datavorous/entrance-exam-dataset huggingface dataset:
Let f:1, 3→R, be a function defined by fx=xx1+x2, where x, denotes the greatest integer ≤x. Then the range of f, is

Options:
A 25, 35∪34, 45
B 25, 12∪35, 45
C 25, 45
D 35, 45

The correct answer is:
25, 12∪35, 45

Solution:
f :1,3→R, fx=xx1+x2

fx=x1+x2,x∈1, 22x1+x2,x∈[2, 3)

f'x=1+x21-x2x1+x22,x∈1, 21+x22-2x2x1+x22,x∈[2,3)

f'x=1-x21+x2,x∈1, 21-2x21+x2,x∈[2,3)

∴ fx is decreasing function.
∴ Rf ∈25, 12∪610, 45

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let $f:[1,3] \rightarrow R$ be a function satisfying $\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}$, for all $x \neq 2$ and $f(2)=1$, where $R$ is the set of all real numbers and $[x]$ denotes the largest integer less than or equal to $x$.
Statement 1: $\lim _{x \rightarrow 2^{-}} f(x)$ exists.
Statement 2: $f$ is continuous at $x=2$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is false, Statement 2 is true.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

D
Statement 1 is true, Statement 2 is false.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
$$
\begin{aligned}
& \text { Consider } \frac{x}{[x]} \leq f(x) \leq \sqrt{6-x} \\
& \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{x}{[x]}=\frac{2}{1}=2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 2^{-}} \sqrt{6-x}=2 \\
& \therefore \lim _{x \rightarrow 2^{-}} f(x)=2
\end{aligned}
$$
[By Sandwich theorem]
Now $\lim _{x \rightarrow 2^{+}} \frac{x}{[x]}=1, \lim _{x \rightarrow 2^{+}} \sqrt{6-x}=2$
Hence by Sandwich theorem $\lim _{x \rightarrow 2^{+}} f(x)$ does not exists.
Therefore $f$ is not continuous at $x=2$. Thus statement-1 is true but statement- 2 is not true

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f:[1,3] \rightarrow R$ be a function satisfying $\frac{x}{[x]} \leq f(x) \leq \sqrt{6-x}$, for all $x \neq 2$ and $f(2)=1$, where $R$ is the set of all real numbers and $[x]$ denotes the largest integer less than or equal to $x$.
Statement 1: $\lim _{x \rightarrow 2^{-}} f(x)$ exists.
Statement 2: $f$ is continuous at $x=2$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is false, Statement 2 is true.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

D
Statement 1 is true, Statement 2 is false.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
$$
\begin{aligned}
& \text { Consider } \frac{x}{[x]} \leq f(x) \leq \sqrt{6-x} \\
& \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{x}{[x]}=\frac{2}{1}=2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 2^{-}} \sqrt{6-x}=2 \\
& \therefore \lim _{x \rightarrow 2^{-}} f(x)=2
\end{aligned}
$$
[By Sandwich theorem]
Now $\lim _{x \rightarrow 2^{+}} \frac{x}{[x]}=1, \lim _{x \rightarrow 2^{+}} \sqrt{6-x}=2$
Hence by Sandwich theorem $\lim _{x \rightarrow 2^{+}} f(x)$ does not exists.
Therefore $f$ is not continuous at $x=2$. Thus statement-1 is true but statement- 2 is not true

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2012 (19 May Online)
Let f:-1,∞→R be defined by f0=1 and fx=1xloge1+x, x≠0. Then the function f

Options:
A Decreases in-1,0 and increases in 0,∞
B Increases in -1,∞
C Increases in -1,0 and decreases in 0,∞
D Decreases in -1,∞

The correct answer is:
Decreases in -1,∞

Solution:
fx=1xloge1+x, x≠0

f'x=x1+x-ln1+xx2=x-1+xln1+xx21+x

Let gx=x-1+xln1+x

g'x=-ln1+x

⇒g'x>0∀x∈-1,0<0∀x∈0,∞

gmax at x=0⇒g0=0 is the maximum value

So, gx<0∀x∈-1,∞ ⇒ f'x<0∀x∈-1,∞

Hence, fx is decreasing ∀x∈-1,∞

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let f:0,1→R be such that fxy=fx.fy, for all x,y∈0,1, and f(0)≠0. If y=y(x) satisfies the differential equation, dydx=fx with y0=1 then y14+y34 is equal to:

Options:
A 5
B 2
C 3
D 4

The correct answer is:
3

Solution:
Given relation is fxy=fx. fy ∀ x, y∈R .....(i)
On putting x= y=0, we get f0=f20⇒f0=0, 1 but

f0≠0
⇒f0=1,
Now if we put y=0 in (i), we get
fx=1
Hence dydx=1⇒y=x+c
⇒y=x+1 since y0=1
⇒y14+y34

=14+1+34+1=3.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let f1x=3x+22x+3,x∈R--32. For n≥2, define fnx=f1ofn-1x. If f5x=ax+bbx+a, gcda,b=1, then a+b is equal to ________

Options:

The correct answer is:
3125

Solution:
Given:

f1x=3x+22x+3

Now,

f2x=f1of1x

⇒f2x=33x+22x+3+223x+22x+3+3

⇒f2x=9x+6+4x+66x+4+6x+9

⇒f2(x)=13x+1212x+13

⇒f3x=313x+1212x+13+2213x+1212x+13+3

⇒f3x=63x+6262x+63

⇒f4x=363x+6262x+63+2263x+6262x+63+3

⇒f4x=313x+312312x+313

∴f5x=1563x+15621562x+1563=ax+bbx+a

∴a+b=1563+1562=3125

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Let f:-π2,π2→R be a differentiable function such that f0=12, If limx→0x∫0xf(t)dtex2-1=α, then 8α2 is equal to :

Options:
A 16
B 2
C 1
D 4

The correct answer is:
2

Solution:
Let, y=limx→0x∫0xf(t)dtex2-1

⇒y=limx→0x∫0xf(t)dtex2-1x2×x2

We know that, limx→0ex2-1x2=1

⇒y=limx→0∫0xf(t)dtx

Applying L-hospital's rule and Newton Leibnitz Theorem we get,

⇒y=limx→0f(x)1

⇒y=f0

It is given that, f(0)=12

⇒y=12

⇒α=12

⇒8α2=2

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let $f:[-2,3] \rightarrow[0, \infty)$ be a continuous function such that $f(1-x)=f(x)$ for all $x \in[-2,3]$.
If $\mathrm{R}_1$ is the numerical value of the area of the region bounded by $y=f(x), x=-2, x=3$ and the axis of $x$ and $R_2=\int_{-2}^3 x f(x) d x$, then :

Options:
A
$3 R_1=2 R_2$

B
$2 \mathrm{R}_1=3 \mathrm{R}_2$

C
$\mathrm{R}_1=\mathrm{R}_2$

D
$\mathrm{R}_1=2 \mathrm{R}_2$

The correct answer is:
$\mathrm{R}_1=2 \mathrm{R}_2$

Solution:
We have
$$
\begin{aligned}
\mathrm{R}_2 &=\int_{-2}^3 x f(x) d x=\int_{-2}^3(1-x) f(1-x) d x \\
& {\left[\mathrm{U} \operatorname{sing} \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right] } \\
\Rightarrow \mathrm{R}_2 &=\int_{-2}^3(1-x) f(x) d x \\
\therefore \mathrm{R}_2+\mathrm{R}_2 &=\int_{-2}^3 x f(x) d x+\int_{-2}^3(1-x) f(x) d x \\
\Rightarrow 2 \mathrm{R}_2=\mathrm{R}_1
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2013 (25 Apr Online)
Let F:3,5→R be a twice differentiable function on 3,5 such that Fx=e-x∫3x3t2+2t+4F'tdt. If F'4=αeβ-224eβ-42, then α+β is equal to _____.

Options:

The correct answer is:
16

Solution:
Given,

Fx=e-x∫3x3t2+2t+4F'tdt.

So, F3=e-3∫333t2+2t+4F'tdt.

