datavorous/entrance-exam-dataset huggingface dataset:
Let $e_1$ be the eccentricity of a hyperbola for which distance between its focii is 2 times the distance between its directrices and $e_2$ be the eccentricity of another hyperbola for which the length of its transverse axis is twice the length of its the conjugate axis. Then $e_1 e_2=$

Options:
A 1
B $\frac{\sqrt{10}}{2}$
C $\sqrt{5}$
D $\frac{\sqrt{5}}{2}$

The correct answer is:
$\frac{\sqrt{10}}{2}$

Solution:
Given $e_1$ is the eccentricity of hyperbola and eccentricity $\mathrm{e}_2$ of another hyperbola.
According to question,
$$
\begin{aligned}
& 2 \mathrm{a}_1 \mathrm{e}_1=2 \times \frac{2 \mathrm{a}_1}{\mathrm{e}_1} \\
& \mathrm{e}_1^2=2 \\
& \mathrm{e}_1= \pm \sqrt{2}
\end{aligned}
$$
Here, $e_1>1$ then $e_1=\sqrt{2}$
Now, $2 \mathrm{a}_2=4 \mathrm{~b}_2$
$$
\begin{aligned}
& \mathrm{a}_2=2 \mathrm{~b}_2 \\
& \mathrm{~b}_2=\frac{\mathrm{a}_2}{2}
\end{aligned}
$$
So, $c_2^2=a_2^2+b_2^2$
$$
\begin{aligned}
& \mathrm{c}_2^2=\mathrm{a}_2^2+\frac{\mathrm{a}_2^2}{4}=\frac{5 \mathrm{a}_2^2}{4} \\
& \mathrm{c}_2=\frac{\sqrt{5}}{2} \mathrm{a}_2 \\
& \frac{\mathrm{c}_2}{\mathrm{a}_2}=\frac{\sqrt{5}}{2} \\
& \mathrm{e}_2=\frac{\sqrt{5}}{2}
\end{aligned}
$$
Then, $\mathrm{e}_1 \mathrm{e}_2=\sqrt{2} \times \frac{\sqrt{5}}{2}=\frac{\sqrt{10}}{2}$
So, option (b) is correct.

Topic: Mathematics, Hyperbola, JEE Main
Let E1,E2,E3 be three mutually exclusive events such that PE1=2+3p6,PE2=2-p8 and PE3=1-p2. If the maximum and minimum values of p are p1 and p2 then p1+p2 is equal to:

Options:
A 23
B 53
C 54
D 1

The correct answer is:
53

Solution:
We know that,

0≤PEi≤1for i=1,2,3

0≤PE1≤1

⇒0≤2+3P6≤1

⇒P∈-23,43 ........i

Now for 0≤PE2≤1

⇒0≤2-P8≤1

⇒P∈-6,2........ii

Now for 0≤PE3≤1

⇒0≤1-P2≤1

⇒P∈-1,1.......iii

Also, E1 and E2 and E3 are mutually exclusive

PE1+PE2+PE3≤1

⇒0≤1312-P8≤1

⇒P∈23,263...........iv

Now taking intersection of all we get,

23≤P≤1

So maximum value p1=1 and minimum value p2=23

So, p1+p2=53

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let E1:x2a2+y2 b2=1, a>b. Let E2 be another ellipse such that it touches the end points of major axis of E1 and the foci of E2 are the end points of minor axis of E1. If E1 and E2 have same eccentricities, then its value is:

Options:
A -1+52
B -1+82
C -1+32
D -1+62

The correct answer is:
-1+52

Solution:
Given ellipse is E1:x2a2+y2b2=1, a>b and E2 is another ellipse which touches the end points of the major axis of E1 and foci of E2 are at the minor axis of E1, thus the two ellipses are given by the following diagram.

Since, the ellipse E2 touches the major axis of the ellipse E1, hence, the minor axis of the ellipse E2 is a1=a.

