limx→0-fx=1e-1
C fx is differentiable at x=0
D fx is discontinuous at x=0

The correct answer is:
fx is discontinuous at x=0

Solution:
limxâ0+fx=limxâ0+f0+h
=limhâ0+e2h+2h-12h+2h
=limhâ0+e2h-12h=1
and limxâ0-fx=limhâ0+f0-h
=limhâ0e-2h-2h-1-2h-2h
=limhâ0e-1-2h-1-1-2h=e-1-1-1
=1-1e
Since, limxâ0+fxâ limxâ0-fx, therefore, fx is not continuous at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=e2x2+x: x>0ax+b: x≤0 is differentiable at x=0, then

Options:
A a=1,b=-1
B a=-1,b=1
C a=1,b=1
D a=-1,b=-1

The correct answer is:
a=1,b=1

Solution:
f ′ 0 − = lim h→ 0 + f 0−h −f 0 −h ⇒ lim h→ 0 + f −ah+b − b −h =a
f'0+=limh→0+f0+h−f0h⇒limh→0+e2h2+h−bh ... 1
For limit to exists, it must be 00 form
∴e20+0-b=0⇒b=1 ... 2
From 1, f'0+=limh→0+e2h2+h−1h
=limh→0+e2h2+h−12h2+h×2h2+hh=1×1=1 ... 3
∵ fx is differentiable at x=0
∴ f'0−=f'0+
⇒a=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=e2x3+xx>0ax+bx≤0 is differentiable at x=0, then

Options:
A a=1,b=-1
B a=-1,b=1
C a=1,b=1
D a=-1,b=-1

The correct answer is:
a=1,b=1

Solution:
f'0â=limhâ0+f0âhâf0âhâlimhâ0+âah+bâbâh=a
f'0+=limhâ0+f0+hâf0hâlimhâ0+e2h3+hâbhâ¦â¦i
For this limit to exist, it must be of 00 form.
â´e20+0-b=0âb=1â¦â¦..ii
From i

f'0+=limhâ0+e2h3+hâ1hâlimhâ0+e2h3+hâ12h3+hÃ2h3+hh
â1Ã1=1
âµf'0â=f'0+
âa=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=\left\{\begin{aligned} e^x+a x, & x \lt 0 \\ b(x-1)^2, & x \geq 0\end{aligned}\right.$ is differentiable at $x=0$, then $(a, b)$ is

Options:
A $(-3,-1)$
B $(-3,1)$
C $(3,1)$
D $(3,-1)$

The correct answer is:
$(-3,1)$

Solution:
Given $f(x)$ is differentiable at $x=0$. Hence, $f(x)$ will be continuous at $x=0$.
$\begin{aligned} & \lim _{x \rightarrow 0^{-}}\left(e^x+a x\right)=\lim _{x \rightarrow 0^{+}} b(x-1)^2 \\ \Rightarrow & e^0+a \times 0=b(0-1)^2 \Rightarrow b=1...(i)\end{aligned}$
But $f(x)$ is differentiable at $x=0$, then
$\begin{aligned} & L f^{\prime}(x)=R f^{\prime}(x) \Rightarrow \frac{d}{d x}\left(e^x+a x\right)=\frac{d}{d x} b(x-1)^2 \\ & \Rightarrow e^x+a=2 b(x-1) \\ & \text { At } x=0, e^0+a=-2 b \Rightarrow a+1=-2 b \Rightarrow a=-3 \\ & \Rightarrow(a, b)=(-3,1) .\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=e^{x} g(x), g(0)=4, g^{\prime}(0)=2$, then $f^{\prime}(0)=$

Options:
A 4
B 6
C 1
D 2

The correct answer is:
6

Solution:
Given $f(x)=e^{x} g(x)$
$$
\begin{aligned}
\therefore f^{\prime}(x) &=e^{x} g^{\prime}(x)+g(x) \cdot e^{x} \\
\therefore f^{\prime}(0) &=e^{0} g^{\prime}(0)+g(0) \cdot e^{0} \\
&=2+4=6
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\left\{\begin{array}{r}e^x ; \quad x \leq 0 \\ |1-x| ; \quad x\gt0\end{array}\right.$, then

