datavorous/entrance-exam-dataset huggingface dataset:
Let for the 9th term in the binomial expansion of 3+6xn, in the increasing powers of 6x, to be the greatest for x=32, the least value of n is n0. If k is the ratio of the coefficient of x6 to the coefficient of x3, then k+n0 is equal to

Options:

The correct answer is:
24

Solution:
Given,

3+6xn=3n1+2xn

Now if T9 is numerically greatest term, then T8≤T9≥T10

So, C7n3n-76x7≤C8n3n-86x8≥C9n3n-96x9

⇒ n!n-7!7!9≤n!n-8!8!3·6x≥n!n-9!9!6x2

⇒9n-7n-8⏟≤1832n-88≥369.894⏟

⇒72≤27n-7 and 27≥9n-8

⇒293≤n and n≤11

So, n0=10

For 3+6x10

Now Tr+1=Cr10310-r6xr

For coefficient of x6

r=6⇒C61034·66

For coefficient of x3

r=3⇒C31037·63

So, k=C610C310·34·6637·63=10!7!3!6!4!10!·8

⇒k=14

∴ k+n0=24

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Let for x∈R, fx=x+x2 and gx=x,x2,x<0x≥0. Then area bounded by the curve y=fogx and the lines y=0,2y-x=15 is equal to _____ .

Options:

The correct answer is:
72

Solution:
We have,

fx=x+x2=x;x≥00; x<0

gx=x2;x≥0x; x<0

So,

fogx=fgx=gx;gx≥00;gx<0

⇒fogx=x2;x≥00; x<0

And,

2y-x=15

Solving 2y-x=15 and y=x2, we get

2x2-x-15=0

⇒x=1±114

⇒x=3, -52

Required area

A=∫03x+152-x2dx+12×152×15

⇒A=x24+152x-x3303 + 2254

⇒A=94+452-9+2254

⇒A=99-36+2254

⇒A=72

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let four circles having radii r1=5 units, r2=5 units, r3=8 units and r4 unitsr4<5 are mutually touching each other externally, then the value of 2r4 is equal to

Options:

The correct answer is:
2.25

Solution:
Let, the centres of circles with radii r1,r2,r3,r4 are C1,C2,C3 and C4 respectively.

The foot of C3 on C1C2 is E, then C3E=132-52=12, C4E=12-8-r4=4-r4.
Now, C4C22=C4E2+52â5+r42=4-r42+52

âr4=89â2r4=94=2.25

Topic: Mathematics, Circle, JEE Main
Let from a point Ah,k chord of contacts are drawn to the ellipse x2+2y2=6 such that all these chords touch the ellipse x2+4y2=4, then locus of the point A is

Options:
A 4x2+9y2=36
B x2+y2=4
C x2-y2=9
D x2+y2=9

The correct answer is:
x2+y2=9

Solution:
Let chord of contact is PQ that touches x2+4y2=4 at R
Now, assume R=2cosâ¡Î¸,sinâ¡Î¸
Equation of PQ is hx+2yk=6 â¦ 1
T=0 for point Ah,k
Again, the equation of PQ is xcosÎ¸2+ysinÎ¸1=1 â¦ 2
From 1 & 2 we get 2hcosâ¡Î¸=2ksinâ¡Î¸=6
cosâ¡Î¸=h3,sinâ¡Î¸=k3âx2+y2=9

Topic: Mathematics, Ellipse, JEE Main
Let from a point Ah,k chords of contact are drawn to the ellipse x2+2y2=6 where all these chords touch the ellipse x2+4y2=4. Then, the perimeter (in units) of the locus of point A is

Options:
A 2π
B 3π
C 4π
D 6π

The correct answer is:
6π

Solution:
Let the chord of contact is PQ which touches x2+4y2=4 at R
Now, assume R=2cos⁡θ,2sin⁡θ
Equation of the chord of contact PQ is hx+2yk=6…..i
Again, the equation of the tangent PQ is xcos⁡θ2+ysin⁡θ1=1…..ii
From i and ii, we get,

h2cos⁡θ=2ksinθ=6
cos⁡θ=h3,sin⁡θ=k3⇒x2+y2=9

Hence, the perimeter of the circle is 6π units

Topic: Mathematics, Ellipse, JEE Main
Let \(f(x)\) be a polynomial and \(a, b\) be distinct real numbers. Then the remainder in the division of \(f(x)\) by \((x-a)(x-b)\) is