⇒F3=0

Again, exFx=∫3x3t2+2t+4 F'tdt

Differentiate both sides

⇒exFx+exF'x=3x2+2x+4 F'x (Newton Leibnitz Theorem for Definite Integral)

⇒ex-4dydx+exy=3x2+2x (Use y=Fx)

⇒dydx+exex-4y=3x2+2xex-4

Here, integrating factor is e∫exex-4dx=elnex-4=ex-4

So, the solution is

y·ex-4=∫3x2+2xdx+c

⇒y·ex-4=x3+x2+c

Put x=3, y=0⇒c=-36

So, Fx=x3+x2-36ex-4

⇒F'x=3x2+2xex-4-x3+x2-36exex-42

Now put value of x=4 we will get

F'4=342+24e4-4-43+42-36e4e4-42=12e4-224e4-42

So, α=12 & β=4

Hence, α+β=16

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let f:-π4,π4→R be defined as, fx=1+sinx3asinx, -π4<x<0b, x=0ecot4x/cot2x, 0<x<π4

If f is continuous at x=0 then the value of 6a+b2 is equal to:

Options:
A 1-e
B e-1
C 1+e
D e

The correct answer is:
1+e

Solution:
If fx is continuous at point x=a, then limx→0fx=limx→0+fx=limx→0-fx

From the question, we can say that limx→0fx=b

And limx→0+ecot4xcot2x=limx→0+etan2xtan4x=limx→0+etan2x2×2xtan4x4x=e12=b

And limx→0-1+sinx3asinx=elimx→0-sinx3asinx=e3a

From the above we can say that

e3a=e12

⇒a=16⇒6a=1

So, 6a+b2=1+e

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let f:-1,3→R be defined as
fx=x+x,x+x,x+x,-1≤x<11≤x<22≤x≤3,
Where t denotes the greatest integer less than or equal to t. Then, f is discontinuous at:

Options:
A Only one point
B Only two points
C Four or more points
D Only three points

The correct answer is:
Only three points

Solution:
fx=x+x;-1≤x<1x+x;1≤x<2x+x;2≤x≤3

=-x-1;-1≤x<0x+0;0≤x<12x;1≤x<2x+2;2≤x<36;x=3

At x=0, 1, 2, 3 f changes its definition.

∴ At x=0 LHL=-1, RHL=0⇒f is discontinuous at

x=0

x=1 LHL=1, RHL=2⇒f is discontinuous at x=1

x=2 LHL=RHL=f2=4⇒f is continuous at x=2

x=3 LHL=5, f3=6⇒f is discontinuous at x=1

∴ Points of discontinuity are 0, 1, 3.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f:-1,3→R be defined as
fx=x+x,x+x,x+x,-1≤x<11≤x<22≤x≤3,
Where t denotes the greatest integer less than or equal to t. Then, f is discontinuous at:

Options:
A Only one point
B Only two points
C Four or more points
D Only three points

The correct answer is:
Only three points

Solution:
fx=x+x;-1≤x<1x+x;1≤x<2x+x;2≤x≤3

=-x-1;-1≤x<0x+0;0≤x<12x;1≤x<2x+2;2≤x<36;x=3

At x=0, 1, 2, 3 f changes its definition.

∴ At x=0 LHL=-1, RHL=0⇒f is discontinuous at

x=0

x=1 LHL=1, RHL=2⇒f is discontinuous at x=1

x=2 LHL=RHL=f2=4⇒f is continuous at x=2

x=3 LHL=5, f3=6⇒f is discontinuous at x=1

∴ Points of discontinuity are 0, 1, 3.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Let f:6,8→9,11 be a function defined as fx=x+x2 (where, . denotes the greatest integer function), then f-1x is equal to

Options:
A x-x2
B –x-3
C x-3
D 1x+x2

The correct answer is:
x-3

Solution:
Since f:6,8→9,11
∴fx=x+3
∴f-1x=x-3

Topic: Mathematics, Functions, JEE Main
Let $f: \mathrm{A} \rightarrow$ B be a function defined as $f(x)=$ $\frac{x-1}{x-2}$, where $A=R-\{2\}$ and $B=R-\{1\}$. Then $f$ is

Options:
A
invertible and $f^{-1}(y)=\frac{2 y+1}{y-1}$

B
invertible and $f^{-1}(y)=\frac{3 y-1}{y-1}$

C
no invertible

D
invertible and $f^{-1}(y)=\frac{2 y-1}{y-1}$

The correct answer is:
invertible and $f^{-1}(y)=\frac{2 y-1}{y-1}$

Solution:
Suppose $y=f(x)$
$$
\begin{aligned}
&\Rightarrow y=\frac{x-1}{x-2} \\
&\Rightarrow y x-2 y=x-1 \\
&\Rightarrow(y-1) x=2 y-1 \\
&\Rightarrow x=f^{-1}(y)=\frac{2 y-1}{y-1}
\end{aligned}
$$
As the function is invertible on the given domain and its inverse can be obtained as above.