Options:
A $f(x)$ is differentiable at $x=0$
B $f(x)$ is continuous at $x=0$
C $f(x)$ is differentiable at $x=1$
D $f(x)$ is continuous at $x=1$

The correct answers are:
$f(x)$ is continuous at $x=0$, $f(x)$ is continuous at $x=1$

Solution:
$f(x)=\left\{\begin{array}{l}e^x ; x \leq 0 \\ 1-x ; 0 \lt x \leq 1 \\ x-1 ; x\gt1\end{array}\right.$
$\begin{aligned}& R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{1-h-1}{h}=-1 \\
& L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{-h}=1\end{aligned}$
So, it is not differentiable at $x=0$.
Similarly, it is not differentiable at $x=1$.
But it is continous at $x=0,1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=ex+x-1x+x:xâ 0-1:x=0 (where . denotes the greatest integer function), then

Options:
A fx is continuous at x=0
B limx→0+fx=-1
C limx→0-fx=1
D limx→0+fx=1

The correct answer is:
limx→0+fx=1

Solution:
fx=ex+x-1x+xx≠0-1x=0
limx→0-fx=limx→0-ex+x-1x+x=e-1-1-1=e-1e
limx→0+fx=limx→0+ex+x-1x+x=limx→0+ex-1x=1
∵LHL≠RHL at x=0⇒fx is discontinuous at x=0

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)=f(a-x)$ then $\int_0^a x f(x) d x$ is equal to

Options:
A $\int_0^a f(x) d x$
B $\frac{a^2}{2} \int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$
C $\frac{a}{2} \int_0^a f(x) d x$
D $-\frac{a}{2} \int_0^{\mathrm{a}} f(x) d x$

The correct answer is:
$\frac{a}{2} \int_0^a f(x) d x$

Solution:
Hints : $f(x)=f(a-x), I=\int_0^a x f(x) d x=\int_0^a(a-x) f(a-x) d x$
$$
\begin{aligned}
& =\int_0^a(a-x) f(x) d x=a \int_0^a f(x) d x-I \\
& \therefore 2 I=a \int_0^a f(x) d x \Rightarrow I=\frac{a}{2} \int_0^a f(x) d x
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
If fx=gx–1x–1;x≠1 1;x=1 and g'1=2, g1=1, then limx→1⁡fx is equal to

Options:
A 1
B 3
C 2
D 4

The correct answer is:
2

Solution:
limx→1⁡fx= limx→1g'(x)2 gx12x
= limx→1g'x .xgx= g'1 .1g1=2

Topic: Mathematics, Limits, JEE Main
If fx is a continuous function and ∫x2x4t3ftdt=sin2πx, then f1 is equal to

Options:
A 1
B –1
C π
D -π

The correct answer is:
π

Solution:
Differentiating using Leibnitz rule, we get,
x12fx44x3-x6f(x2)2x=2πcosπx
Putting x=1
⇒4f1-2f1=2π
⇒2f1=2π⇒f1=π

Topic: Mathematics, Definite Integration, JEE Main
If fx is a continuous function satisfying fx=f2-x, then the value of the integral I=∫-33f1+xln⁡2+x2-xdx is equal to

Options:
A 3π
B 6π
C 0
D 9π

The correct answer is:
0

Solution:
As fx=f2-x
Puting x→x+1, we get,

fx+1=f1-x
Applying a+b-x property in I, we get,

I=∫-33f1-xln⁡2-x2+xdx
⇒I=-∫-33f1+xln⁡2+x2-xdx
⇒I=-I⇒I=0

Topic: Mathematics, Definite Integration, JEE Main
If fx is a continuous function such that its value ∀x∈R is a rational number and f1+f2=6, then the value of f3 is equal to