Now, let the major axis of the ellipse E2 is b1=c and it is obvious from the diagram and the given conditions that b1>a1.

We know that the eccentricity of an ellipse x2A2+y2B2=1, A>B is 1-B2A2.

Thus, for the ellipse E1, we have e=1-b2a2 and for the ellipse E2, we have e1=1-a2c2.

Given, e=e1

1-b2a2=1-a2c2

⇒ 1-b2a2=1-a2c2

⇒ b2a2=a2c2

⇒ c2=a4b2

⇒ c=a2b ...i

Also, given that the foci of E2 are the end points of the minor axis of E1, thus b=ce

⇒ c=be ...ii

From the above two equations, we get be=a2 b

⇒ e=b2a2

Now, using the definition of eccentricity, we get e=b2a2=1-e2

⇒ e2+e-1=0

Now, applying the Sridharacharya's formula for the roots of a quadratic equation, i.e. if ax2+bx+c=0, a≠0, then x=-b±b2-4ac2a, we get

e=-1±12-4×1×-12×1

⇒ e=-1±1+42

⇒ e=-1±52

But, eccentricity can never be negative, hence e=-1+52.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let e and l are the eccentricity and length of the lactus rectum respectively of the conic described parametrically by x=t2+t+1,y=t2-t+1, then the value of el2 is equal to

Options:

The correct answer is:
0.5

Solution:
x+y2=t2+1 and x-y2=t
⇒x+y-22=x-y22
⇒x-y22=2x+y-22
Comparing it with y2=4ax, we get,
l=2 and e=1
⇒el2=12=0.5

Topic: Mathematics, Parabola, JEE Main
Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its centre at 3,-4, one focus at 4,-4 and one vertex at 5,-4. If mx-y=4,m>0 is a tangent to the ellipse E, then the value of 5m2 is equal to _____ .

Options:

The correct answer is:
3

Solution:
We know that, If the standard ellipse is shifted to have center h, k, then it's equation is x-h2a2+y-k2b2=1

The center is Ch, k
The vertices on the major axis are Ah+a, k and A'h-a, k
The foci are Sh+ae, k and S'h-ae, k

The eccentricity e=a2-b2a2

Given C3,-4,S4,-4 and A5,-4

Therefore, h, k=3, -4

h+a, k=5, -4

h+ae, k=4, -4

Clearly, a=2 &ae=1

⇒e=12

We know that, b2=a21-e2

b2=a2-a2e2

b2=22-1

⇒b2=3

So, equation of ellipse x-324+y+423=1

Intersecting with given tangent

x2-6x+94+m2x23=1

⇒3x2-6x+9+4m2x2=12

⇒3x2-18x+27+4m2x2=12

⇒3+4m2x2-18x+15=0

Now, D=0 (as it is tangent)

⇒182=43+4m2×15

So, 5m2=3.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let $E$ be the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and $C$ be the circle $x^2+y^2=9$. Let $P$ and $Q$ be the points (1, $2)$ and $(2,1)$ respectively. Then

Options:
A $Q$ lies inside $C$ but outside $E$
B $Q$ lies outside both $C$ and $E$
C $P$ lies inside both $C$ and $E$
D $P$ lies inside $C$ but outside $E$

The correct answer is:
$P$ lies inside $C$ but outside $E$

Solution:
The given ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$. The value of the expression $\frac{x^2}{9}+\frac{y^2}{4}-1$ is positive for $x=1, y=2$ and negative for $x=2, y=1$. Therefore $P$ lies outside $E$ and $Q$ lies inside $E$. The value of the expression $x^2+y^2-9$ is negative for both the points $P$ and $Q$. Therefore $P$ and $Q$ both lie inside $C$. Hence $P$ lies inside $C$ but outside $E$.