Topic: Mathematics, Functions, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Let $f: \mathrm{A} \rightarrow \mathrm{B}$ be defined as $f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be defined $g(x)=\sqrt{3+4 x-4 x^2}$. If A, B, C are subsets of $\mathbb{R}$ and $f$ is an onto function then the range of the function $f(x)$ is

Options:
A $(-\infty, \infty)$
B $[0, \infty)$
C $\left[-\frac{1}{2}, \frac{3}{2}\right]$
D $[-1,1]$

The correct answer is:
$\left[-\frac{1}{2}, \frac{3}{2}\right]$

Solution:
Since $f$ is an onto function
Thus domain of $\mathrm{g}(\mathrm{x})$ will be range of $\mathrm{f}(\mathrm{x})$
Now, $3+4 \mathrm{x}-4 \mathrm{x}^2 \geq 0$ to define $\mathrm{g}(\mathrm{x})$
$$
\Rightarrow 4 \mathrm{x}^2-4 \mathrm{x}-3 \mathrm{x} 0
$$
On solving we get $\mathrm{x} \in\left[-\frac{1}{2}, \frac{3}{2}\right]$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2022 (19 Jul Shift 2)
Let f:A→B is a function defined by fx=2x1+x2. If the function fx is a bijective function, then the correct statement can be

Options:
A A=B=-1,1
B A=B=-2,2
C A=-1,1,B=-2,2
D A=-2,2,B=-1,1

The correct answer is:
A=B=-1,1

Solution:
∵fx=2x1+x2
Now, f'x=1+x22−2x2x1+x22⇒21−x21+x22
Clearly, f ′ x is injective either for x∈-1,1 or for x∈-∞,-1∪1,∞
Now, for x∈-1,1
Range of fx∈f-1,f1≡-1,1

Topic: Mathematics, Functions, JEE Main
Let $\mathrm{f}(\mathrm{a})=\mathrm{g}(\mathrm{a})=\mathrm{k}$ and their $n$th derivatives $\mathrm{f}^{\mathrm{n}}(\mathrm{a}), \mathrm{g}^{\mathrm{n}}(\mathrm{a})$ exist and are not equal for some $\mathrm{n}$. Further if $\lim _{x \rightarrow a} \frac{f(a) g(x)-f(a)-g(a) f(x)+f(a)}{g(x)-f(x)}=4$ then the value of $k$ is

Options:
A
0

B
4

C
2

D
1

The correct answer is:
4

Solution:
$\lim _{x \rightarrow a} \frac{k 9(x)-k f(x)}{9(k)-f(x)}=4$ (By L'Hospital rule)
$\lim _{x \rightarrow a} k \frac{9^{\prime}(x)-f^{\prime}(x)}{9^{\prime}(x)-f^{\prime}(x)}=4$ or $k=4$

Topic: Mathematics, Limits, JEE Main, JEE Main 2003
Let f:AâR, where fx=Î±x2+6x-8Î±+6x-8x23. If f is onto, then the values of Î± lie in (given, A is the natural domain of fx)

Options:
A 2,14
B 2,14
C 2,14
D 2,14

The correct answer is:
2,14

Solution:
Let, fx=y3
Where, y=αx2+6x-8α+6x-8x2
i.e., α+8yx2+61-yx-αy+8=0
According to the given condition, y takes all real values, for real x .

In other words, the above quadratic equation in x should have real roots for every real y
i.e., D≥0,∀y∈R
i.e., 361-y2+4αy+8α+8y≥0,∀y∈R
i.e., 9+8αy2+α2+46y+9+8α≥0,∀y∈R
i.e., D<0 and coefficient of y2>0
i.e., α2+462<49+8α2 and 9+8α>0
i.e., α2+46<29+8α and α>-9/8
i.e., α2-16α+28<0 and α>-9/8
i.e., 2<α<14 and α>-9/8
i.e., 2<α<14
Hence, f:R→R is onto for 2<α<14

Topic: Mathematics, Functions, JEE Main
Let f and g be continuous functions on 0,a such that fx=fa-x and gx+ga-x=4, then ∫0afxgxdx is equal to