Options:
A 3
B 9
C 2
D 4

The correct answer is:
3

Solution:
Since fx is continuous and attains only rational values, then by intermediate value theorem, fx is a constant function.

i.e. f1=f2=3

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)$ is a continuous periodic function with period $T$, then the integral $I=\int_a^{a+\tau} f(x) d x$ is

Options:
A Equal to 2a
B Equal to 3a
C Independent of a
D None of these

The correct answer is:
Independent of a

Solution:
Consider the function $g(a)=\int_a^{a+T} f(x) d x$
$=\int_a^0 f(x) d x+\int_0^T f(x) d x+\int_T^{a+T} f(x) d x$
Putting $x-T=y$ in last integral, we get $\int_T^{a+T} f(x) d x=\int_0^a f(y+T) d y=\int_0^a f(y) d y$
$\Rightarrow g(a)=\int_a^0 f(x) d x+\int_0^T f(x) d x+\int_T^{a+T} f(x) d x=\int_0^T f(x) d x$
Hence $g(a)$ is independent of $a$.

Topic: Mathematics, Definite Integration, JEE Main
If fx is a differentiable function in the interval 0, ∞ such that f1=1 and limt→x⁡t2 fx-x2ftt-x=1,for each x>0, then f32 is equal to

Options:
A 2318
B 136
C 259
D 3118

The correct answer is:
3118

Solution:
Let L= limt→x⁡t2fx-x2ftt-x=1
Applying L'Hospital Rule
L= limt→x⁡2t fx-x2f′t1=1
Or 2x fx-x2f′x=1 ⇒ dydx+-2xy=-1x2..........(1)
Solving the above linear differential equation we get

Integrating factor =e∫-2xdx=e-2lnx=1x2

After multiplying the equation 1 by 1x2, and simplifying the equation 1 becomes,

ddxyx2=-1x4

⇒yx2=∫-1x4dx=13x3+c

⇒y=13x+cx2

Putting x=1, y=1, we get

c=23.
∴fx=23x2+13x
Put x=32
f32=23 ×322+23×3
=32+29=27+418=3118.

Topic: Mathematics, Limits, JEE Main, JEE Main 2016 (09 Apr Online)
If fx is a differentiable function satisfying f'xâ¤2 âxâ0,4 and f0=0, then

Options:
A fx=18 has no solution in xâ0,4
B fx=18 has more than 2 solutions in x∈0,4
C fx=14 has 3 solutions in xâ0,4
D fx=20 has 2 solutions in xâ0,4

The correct answer is:
fx=18 has no solution in x∈0,4

Solution:
Applying LMVT in x∈0,t for fx, we get,

f'c=ft−f0t−0⇒f'c=ftt≤2
⇒ft≤2t as t∈0,4

∴ft≤8⇒-8≤ft≤8

Topic: Mathematics, Application of Derivatives, JEE Main
If fx is a differentiable function satisfying f'x≤4 ∀x∈0,4 and f0=0, then

Options:
A fx=18 has no solution in x∈0,4
B fx=18 has more than 2 solutions in x∈0,4
C fx=14 has no solution in x∈0,4
D fx=20 has 2 solution in x∈0,4

The correct answer is:
fx=18 has no solution in x∈0,4

Solution:
Applying LMVT in x∈0,t for fx, we get
f ′ c = f t −f 0 t−0
⇒ f ′ c = f t t ≤4
⇒ft≤4t
As t∈0,4
∴ft≤16

Topic: Mathematics, Application of Derivatives, JEE Main
If fx is a differentiable function such that ∫fxdx=2fx2+C, (where, C is the constant of integration) and f1=1/4, then fπ equals

Options:
A π
B π 4
C 23
D π6

The correct answer is:
π 4

Solution:
Differentiating w.r.t. ‘x’ we get f x =4f x · f ′ x
f ′ x =0 or 14=fx
Integrating, we have
∫ 1 4 dx=∫ f ′ x dx
⇒x4=fx+C
As f1=14→C=0
⇒fx=x4
⇒fπ=π4