Topic: Mathematics, Hyperbola, JEE Main
Let E denote the set of letters of the English alphabet, V=a,e,i,o,u and C be the complement of V in E. Then, the number of four-letter words (where repetitions of letters are allowed) having at least one letter from V and at least one letter from C is

Options:
A 261870
B 314160
C 425880
D 851760

The correct answer is:
261870

Solution:
The number of all four-letter words =264

The number of all four-letter words from set V=a,e,i,o,u i.e vowels =54

Similarly, the number of all four-letter words from set of consonants =215

∴ Number of words which contains atleast one vowel and atleast one consonant

=264-214-54

=262+212262-212-54

=676+441547-54

=51117×47-125

=5×52374=261870

Topic: Mathematics, Permutation Combination, JEE Main
Let E & F be two independent events. The probability that E & F happen is 112 and the probability that neither E nor F happens is 12 , then a value of PEPF is:

Options:
A 43
B 13
C 32
D 512

The correct answer is:
43

Solution:
PE∩F=PE. PF=112

⇒xy=112
⇒PE-∩F-=PE- . PF-=12
⇒ 1-PE 1-PF=12
⇒1-x-y+xy=12
⇒1-x-y+112=12
⇒1-x-y=12-112=512
⇒x+y=712
⇒x+112x=712
⇒⇒12x2+112x=712
⇒12x2-7x+1=0
⇒12x2-4x-3x+1=0
⇒4x3x-1-13x-1=0
⇒x=13, x=14
⇒y=14, y=13
⇒xy=1314 & 1413

⇒xy=43 & 34

Topic: Mathematics, Probability, JEE Main, JEE Main 2017 (09 Apr Online)
Let â«exâx2dx=fxex+C (where, C is the constant of integration). The range of fx as xâR is a,â. The value of a4 is

Options:

The correct answer is:
0.25

Solution:
As we know, ∫exfx+f'xdx=ex⋅fx+C,
Thus, ∫ex⋅x2dx=∫exx2+2x-2x-2+2dx
=∫exx2-2x+2+ex2x-2dx
=exx2-2x+2+C
i.e. fx=x2-2x+2=x-12+1
Hence, range is 1,∞⇒a=1
∴a4=0.25

Topic: Mathematics, Indefinite Integration, JEE Main
Let electronegativity, ionisation energy and electron affinity be represented as EN, IP and EA respectively. Which one of the following equation is correct according to Mulliken?

Options:
A $\mathrm{EN}=\mathrm{IP} \times \mathrm{EA}$
B $\mathrm{EN}=\frac{\mathrm{IP}}{\mathrm{EA}}$
C $\mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2}$
D $\mathrm{EN}=\mathrm{IP}-\mathrm{EA}$

The correct answer is:
$\mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2}$

Solution:
Mulliken suggested that EN of an element is the average of its IP and EA.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Let f:(0,∞)→(0,∞) be a differentiable function such that f(1)=e and limt→xt2f2(x)-x2f2(t)t-x=0. If f(x)=1, then x is equal to:

Options:
A 1e
B 2e
C 12e
D e

The correct answer is:
1e

Solution:
limt→xt2f2(x)-x2f2(t)t-x=0
Using L'Hospital

limt→x2tf2(x)-x22f(t)f'(t)1=0
⇒2xf2(x)-x22f(x)f'(x)=0
⇒2 x f(x)[ f(x)-xf'(x)]=0
But f(x)≠0

So, xf'(x)=f(x)
⇒xdydx=y
⇒1ydy=1xdx
Integration gives

lny=lnx+lnc
⇒y=cx⇒f(x)=cx
Now

f(1)=c=e (given)

So, f(x)=e x
Now if f(x)=1, then ex=1⇒x=1e

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Let f:[0,∞)→[0,∞) be defined as fx=∫0xydy where [x] is the greatest integer less than or equal to x. Which of the following is true?

Options:
A f is continuous at every point in [0,∞) and
differentiable except at the integer points.
B f is both continuous and differentiable except at the integer points in [0,∞).
C f is continuous everywhere except at the integer points in [0,∞).
D f is differentiable at every point in [0,∞).