Options:
A ∫0afxdx
B -3∫0afxdx
C 4∫0afxdx
D 2∫0afxdx

The correct answer is:
2∫0afxdx

Solution:
Let I=∫0afxgxdx

⇒I=∫0afa-xga-xdx (Using properties of definite integration)

⇒I=∫0afx4-gxdx ∵ fx=fa-x & gx+ga-x=4

⇒I=4∫0afxdx-∫0afxgxdx

⇒I=4∫0afxdx-I

⇒2I=4∫0afxdx

⇒I=2∫0afxdx

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let f and g be differentiable functions on R such that fog is the identity function. If for some a,b∈R,g'a=5 and ga=b, then f'b is equal to:

Options:
A 15
B 1
C 5
D 25

The correct answer is:
15

Solution:
fgx=x

⇒f'gx.g'x=1

Put x=a

⇒f'gag'a=1

⇒f'b×5=1

⇒f'b=15

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Let $f$ and $g$ be functions defined by $f(x)=\frac{x}{x+1}, g(x)=\frac{x}{1-x}$, then $(f \circ g)(x)$ is

Options:
A $\frac{1}{x}$
B $\frac{1}{x-1}$
C $x-1$
D $x$

The correct answer is:
$x$

Solution:
$(f \circ g)(x)=f(g(x))=f\left(\frac{x}{1-x}\right)=\frac{\frac{x}{1-x}}{\frac{x}{1-x}+1}=\frac{x}{x+1-x}=x$

Topic: Mathematics, Functions, JEE Main
Let $f$ and $g$ be periodic functions with the periods $T_{1}$ and $T_{2}$ respectively. Then $f+g$ is

Options:
A periodic with period $\mathrm{T}_{1}+\mathrm{T}_{2}$
B non-periodic
C periodic with the period $\mathrm{T}_{1}$
D periodic when $T_{1}=T_{2}$

The correct answers are:
periodic when $T_{1}=T_{2}$

Solution:
$\mathrm{f}+\mathrm{g}$ is a periodic function.

Topic: Mathematics, Functions, JEE Main
Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq 0$ and $f(x)=g(x) \sin x$.
Statement $1 \lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.
Statement $2 f^{\prime}(0)=g(0)$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
We have,
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \\
& =\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0
\end{aligned}
$$
Since, $f(x)=g(x) \sin x$
$$
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x \\
& \Rightarrow \quad f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x \\
& \Rightarrow \quad f^{\prime \prime}(0)=0
\end{aligned}
$$
Thus, $\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)$
$\Rightarrow$ Statement 1 is true.
Statement 2. $f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$
$$
\Rightarrow \quad f^{\prime}(0)=g(0)
$$
Statement 2 is not a correct explanation of Statement 1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f$ and $g$ be real-valued functions. If $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{[f(x)+7)]^{2 / 3}}=\frac{7}{4}, \lim _{x \rightarrow 0} f(x)=1$ and $\lim _{x \rightarrow 0} g(x)=\alpha$, then $h(x)= \begin{cases}\sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5}\end{cases}$

Options:
A continuous at $x=\frac{\pi}{10}$ only
B discontinuous on $\left[0, \frac{\pi}{5}\right]$
C discontinuous at $x=\frac{\pi}{10}$
D continuous on $\left[0, \frac{\pi}{5}\right]$

The correct answer is:
continuous on $\left[0, \frac{\pi}{5}\right]$

Solution:
It is given that $\lim _{x \rightarrow 0} f(x)=1, \lim _{x \rightarrow 0} g(x)=\alpha$
and $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{2 / 3}}=\frac{7}{4}$
$\Rightarrow \quad \frac{2-\alpha}{(8)^{2 / 3}}=\frac{7}{4} \Rightarrow \alpha=-5$
For $\quad h(x)= \begin{cases}\sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5}\end{cases}$
$\because \alpha=-5$, so $\sin (\alpha x)$ is continuous for $x \in\left[0, \frac{\pi}{10}\right)$
and $\cos (2 \alpha x)$ is also continuous for $x \in\left(\frac{\pi}{10}, \frac{\pi}{5}\right]$
Now, at $x=\pi / 10$
$\begin{aligned} \operatorname{LHL}\left(\text { at } x=\frac{\pi}{10}\right) & =\lim _{h \rightarrow 0} \sin \left(-5\left(\frac{\pi}{10}-h\right)\right) \\ & =\sin \left(-\frac{\pi}{2}\right)=-1\end{aligned}$
and $f\left(\frac{\pi}{10}\right)=\sin \left(-\frac{5 \pi}{10}\right)=\sin \left(-\frac{\pi}{2}\right)=-1$ and RHL $\left(\right.$ at $\left.x=\frac{\pi}{10}\right)=\lim _{h \rightarrow 0} \cos \left(-10\left(\frac{\pi}{10}+h\right)\right)$ $=\lim _{h \rightarrow 0} \cos (\pi+10 h)=\cos \pi=-1$
$\therefore$ The function $h(x)$ is continuous on $\left[0, \frac{\pi}{5}\right]$. Hence, option (d) is correct.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f and g be twice differentiable even functions on -2, 2 such that f14=0, f12=0, f1=1 and g34=0,g1=2 Then, the minimum number of solutions of fxg"x+f'xg''x=0 in -2, 2 is equal to _____.