Options:
A \(\frac{(x-a) f(a)-(x-b) f(b)}{a-b}\)
B \(\frac{(x-a) f(b)-(x-b) f(a)}{a-b}\)
C \(\frac{(x-a) f(b)-(x-b) f(a)}{b-a}\)
D \(\frac{(x-a) f(a)-(x-b) f(b)}{b-a}\)

The correct answer is:
\(\frac{(x-a) f(b)-(x-b) f(a)}{b-a}\)

Solution:
Let \(f(x)=(x-a)(x-b) \cdot q(x)+r(x)\)
Let \(r(x)=\alpha x+\beta \quad[\because \operatorname{deg} r(x) < \operatorname{deg}\). of divisor]
\(\begin{aligned}
\therefore \quad f(x) & =(x-a)(x-b) \cdot q(x)+\alpha x+\beta \\
f(a) & =\alpha a+\beta \quad \ldots (i) \\
f(b) & =\alpha b+\beta \quad \ldots (ii)
\end{aligned}\)
Subtract Eqs. (i) from (ii)
\(\begin{aligned}
f(a)-f(b) & =\alpha(a-b) \\
\alpha & =\frac{f(a)-f(b)}{a-b} \\
\alpha & =\frac{f(b)-f(a)}{b-a}
\end{aligned}\)
Put, \(\alpha\) in Eq. (i)
\(\begin{aligned}
f(a) & =\left(\frac{f(b)-f(a)}{b-a}\right) \times a+\beta \\
f(a)[b-a] & =a f(b)-a f(a)+\beta(b-a) \\
b f(a)-a f(a) & =a f(b)-a f(a)+\beta(b-a) \\
\beta & =\frac{b f(a)-a f(b)}{(b-a)} \\
r(x) & =\alpha x+\beta \\
& =\frac{f(b)-f(a) x}{b-a}+\frac{b f(a)-a f(b)}{(b-a)} \\
& =\frac{x f(b)-x f(a)+b f(a)-a f(b)}{b-a} \\
r(x) & =\frac{(x-a) f(b)+(b-x) f(a)}{b-a} \\
r(x) & =\frac{(x-a) f(b)-(x-b) f(a)}{b-a}
\end{aligned}\)
Hence, option (c) is correct.

Topic: Mathematics, Quadratic Equation, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
Let \( f(x), g(x) \) be two continuous differentiable functions satisfying the relationships \( f^{\prime}(x)=g(x) \) and \( f^{\prime \prime}(x)=-f(x) . \) Let \( h(x)=[f(x)]^{2}+[g(x)]^{2} . \) If \( h(0)=5 \), then \( h(10)= \)

Options:
A \( 10 \)
B \( 5 \)
C \( 15 \)
D None of these

The correct answer is:
\( 5 \)

Solution:
Given fx, gx are two continuous differentiable functions
satisfying f′x=gx, f′′x=g′x and f′′x=-fx
Hence g′x=-fx
Given h(x)=f(x)2+g(x)2

Differentiating the above equation,
h′x=2fx f′x + 2gx g′x
h'(x)=2fx gx + gx -fx
h'(x)=2fx gx- fx gx=0
∴ hx=C , a constant
∴ h0=C i.e. C=5
hx=5 for all x .
Hence h10=5 .

Topic: Mathematics, Differentiation, JEE Main
Let \( f(x)=a x^{3}+5 x^{2}+c x+1 \) be a polynomial function. If \( f(x) \) has extreme at \( x=\alpha \) and \( \beta \) such that \( \alpha \beta < 0 \) and \( f(\alpha) f(\beta) < 0 \), then the equation \( f(x)=0 \) has

Options:
A three distinct real roots.
B one positive root, if \( f(\alpha) < 0 \) and \( f(\beta)>0 \).
C one negative root, if \( f(\alpha)>0 \) and \( f(\beta) < 0 \).
D All of the above

The correct answer is:
All of the above

Solution:
Given,

αβ<0

⇒α and β are of opposite signs.

Let α<0 and β>0.

It is given that fx has extremum at x=α, β.

Therefore, α and β are two distinct real roots of f'x=0.