The correct answer is:
f is continuous at every point in [0,∞) and
differentiable except at the integer points.

Solution:
f:[0,∞)→[0,∞), fx=∫0xydy

Let x=n+f,f∈(0,1)

⇒x=n, x=f.

So

fx=∫01ydy+∫12ydy+∫23ydy+…+∫n-1nydy+∫nn+fydy

fx=∫010dy+∫121dy+∫232dy+…+∫n-1nn-1dy+∫nn+fndy

fx=01-0+12-1+23-2+…+n-1n-n-1+nn+f-n

fx=0+1+2+…+n-1+nf

⇒fx=n(n-1)2+nf

=[x]([x]-1)2+xx

Note limx→n+fx=nn-12

limx→n-fx=(n-1)(n-2)2+n-1

=n(n-1)2

fx=nn-12n∈N0

so f(x) is continuous ∀x≥0 and differentiable except at integer.'

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let f:0,1→0,1 be a continuous function such that x2+fx2≤1 for all x∈0,1 and ∫01fxdx=π4 Then, ∫1212fx1-x2dx equals

Options:
A π12
B π15
C 2-12π
D π10

The correct answer is:
π12

Solution:
For a continuous function

f:0,1⟶0,1 it is given that x2+fx2≤1

⇒fx2≤1-x2

⇒fx≤1-x2 ∵fx∈0,1

∴∫01fxdx≤∫011-x2dx

⇒∫01fxdx≤x21-x2+12sin-1x01

⇒∫01fxdx≤π4

So, fx=1-x2, because it is given that ∫01fx=π4.

Now, ∫1/212fx1-x2dx=∫1/212dx1-x2=sin-1x1/21/2

=π4-π6=π12

Topic: Mathematics, Definite Integration, JEE Main
Let f:0,1→-1,1 and g:-1,1→0,2 be two functions such that g is injective and gof:0,1→0,2 is surjective. Then,

Options:
A f must be injective but need not be surjective
B f must be surjective but need not be injective
C f must be bijective
D f must be a constant function

The correct answer is:
f must be surjective but need not be injective

Solution:
Let hx=gfx where, gx is injective and hx is surjective. hx is surjective.

∴ Codomain of hx= Range of hx

Range of hx=0,2 which is also codomain of g. So, must be surjective.

Now, domain of g=-1,1 which must be range of f. But codomain of f=-1,1

So, it must be surjective.

Topic: Mathematics, Functions, KVPY, KVPY 2018 (SB/SX)
Let f:0,1→ℝ be a function defined by fx=11-e-x, and gx=f-x-fx. Consider two statements

(I) g is an increasing function in 0,1

(II) g is one-one in 0,1
Then,

Options:
A Only (I) is true
B Only (II) is true
C Neither (I) nor (II) is true
D Both (I) and (II) are true

The correct answer is:
Both (I) and (II) are true

Solution:
Given,

f:0,1→ℝ be a function defined by fx=11-e-x, and gx=f-x-fx

Now using gx=f-x-fx we get,

g(x)=11-ex-exex-1

⇒gx=1+ex1-ex

Now differentiating the above function with respect to x we get,

⇒g'x=2ex1-ex2>0

Since g'x is always positive

So, gx is one-one in 0,1

And gx is increasing in 0,1

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let f:0,1→R be an injective continuous function that satisifes the condition -1<f0<f1<1 Then, the number of functions g:-1,1→0,1 such that gofx=x for all x∈0,1 is

Options:
A 0
B 1
C more than 1, but finite
D infinite

The correct answer is:
infinite

Solution:
We have,

-1<f0<f1<1

g=-1,1→0,1

gofx=x,∀x∈0,1

Only condition that gx should satisfy for gofx=x,∀x∈0,1 is that gx should attain all values in 0,1 when range of fx a subset of -1,1 is used as image for gx. Thus, there can be infinite such functions gx with domain -1,1 and range 0,1