Topic: Mathematics, Indefinite Integration, JEE Main
If fx is a non-negative function such that the area bounded by y=fx, x-axis and the lines x=0 and x=Î± is 4Î±sinâ¡Î±+2 sq. units âÎ±â0,Ï, then the value of fÏ2 is equal to

Options:
A 0
B 4
C 2π
D 8π

The correct answer is:
4

Solution:
Given, ∫0αfxdx=4αsin⁡α+2
On differentiating with respect to α, we get,

f α =4α⋅cosα+4sinα
∴fπ2=4

Topic: Mathematics, Area Under Curves, JEE Main
If f(x) is a non-zero polynomial of degree four, having local extreme points at x= –1, 0, 1; then the set S={x∈R :fx=f0} contains exactly

Options:
A Two irrational and two rational numbers
B Four rational numbers
C Two irrational and one rational number
D Four irrational numbers

The correct answer is:
Two irrational and one rational number

Solution:
f(x) has extreme points at {-1, 0, 1}

⇒f'x=0 at x= –1, 0, 1

⇒f'x=ax2-1x ∵f x is a polynomial

⇒fx=∫ax3-xdx=ax44-x22+ C

Now the equation,

fx=f(0)

⇒ax44-x22+ C=C
⇒ x44-x22=0 ⇒x2x2-2=0
⇒x=0, ±2

⇒S= {0, 2, -2}.

Hence, the set contains two irrational and one rational number.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If fx is a polynomial satisfying fx f1x=fx+f1x and f2>1, then lim x â 1 f x is

Options:

The correct answer is:
2

Solution:
As given fx .f1x=fx+f1x, f2,
âf(x)f( 1 x )âf(x)âf( 1 x )=0
â( f(x)â1 )( f( 1 x )â1 )=1
âh(x)h( 1 x )=1 as it f( x ) is a polynomial,hence h( x ) will also be a polynomial
âh( x )=Â± x n
âf( x )=Â± x n +1
Hence f( x )=Â± x n +1
Now f( 2 )>1
âf(x)= x n +1
hence lim xâ1 f( x )= lim xâ1 ( x n +1 )=2

Topic: Mathematics, Limits, JEE Main
If $f(x)$ is a quadratic expression such that $f(1)+f$ (2) $=0$, and $-1$ is a root of $f(x)=0$, then the other root of $f(x)=0$ is

Options:
A
$-\frac{5}{8}$

B
$-\frac{8}{5}$

C
$\frac{5}{8}$

D
$\frac{8}{5}$

The correct answer is:
$\frac{8}{5}$

Solution:
If $a$ and $-1$ are the roots of the polynomial, then we get
$$
\begin{aligned}
& f(x)=x^2+(1-a) x-a . \\
\therefore \quad & f(1)=2-2 a \\
\text { and } & f(2)=6-3 a \\
\text { As, } & f(1)+f(2)=0 \\
\Rightarrow & 2-2 a+6-3 a=0 \Rightarrow a=\frac{8}{5}
\end{aligned}
$$
Therefore, the other root is $\frac{8}{5}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If fx is a quadratic polynomial with vertex V1,α, then the integral I=∫02efxefx+ef2-xdx is equal to

Options:

The correct answer is:
1

Solution:
As vertex is V1,α, thus the quadratic is symmetric about x=1

⇒f1+x=f1-x
Replacing x by x-1, we get,

fx=f2-x
Applying a+b-x in I and adding, we get,
2I=∫02efx+ef2-xefx+ef2-xdx
⇒2I=∫021dx=x02
⇒2I=2-0
⇒I=1

Topic: Mathematics, Definite Integration, JEE Main
If fx is a real valued function such that fx+6-fx+3+fx=0, âxâR, then period of fx is

Options:
A 6
B 12
C 18
D 24

The correct answer is:
18

Solution:
In given equation, putting xâx+3
fx+9-fx+6+fx+3=0
On adding both equations, we get,
fx+9+fx=0 â¦(i)
now, putting xâx+9
fx+18+fx+9=0 â¦(ii)
Equation (ii) â Equation (i)
fx+18=fx
â´Period =18