Options:

The correct answer is:
4

Solution:
Let hx=fxg'x roots
∵ fx is even ⇒f14=f12=f-12=f14=0 and gx has two roots as g34=g-34=0 (as gx is even function)

So, hx=fxg'x will have 5 roots, hence its derivative h'x=fxg"x+f'xg'x will have 4 roots.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let $f$ and $g$ be two differentiable functions on $R$ such that $f^{\prime}(x)>0$ and $g^{\prime}(x) < 0$ for all $x \in R$. Then for all $\mathrm{x}$ :

Options:
A
$f(g(x))>f(g(x-1))$

B
$f(g(x))>f(g(x+1))$

C
$g(f(x))>g(f(x-1))$

D
$g(f(x)) < g(f(x+1))$

The correct answer is:
$f(g(x))>f(g(x+1))$

Solution:
Since $f^{\prime}(x)>0$ and $g^{\prime}(x) < 0$, therefore $f(x)$ is increasing function and $g(x)$ is decreasing function.
$\Rightarrow f(x+1)>f(x)$ and $g(x+1) < g(x)$
$\Rightarrow g[f(x+1)] < g[f(x)]$ and $f[g(x+$
1) $] < f[g(x)]$
Hence option (b) is correct.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2014 (12 Apr Online)
Let f and g be two functions defined by fx=x+1, x<0x-1, x≥0and gx=x+1, x<01, x≥0.Then gofx is

Options:
A Continuous everywhere but not differentiable exactly at one point
B Continuous everywhere but not differentiable at x=1
C Differentiable everywhere
D Not continuous at x=1

The correct answer is:
Continuous everywhere but not differentiable exactly at one point

Solution:
Given,

fx=x+1, x<0x-1, x≥0and gx=x+1, x<01, x≥0

⇒fx=x+1,x<01-x,0≤x<1x-1,x>1& gx=1+x,x<01,x≥0

Now, for gofx

gofx=1+fx,fx<0, x<01,fx≥0, x≥0

⇒ gofx=x+2,x<-11,x≥-1

Now plotting the diagram we get,
Now from above diagram we can say that function is continuous but not differentiable at x=-1 due to sharp corner.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let $f$ and $g$ be two real functions given by and $\begin{aligned} f &=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\} \\ g &=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\} \end{aligned}$ and $g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}$ then the domain of $f$. $g$ is given by

Options:

None

Solution:
$\{2,3,4,5\}$

Topic: Mathematics, Relations and Functions (Class 11)
Let fb be the minimum value of the expression y=x2-2x+b3-3b2+4 âxâR. Then, the maximum value of fb as b varies from 0 to 4 is

Options:
A 20
B 19
C 63
D 64

The correct answer is:
19

Solution:
y=x-12+b3-3b2+3
â´ymin=fb=b3-3b2+3
Now, f â² b =3 b 2 â6b
=3bb-2
Also, f0=3
f2=-1
f4=19
â´maxâ¡fb=19

Topic: Mathematics, Application of Derivatives, JEE Main
Let $f$ be a composite function of $x$ defined by $f(u)=\frac{1}{u^2+u-2}, u(x)=\frac{1}{x-1}$.
Then the number of points $x$ where $f$ is discontinuous is :

Options:
A
4

B
3

C
2

D
1

The correct answer is:
3

Solution:
$\mu(x)=\frac{1}{x-1}$, which is discontinous at $x=1$
$$
f(u)=\frac{1}{u^2+u-2}=\frac{1}{(u+2)(u-1)},
$$
which is discontinous at $u=-2,1$
when $u=-2$, then $\frac{1}{x-1}=-2 \Rightarrow x=\frac{1}{2}$
when $u=1$, then $\frac{1}{x-1}=1 \Rightarrow x=2$
Hence given composite function is discontinous at three points, $x=1, \frac{1}{2}$ and 2 .