Topic: Mathematics, Functions, KVPY, KVPY 2018 (SB/SX)
Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x}$, where $b$ is a constant such that $0 < b < 1$. Then,

Options:
A
$f$ if not invertible on $(0,1)$

B
$f \neq f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$

C
$f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$

D
$f^{-1}$ is differentiable on $(0,1)$

The correct answers are:
$f$ if not invertible on $(0,1)$

Solution:
Here, $f(x)=\frac{b-x}{1-b x}$ where, $0 < b < 1,0 < x < 1$ For function to be invertible it should be one-one onto.
$\therefore$ Check range :
Let $\quad f(x)=y \Rightarrow y=\frac{b-x}{1-b x}$
$$
\begin{aligned}
& \Rightarrow y-b x y=b-x \Rightarrow x(1-b y)=b-y \\
& \Rightarrow \quad x=\frac{b-y}{1-b y}
\end{aligned}
$$
where, $0 < x < 1$
$$
\begin{gathered}
\therefore \quad 0 < \frac{b-y}{1-b y} < 1 \\
\frac{b-y}{1-b y}>0 \text { and } \frac{b-y}{1-b y} < 1 \\
\quad+\quad-\quad+ \\
\Rightarrow \quad b \\
y < b \text { or } y>\frac{1}{b} \\
\frac{(b-1)(y+1)}{1-b y} < -1 < y < \frac{1}{b}
\end{gathered}
$$
From Eqs. (i) and (ii), we get $y \in\left(-1, \frac{1}{b}\right) \subset$ Codomain
Thus, $f(x)$ is not invertible.

Topic: Mathematics, Functions, JEE Main
Let f:0, 2→R be a twice differentiable function such that f''x>0, for all x∈0, 2. If ϕx= fx+ f2–x, then ϕ is

Options:
A decreasing on 0,2
B increasing on 0,2
C increasing on (0,1) and decreasing on 1,2
D decreasing on 0,1 and increasing on (1,2)

The correct answer is:
decreasing on 0,1 and increasing on (1,2)

Solution:
fx:0,2→R and f''x> 0 for x∈ [0,2]
⇒f'(x) is increasing for x∈ [0,2]
Now, ϕ(x)=f(x)+f(2-x)
⇒ϕ'x=f'x-f'(2-x)
For x∈ [0,1) , x < 2 – x

⇒f'(x)<f'(2-x)⇒ϕ'(x)<0
For x∈ (1, 2] , x > 2 – x

⇒f'x>f'2-x⇒ϕ'x>0
Hence, ϕ is decreasing on 0,1 and increasing on 1,2.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Let f:0,2→R be defined as fx=log21+tanπx4.

Then, limn→∞2nf1n+f2n+….+f1 is equal to ________.

Options:

The correct answer is:
1

Solution:
Given fx=log21+tanπx4=loge1+tanπx4loge2

E=2limn→∞∑r=1n1nfrn

E=2loge2∫01loge1+tanπx4dx ...(i)

Replacing x→1-x

E=2loge2∫01loge1+tanπ41-xdx

E=2loge2∫01loge1+tanπ4-π4xdx

E=2loge2∫01loge1+1-tanπ4x1+tanπ4xdx

E=2loge2∫01loge21+tanπx4dx

E=2loge2∫01loge2-loge1+tanπx4dx ...(ii)

Adding an equation (i) & (ii) we get,

E=1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let $f:[0,10] \rightarrow[1,20]$ be a function defined
as $f(x)=\left\{\begin{array}{ll}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{array}\right.$ then $f$ is

Options:
A bijective function
B one-one but not onto function
C onto but not one-one function
D neither one-one nor onto function