Topic: Mathematics, Functions, JEE Main
If fx is a twice differentiable function such that f0=f1=f2=0, then

Options:
A fx=0 has exactly 3 roots
B f ′ x =0 for atleast 3 real values of x
C f ″ x =0 for atleast 2 real values of x
D f â³ x =0 for atleast 1 real value of x

The correct answer is:
f ″ x =0 for atleast 1 real value of x

Solution:
fx=0 has atleast 3 real roots
∴ By Rolle’s theorem f ′ x =0 has atleast two real roots and f''x=0 has atleast one real root.

Topic: Mathematics, Application of Derivatives, JEE Main
If fx is a twice differentiable function such that f"x=-fx, f'x=gx, hx=f2x+g2x and h10=10, then h5 is equal to

Options:
A 5
B 15
C 10
D 17

The correct answer is:
10

Solution:
h'x=2f⋅f'+2gg'
=2f⋅f'+2f'f" (∵g=f')
=2f⋅f'+2f'-f=0
∴h'x=0⇒hx is a constant function
∴ hx=10⇒ h5=10

Topic: Mathematics, Differentiation, JEE Main
If fx is continuous and f92=29, then limx→0f1-cos3xx2 equals to

Options:
A 8 9
B 0
C 2 9
D 9 2

The correct answer is:
2 9

Solution:
Given: f92=29 and fx is continuous.
limx→0f1-cos3xx2=flimx→01-cos3xx2
=flimx→02sin23x2x2 ∵ 1-cos2θ=2sin2θ
=flimx→02sin23x2493x22
=f92limx→0sin23x23x22
=f92×12 ∵ limx→0sinxx=1
=f92
=29

Topic: Mathematics, Limits, JEE Main, JEE Main 2014 (09 Apr Online)
If fx is continuous in 0,1 and f13=12, then the value oflimn→∞⁡fn3n+1 is equal to

Options:
A 2
B 3
C 12
D None of these

The correct answer is:
12

Solution:
Since, fx is continuous in 0,1 , therefore,
limnâââ¡fn3n+1=flimnâââ¡n3n+1
=f13=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $\mathrm{f}(x)$ is continuous on its domain $[-2,2]$, where
$$
\mathrm{f}(x)= \begin{cases}\frac{\sin a x}{x}+3 & , \text { for }-2 \leq x < 0 \\ 2 x+7 & , \text { for } 0 \leq x \leq 1 \\ \sqrt{x^2+8}-\mathrm{b}, & \text { for } 1 < x \leq 2\end{cases}
$$
then the value of $2 a+3 b$ is

Options:
A $-12$
B $-10$
C 10
D 12

The correct answer is:
$-10$

Solution:
Since $\mathrm{f}(x)$ is continuous in $[-2,2]$, it is continuous at $x=0$ and $x=1$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(\frac{\sin a x}{x}+3\right)=\lim _{x \rightarrow 0^{+}}(2 x+7)$
$\begin{aligned} & \Rightarrow a+3=0+7 \\ & \Rightarrow a=4\end{aligned}$
Also, $\lim _{x \rightarrow 1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 1^{+}} \mathrm{f}(x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(2 x+7)=\lim _{x \rightarrow 1^{+}}\left(\sqrt{x^2+8}-b\right)$
$\begin{aligned} & \Rightarrow 2(1)+7=\sqrt{1+8}-b \\ & \Rightarrow 9=3-b \\ & \Rightarrow b=-6\end{aligned}$
$\therefore \quad 2 a+3 b=8-18=-10$

Topic: Mathematics, Continuity and Differentiability, MHT CET, MHT CET 2023 (13 May Shift 2)
If $\mathrm{f}(\mathrm{x})$ is differentiable everywhere,then which one of the following is correct?