But we know that between two distinct real roots of a polynomial, there is at least one real root of its derivative.

Therefore, fx has three distinct real roots λ, μ and v (say) such that λ<α<μ<β<v.

Thus, first option is correct.

If fx=0 has exactly one positive root, then it is evident from the figure that v>0 and λ, μ<0.

Therefore, α<μ<0

⇒fαf0<0 [∵μ lies between α and 0]

⇒fα<0 ∵f0=1>0

⇒ fβ>0 ∵fαfβ<0

Thus, second option is also correct.

If fx=0 has exactly one negative real root, then from the figure, we have λ<0 and μ, v>0.

⇒ 0<μ<β

⇒f0fβ<0 [∵μ lies between 0 and β]

⇒fβ<0 ∵f0=1>0

Thus, third option is also correct.

Topic: Mathematics, Application of Derivatives, JEE Main
Let \( f(x)=\left\{\begin{array}{cc}\frac{1+\cos x}{(\pi-x)^{2}} \cdot \frac{\sin ^{2} x}{\log \left(1+\pi^{2}-2 \pi x+x^{2}\right)} & , \quad x \neq \pi \\ k & , \quad x=\pi\end{array}\right. \). If \( f(x) \) is continuous functions at \( x=\pi \), then \( k \) is equal to

Options:
A \( \frac{1}{4} \)
B \( \frac{1}{2} \)
C \( \frac{-1}{2} \)
D \( -\frac{1}{4} \)

The correct answer is:
\( \frac{1}{2} \)

Solution:
The given function is
f x = { 1 + cos x Ï - x 2 Â· sin 2 x log 1 + Ï 2 - 2 Ï x + x 2 , x â Ï k , x = Ï
Since, f(x) continuous at x = Ï.
â´ lim x â Ï f x = f Ï
â lim h â 0 f Ï + h = k â lim h â 0 1 + cos Ï + h Ï - Ï - h 2 Â·
sin 2 Ï + h log 1 + Ï 2 - 2 Ï Ï + h + Ï + h 2 = k
â lim h â 0 1 - cos h Ï 2 Ã sin 2 h log 1 + h 2 = k
â lim h â 0 1 2 sin h/2 h/2 2 Â· h 2 log 1 + h 2 Â· sin h h 2 = k
â 1 2 = k

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let \( f(x)=\left\{\begin{array}{ll}3 x-p & : 0 \leq x \leq 2 \\ 2 x^{2}+q x & : 2 < x \leq 3\end{array}\right. \)
If \( f(x) \) is differentiable at \( x=2,(p, q)= \)

Options:
A \( (8,-5) \)
B \( \left(-\frac{5}{2}, 3\right) \)
C \( (-10,4) \)
D None of these

The correct answer is:
\( (8,-5) \)

Solution:
âµ diff. at x=2â conti. at x=2
â´ f2=f2+â6â p=8+2qâ p+2q=â 2 ....1
Now f â²x =3,0<x<24x+q,2<xâ¤3
â´ f â²2+=f â²2â â 8+q=3 â q=â 5, p=8

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let \( f(x)=\operatorname{sgn}(x) \) and \( g(x)=x\left(x^{2}-5 x+6\right) . \) The function \( f(g(x)) \) is discontinuous at

Options:
A infinitely many points.
B exactly one point.
C exactly three points.
D no point.

The correct answer is:
exactly three points.

Solution:
fx=sgnx=-1, x<0 0, x=01, x>0

gx=xx2-5x+6=xx-2x-3

fgx=sgngx=sgnxx-2x-3=-1 , xx-2x-3<00 , xx-2x-3=01 , xx-2x-3>0

Let hx=fgx

hx=-1 if x∈-∞,0∪2,3

hx=0, if x=0,2,3

hx=1 if x∈0,2∪3,∞

Therefore, fgx is discontinuous at three points 0, 2, 3.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let \( f(x)=x|x|, g(x)=\sin x \) and \( h(x)=(g o f)(x) \). Then

Options:
A \( h(x) \) is differentiable at \( x=0 \), but \( h^{\prime}(x) \) is not continuous at \( x=0 \)
B \( h(x) \) is not differentiable at \( x=0 \)
C \( h^{\prime}(x) \) is differentiable at \( x=0 \)
D \( h^{\prime}(x) \) is continuous at \( x=0 \) but is not differentiable at \( x=0 \)