The correct answer is:
onto but not one-one function

Solution:
We have, $f(x)= \begin{cases}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{cases}$
Now, $f(x)=10 \forall x \in[6,7]$
So, $f(x)$ is not one-one.
Again,
$\begin{aligned}
& 0 \leq x \leq 6 \\
& \Rightarrow \quad-30 \leq-5 x \leq 0 \\
& \Rightarrow \quad 60-30 \leq 60-5 x \leq 60 \\
& \Rightarrow \quad \frac{30}{3} \leq \frac{60-5 x}{3} \leq \frac{60}{3} \\
& \Rightarrow \quad 10 \leq \frac{60-5 x}{3} \leq 20 \text { and } 7 \leq x \leq 10 \\
& \Rightarrow \quad-30 \leq-3 x \leq-21 \Rightarrow 1 \leq 31-3 x \leq 10 \\
& \therefore \text { Range of } f(x)=[1,20] \\
&
\end{aligned}$
It is given that co-domain of $f(x)=[1,20]$
$\therefore$ Range of $f(x)=$ co-domain of $f(x)$
So, $f(x)$ is onto.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 1)
Let $f:(0, \pi) \rightarrow \mathbf{R}$ be a function given by $f(x)=$
\(\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+|\cot x|)^{\frac{\mathrm{b}}{}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.\)
where $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to

Options:

The correct answer is:
81

Solution:
LHL at $\mathrm{x}=\frac{\pi}{2}$
$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}=\left(\frac{8}{7}\right)^0=1$
RHL at $\mathrm{x}=\frac{\pi}{2}$
$\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(1+|\cot x|)^{\frac{b}{\operatorname{lan}} \tan x} \\
& =e^{\lim _{x-\frac{\pi}{2}}|\cot x| \frac{b}{a}|\tan x|}=e^{\frac{b}{a}}
\end{aligned}$
$\begin{aligned} & \Rightarrow 1=a-8=e^{\frac{b}{a}} \\ & \Rightarrow a=9, b=0 \\ & \Rightarrow a^2+b^2=81\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let $f:[0, \pi] \rightarrow R$ be defined as $f(x)=\left\{\begin{array}{ll}\sin x, & \text { if } x \text { is irrational and } x \in[0, \pi] \\ \tan ^{2} x, & \text { if } x \text { is rational and } x \in[0, \pi]\end{array}\right\}$
The number of points in $[0, \pi]$ at which the function $f$ is continuous is :

Options:
A 6
B 4
C 2
D 0

The correct answer is:
4

Solution:
$\sin x=\tan ^{2} x$
$\sin ^{2} x=\sin x\left(1-\sin ^{2} x\right)$
$\sin x\left(\sin ^{2} x+\sin -1\right)=0$
$\sin x=0 \quad \sin x=\frac{\sqrt{5}-1}{2}$
$x=0, \pi \quad \sin ^{-1} \frac{\sqrt{5}-1}{2}, \pi-\sin ^{-1} \frac{\sqrt{5}-1}{2}$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f:-1,1âB be a function defined as fx=cot-1â¡2x31+x2. If f is both one-one and onto, then B is the interval

Options:
A 0,x3
B 0,2π3
C π3,2π3
D π3,π

The correct answer is:
π3,2π3

Solution:
Let, y=2x31+x2

â3yx2-2x+3y=0
âµxâRâDâ¥0

â4-12y2â¥0âyâ-13,13
â-13â¤2x31+x2â¤13

âcot-1â¡-13â¥cot-1â¡2x31+x2â¥cot-1â¡13
2Ï3â¥fxâ¥Ï3

Topic: Mathematics, Functions, JEE Main
Let $f:(-1,1) \rightarrow B$, be a function defined by $f(x)=\tan ^{-1} \frac{2 x}{1-x^2}$, then $f$ is both one-one and onto when $B$ is the interval