Options:
A $|\mathrm{f}|$ is differentiable everywhere
B $|\mathrm{f}|^{2}$ is differentiable everywhere
C $\mathrm{f}|\mathrm{f}|$ is not differentiable at some points
D $\text { None of the above }$

The correct answer is:
$\mathrm{f}|\mathrm{f}|$ is not differentiable at some points

Solution:
If $\mathrm{f}(\mathrm{x})$ is differential everywhere then $\mid \mathrm{f}$ | is not differentiable at some point, so, $\mathrm{f}|\mathrm{f}|$ is not differentiable at some point. [Example: $\mathrm{f}(\mathrm{x})=\mathrm{x}$ is differentiable everywhere but $|\mathrm{f}(\mathrm{x})|=|\mathrm{x}|$ is not differentiable at $\mathrm{x}=0]$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $f(x)$ is periodic function with period $T$ then the function $f(a x+b)$ where $a\gt0$, is periodic with period

Options:
A $T / b$
B $a T$
C $b T$
D $T / a$

The correct answer is:
$T / a$

Solution:
It is a fundamental concept

Topic: Mathematics, Functions, JEE Main
If f(x) is the antiderivative of 1+2tanx(tanx+secx)12 and fπ6=log2, then the value of f(0) is

Options:

The correct answer is:
0

Solution:
fx=∫1+2tan⁡xtan⁡x+sec⁡x12dx‍
=∫1+2tan2⁡x+2tan⁡xsec⁡x12dx‍
=∫1+tan2⁡x+tan2⁡x+2tan⁡xsec⁡x12dx‍
=∫sec2⁡x+tan2⁡x+2tan⁡xsec⁡x12dx‍
=∫secx+tanx212dx‍
=log⁡secx+tanx+log⁡sec⁡x+c
=logsecxsecx+tanx+c
fπ6=log2⇒log2323+13+c=log2
⇒c=0
fx=log⁡sec⁡xsec⁡x+tan⁡x
⇒f0=log11+0=0

Topic: Mathematics, Indefinite Integration, JEE Main
If $f(x)$ is twice differentiable function such that $f(a)=0, f(b)=2, f(c)=1, f(d)=2$, $f(e)=0$, where $a < b < c < d < e$, then the minimum number of zeros of $g(x)=\left\{f^{\prime}(x)\right\}^2+f^{\prime \prime}(x) \cdot f(x)$ in the interval $[a, e]$ is

Options:

The correct answer is:
6

Solution:
Let, $g(x)=\frac{d}{d x}\left[f(x) \cdot f^{\prime}(x)\right]$ to get the zero of $g(x)$ we take function $h(x)=f(x) \cdot f^{\prime}(x)$ between any two roots of $h(x)$ there lies atleast one root of $h^{\prime}(x)=0$
$$
\begin{aligned}
& \Rightarrow \quad g(x)=0 \\
& \Rightarrow \quad h(x)=0 \\
& \Rightarrow \quad f(x)=0 \text { or } f^{\prime}(x)=0 \\
& \text { If } f(x)=0 \text { has } 4 \text { minimum solutions, } \\
& f^{\prime}(x)=0 \text { has } 3 \text { minimum solutions, } \\
& h(x)=0 \text { has } 7 \text { minimum solutions, then } \\
& h^{\prime}(x)=g(x)=0 \text { has } 6 \text { minimum solutions. } \\
&
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
If $f(x)$ is twice differentiable polynomial function such that $f(1)=1, f(2)=-4, f(3)=9$, then

Options:
A $f^{\prime \prime}(x)=2, \forall x \in R$
B There exist at least one $x \in(1,3)$ such that $f^{\prime \prime}(x)=2$
C There exist at least one $x \in(2,3)$ such that $f^{\prime}(x)=5=f^{\prime \prime}(x)$
D There exist at least one $x \in(1,2)$ such that $f(x)=3$

The correct answer is:
There exist at least one $x \in(1,3)$ such that $f^{\prime \prime}(x)=2$

Solution:
Let a function be $g(x)=f(x)-x^2$
$\Rightarrow g(x)$ has at least 3 real roots which are $x=1,2,3$
$\Rightarrow g^{\prime}(x)$ has at least 2 real roots in $x \in(1,3)$
$\Rightarrow g^{\prime \prime}(x)$ has at least 1 real roots in $x \in(1,3)$
$\Rightarrow f^{\prime \prime}(x)=2$ for at least one $x \in(1,3)$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If fx=∏k=1999x2-47x+k, then product of all real roots of fx=0 is

Options:
A 550!
B 551!
C 552!
D 999!