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2013 (23 Apr Online)
Let $f$, be a continuous function in $[0,1]$, then $\lim_{n \rightarrow \infty} \sum_{j=0}^n \frac{1}{n} f\left(\frac{j}{n}\right)$ is

Options:
A $\frac{1}{2} \int_{0}^{\frac{1}{2}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$
B $\int_{\frac{1}{2}}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}$
C $\int_{0}^{1} f(x) d x$
D $\int_{0}^{\frac{1}{2}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$

The correct answer is:
$\int_{0}^{1} f(x) d x$

Solution:
\(\lim _{n \rightarrow \infty} \sum_{i=0}^{n} \frac{1}{n} f\left(\frac{i}{n}\right)\)
Let \(1 / n \rightarrow d x\)

Topic: Mathematics, Definite Integration, JEE Main
Let f be a continuous function satisfying ∫0t2f(x)+x2dx=43t3,∀t>0 . Then fπ24 is equal to

Options:
A π21-π216
B -π1+π316
C π1-π316
D -π21+π216

The correct answer is:
π1-π316

Solution:
Given equation is ∫0t2f(x)+x2dx=43t3,∀t>0

According to Newton Leibnitz theorem we haveddx∫uxvxftdt=fvx×v'x-fux×u'x

Apply Newtons Leibnitz theorem in the given equation.

⇒ft2+t42t-0=4t2

⇒ft2+t4=2t

⇒fx2=-x4+2x

⇒f(x)=-x2+2x

⇒fπ24=-π442+2×π2

=-π416+π

=π1-π316

Hence this is the correct option.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let f be a differentiable function from R to R such that fx-fy≤2x-y3/2, for all x,y∈R. If f0=1 then ∫01f2xdx is equal to

Options:
A 0
B 1
C 2
D 12

The correct answer is:
1

Solution:
Given functional equality is
fx-fy≤2 x-y3/2
Put x=y+h, we get
fy+h-fy≤2. |h3/2
⇒fy+h-f(y)h≤2 h1/2
⇒limh→0⁡fy+h-fyh≤limh→0⁡2 h1/2
⇒f′y≤0
⇒f′y=0
⇒f′y=0
⇒fy=c
⇒fy=1 (since f0=1 )
Now ∫01f2xdx= ∫011 dx=1.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let f be a differentiable function in 0,π2. If ∫cosx1t2ftdt=sin3x+cosx, then 13f'13 is equal to

Options:
A 6-92
B 6+92
C 6-92
D 3+2

The correct answer is:
6-92

Solution:
∫cosx1t2ftdt=sin3x+cosx

On differentiating, we get

fcosxsinx·cos2x=3sin2xcosx-sinx

fcosx=3tanx-sec2x

Again differentiating, we get

-sinxf'cosx=3sec2x-2sec2xtanx

When cosx=13 then secx=3, tanx=2 & sinx=23

Then -23f'13=3×3-2×32

⇒13f'13=6-92

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let f be a differentiable function satisfying fx=23∫03fλ2x3dλ,x>0 and f1=3. If y=fx passes through the point α,6, then α is equal to _______.

Options:

The correct answer is:
12

Solution:
Given,

fx=23∫03fλ2x3dλ

Now let λ2x3=t

⇒2λx3 dλ=dt

⇒dλ=32·1xx·3tdt

⇒dλ=32·1x·dtt

So, fx=1x∫0xfttdt

xfx=∫0xfttdt

Now differentiating both side we get,

⇒x·f'x+fx2x=fxx

⇒x·f'x=fx2x

⇒dyy=dx2x

Now integrating both side we get,

⇒∫dyy=∫dx2x

⇒lny=12lnx+lnc

Given f1=3, so ⇒ln3=0+lnc

⇒c=3

So, fx=3x

Now given fx is passing through α,6,

So fα=6⇒36=3α

⇒α=12

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (27 Jul Shift 2)