Options:
A
$\left(0, \frac{\pi}{2}\right)$

B
$\left[0, \frac{\pi}{2}\right)$

C
$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

D
$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$

The correct answer is:
$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$

Solution:
Given $f(x)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ for $x \in(-1,1)$
clearly range of $f(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$\therefore$ co-domain of function $=B=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Topic: Mathematics, Functions, JEE Main, JEE Main 2005
Let f:(-1, 1)→R be a continuous function. If ∫0sin⁡xf(t) dt=32x, then f32 is equal to:

Options:
A 32
B 3
C 12
D 32

The correct answer is:
3

Solution:
∫0sinxftdt= 32 x.
Differentiating w.r.t x both the sides,
fsin⁡x·cos⁡x= 32
Put x= π 3.
12·f 32=32
f32=3.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2015 (11 Apr Online)
Let f:-1,1→R be a differentiable function satisfying f'x4=16fx2 for all x∈-1,1 f0=0 The number of such functions is

Options:
A 2
B 3
C 4
D more than 4

The correct answer is:
more than 4

Solution:
It is given that,

f'x4=16fx2,∀x∈-1,1

⇒f'x2=±4fx

Case I

If fx≥0∀x∈-1,1

Then, f'x=±2fx

⇒f'xfx=±2

⇒2fx=±2x, ∵f0=0

⇒fx=x2∀x∈-1,1

Case II

If fx<0∀x∈-1,1

Then, f'x=±2-fx

∴ fx=-x2∀x∈-1,1

∴ f1x=x2,0≤x<1x2,-1<x<0

f2x=x2,0≤x<1-x2-1<x<0

f3x=-x2,0≤x<1x2,-1<x<0

f4x=-x2,0≤x<1-x2,-1<x<0

f5x=0

f6x=0,0≤x<1x2,-1<x<0,… and so on.

Hence, there are more than 4 functions possible.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $f:(-1,1) \rightarrow R$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. Let $g(x)=[f(2 f(x)+2)]^2$. Then $g^{\prime}(0)=$

Options:
A
$-4$

B
0

C
$-2$

D
4

The correct answer is:
$-4$

Solution:
No Solution Available

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2010
Let f:-1, 1→R be a function defined by fx=max-x, -1-x2. If K be the set of all points at which f is not differentiable, then K has exactly

Options:
A two elements
B one element
C three elements
D five elements

The correct answer is:
three elements

Solution:
From the graph, it can be concluded that fx is non-differentiable at x=0,±12 in -1,1.

∴ At 3 points in -1,1.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let f:-1, 1âR be a function defined by fx=maxâ¡-x, -1-x2. If K be the set of all points at which f is not differentiable, then K has exactly

Options:
A two elements
B one element
C three elements
D five elements

The correct answer is:
three elements

Solution:
From the graph it can be concluded that f(x) is non-differentiable at x=0,ââÂ±12 in (-1,â1)
i.e., at 3 points in -1,â1

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f:-1,1→R be a function defined by fx=min⁡x,1-x2. If K be the set of all points at which f is not differentiable, then K has exactly

Options:
A one element
B two elements
C five elements
D three elements

The correct answer is:
five elements

Solution:
Graph of given function fx=min⁡x,1-x2, is as shown
Clearly, fx is not differentiable at 5 points in -1,1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let f:[-1,1]→R be defined as f(x)=ax2+bx+c for all x∈[-1, 1], where a, b, c∈R such that f(-1)=2, f'(-1)=1 and for x∈(-1, 1) the maximum value of f"(x) is 12. If f(x)≤α, x∈-1, 1, then the least value of α is equal to

Options:

The correct answer is:
5

Solution:
Given f:[-1, 1]→R, f(x)=ax2+bx+c, f'(x)=2ax+b and f''(x)=2a

⇒f(-1)=a-b+c=2 ...1,

⇒f'(-1)=-2a+b=1 ...2 and

f''(x)=2a

Given the maximum value of f''(x)=2a=12 ⇒a=14,

From the equations 1 and 2 we get b=32 and c=134.