The correct answer is:
552!

Solution:
Consider x2-47x+k=0
For real roots, 472-4kâ¥0âkâ¤552
â´ k=1, 2, 3â¦.,552
Product of real roots =1Ã2Ã3Ã4Ãâ¦..Ã552=552!

Topic: Mathematics, Quadratic Equation, JEE Main
If $f(x)=k x^3-9 x^2+9 x+3$ is monotonically increasing in each interval, then

Options:
A $k \lt 3$
B $k \leq 3$
C $k\gt3$
D None of these

The correct answer is:
$k\gt3$

Solution:
$f^{\prime}(x)=3 k x^2-18 x+9=3\left[k x^2-6 x+3\right]\gt0, \forall x \in R$
$\therefore \Delta=b^2-4 a c \lt 0, k\gt0$ i.e., $36-12 k \lt 0$ or $k\gt3$

Topic: Mathematics, Application of Derivatives, JEE Main
If $f(x)=k x^{3}-9 x^{2}+9 x+3$ is monotonically increasing in every interval, then which one of the following is correct?

Options:
A $k < 3$
B $k \leq 3$
C $k>3$
D $k \geq 3$

The correct answer is:
$k>3$

Solution:
Given $f(x)=k x^{3}-9 x^{2}+9 x+3$ On differentiating w.r.t. $x$, we get $f^{\prime}(x)=3 k x^{2}-18 x+9$
For a function to be monotonically increasing. $b^{2}-4 a c < 0$
Here, $a=3 k, b=-18, c=9$
$\therefore^{2}-4 a c=(-18)^{2}-4(3 k)(9)=(-18)(-18)-(3 k) 18 \times 2$
$\Rightarrow 36-12 k < 0$
$\Rightarrow k>3$

Topic: Mathematics, Application of Derivatives, JEE Main
If $f(x)=\log _{5} \log _{3} x,$ then $f^{\prime}(e)$ is equal to

Options:
A $e \log _{e} 5$
B $e \log _{4} 3$
C $\frac{1}{e \log _{e} 5}$
D $\frac{1}{e \log _{e} 3}$

The correct answer is:
$\frac{1}{e \log _{e} 5}$

Solution:
We have,
$\therefore$
$\begin{aligned} f(x) &=\log _{5} \log _{3} x \\ &=\frac{\log \log _{3} x}{\log 5} \\ &=\frac{\log \left(\frac{\log x}{\log 3}\right)}{\log 5} \\ &=\frac{\log \log x-\log \log 3}{\log 5} \\ f^{\prime}(x)=\frac{1}{\log 5} \cdot \frac{1}{\log x} \cdot \frac{1}{x} \end{aligned}$
$\begin{aligned} f^{\prime}(e) &=\frac{1}{e \log 5 \cdot \log e} \\ &=\frac{1}{e \log 5} \quad[\because \log e=1] \\ &=\frac{1}{e \log _{n} 5} \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If $f(x)=\log _a x$ and $F(x)=a^x$, then $F[f(x)]$ is

Options:
A $f[F(x)]$
B $f[F(2 x)]$
C $F[f(2 x)]$
D $F[(x)]$

The correct answer is:
$f[F(x)]$

Solution:
$\begin{aligned} & F[f(x)]=F\left(\log _a x\right)=a^{\log _a x}=x \\ & f[F(x)]=f\left(a^x\right)=\log _a a^x=x \log _a a=x\end{aligned}$

Topic: Mathematics, Functions, JEE Main