∴ f(x)=x24+3x2+134

We know that, the vertex of the quadratic Ax2+Bx+C is at -B2A, -B2-4AC4A

Thus, the vertex of x24+3x2+134 is at -3, 1, hence both the numbers ±1 are on the same side of the vertex and the graph of fx is given below.

For, x∈[-1, 1], we have f-1=2 & f1=5

⇒2≤f(x)≤5

∴ Least value of α is 5.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let fθ=11+tan⁡θ2021, then the value of ∑θ=1°89°fθ is equal to

Options:
A 45
B 44
C 892
D 912

The correct answer is:
892

Solution:
fθ=cos⁡θ2021sin⁡θ2021+cos⁡θ2021
f90°-θ=sin⁡θ2021cos⁡θ2021+sin⁡θ2021
⇒fθ+f90°-θ=1 and f45°=12
Hence, ∑ θ= 1 o 89 o f θ =f 1 o +f 2 o +…+f 44 o +f 46 o +…+f 88 o +f 89 o +f 45 o
=f1o+f89o+f2o+f88o+…+f44o+f46o+f45o
=1×44+12=892

Topic: Mathematics, Trigonometric Equations, JEE Main
Let fθ=11+tan⁡θ2021, then the value of ∑θ=1°89°fθ is equal to

Options:
A 45
B 44
C 892
D 912

The correct answer is:
892

Solution:
fθ=cos⁡θ2021sin⁡θ2021+cos⁡θ2021
f90°-θ=sin⁡θ2021cos⁡θ2021+sin⁡θ2021
⇒fθ+f90°-θ=1 and f45°=12
Hence, ∑ θ= 1 o 89 o f θ =f 1 o +f 2 o +…+f 44 o +f 46 o +…+f 88 o +f 89 o +f 45 o
=f1o+f89o+f2o+f88o+…+f44o+f46o+f45o
=1×44+12=892

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let $f(1)=-2$ and $f^{\prime}(x) \geq 4.2$ for $1 \leq x \leq 6$. The possible value of $f(6)$ lies in the interval :

Options:
A
$[15,19)$

B
$(-\infty, 12)$

C
$[12,15)$

D
$[19, \infty)$

The correct answer is:
$[19, \infty)$

Solution:
Given $f(1)=-2$ and $f^{\prime}(x) \geq 4.2$ for $1 \leq x \leq 6$
Consider $f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}$ $\Rightarrow f(x+h)-f(x)=f^{\prime}(x) \cdot h \geq(4.2) h$
So, $f(x+h) \geq f(x)+(4.2) h$
put $x=1$ and $h=5$, we get
$f(6) \geq f(1)+5(4.2) \Rightarrow f(6) \geq 19$
Hence $f(6)$ lies in $[19, \infty)$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (25 Apr Online)
Let f [ 1 , â ) â [ 2 , â ) be a differentiable function such that f(1)= 1 3 . If 6 â« 1 x f t dt = 3 x f x - x 3 for all x â¥ 1 , then the value of 3 f 2 is

Options:

The correct answer is:
8

Solution:
Given, f 1 = 1 3 and 6 â« 1 x fâ¡ t dt = 3 x fâ¡ x - x 3 for all x â¥ 1
Using Newton- Leibnitz formula.
Differentiating both sides,
â 6 fâ¡ x Â· 1 - 0 = 3 f x + 3 x f â² x - 3 x 2
â 3 x f â² x - 3 f x = 3 x 2
â f â² x - 1 x f x = x
â x f â² x - f x x 2 = 1
â d dx f x x = 1
Integrating both sides,
â f x x = x + C âµ f 1 = 1 3
1 3 = 1 + C
â C = - 2 3
f x = x 2 - 2 3 x
â f 2 = 4 - 4 3 = 8 3
â´â3f 2 =8

Topic: Mathematics, Differential Equations, JEE Main