The correct answer is:
\( h^{\prime}(x) \) is continuous at \( x=0 \) but is not differentiable at \( x=0 \)

Solution:
hx=sin⁡x2 x≥0-sin⁡x2 x<0
h'0+=limx→0+⁡sin⁡x2x=0
h'0-=limx→0-⁡-sin⁡x2x=0
Hence, h'0=0
So h'x2xcos⁡x2 x>00 x=0-2xcos⁡x2 x<0
Now, h''0+=limx→0+2xcosx2x=2
h''(0-)=limx→0-⁡-2xcos⁡x2x=-2
Hence, h'x is non-derivable at x=0.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let g: -2, 2→ℝ and f: -2, 2→ℝ are two functions defined as
gx=-1, if -2≤x<0x2-1,if 0≤x≤2 and f(x)=|g(x)|+g(|x|)+2. In the interval (-2, 2), f is not differentiable at x=

Options:
A 0
B 1
C 12
D -1

The correct answer is:
1

Solution:
gx=-1-2≤x≤0x2-10≤x≤2

=1-2≤x≤01-x20≤x<1x2-11≤x≤2

And gx=-1-2≤x<0x2-10≤x≤2

So, gx=x2-1, -2≤x≤2

Hence, fx=x2-2≤x<000≤x<12x2-11≤x≤2

f'x=2x,-2<x<00,0<x<14x,1<x<2

This clearly shows fx is not differentiable at x=1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $\mathrm{G}$ be the geometric mean of two positive numbers $\mathrm{a}$ and $\mathrm{b}$, and $\mathrm{M}$ be the arithmetic mean of $\frac{1}{\mathrm{a}}$ and $\frac{1}{\mathrm{~b}}$. If $\frac{1}{\mathrm{M}}: \mathrm{G}$ is $4: 5$, then $\mathrm{a}: \mathrm{b}$ can be:

Options:
A
$1: 4$

B
$1: 2$

C
$2: 3$

D
$3: 4$

The correct answer is:
$1: 4$

Solution:
$\mathrm{G}=\sqrt{a b}$
$$
\begin{aligned}
&\mathrm{M}=\frac{\frac{1}{a}+\frac{1}{b}}{2} \\
&\mathrm{M}=\frac{a+b}{2 a b}
\end{aligned}
$$
Given that $\frac{1}{\mathrm{M}}: \mathrm{G}=4: 5$
$$
\begin{aligned}
& \frac{2 a b}{(a+b) \sqrt{a b}}=\frac{4}{5} \\
\Rightarrow & \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4} \\
\Rightarrow & \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{5+4}{5-4}
\end{aligned}
$$
\{Using Componendo \& Dividendo
$$
\begin{aligned}
&\Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2 \sqrt{a b}}{(\sqrt{a})^2+(\sqrt{b})^2-2 \sqrt{a b}}=\frac{9}{1} \\
&\Rightarrow\left(\frac{\sqrt{b}+\sqrt{a})^2}{\sqrt{b}-\sqrt{a}}\right)^2=\frac{9}{1} \Rightarrow \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{3}{1} \\
&\Rightarrow \frac{\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}-\sqrt{b}+\sqrt{a}}=\frac{3+1}{3-1}
\end{aligned}
$$
\{Using Componendo \& Dividendo
$$
\sqrt{\frac{b}{a}}=\frac{4}{2}=2
$$
$$
\begin{aligned}
\frac{b}{a} &=\frac{4}{1} \\
\frac{a}{b} &=\frac{1}{4} \Rightarrow a: b=1: 4
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (12 Apr Online)
Let gt=∫-π/2π/2cosπ4t+fxdx, where fx=logex+x2+1,x∈R. Then which one of the following is correct?

Options:
A g1=g0
B 2 g1=g0
C g1=2 g0
D g1+g0=0

The correct answer is:
2 g1=g0

Solution:
We have,

fx=logex+x2+1,x∈R

⇒fx=logex+x2+1x-x2+1x-x2+1

⇒fx=loge-1x-x2+1

⇒fx=loge1x2+1-x

⇒f-x=loge1x2+1+x

⇒f-x=logex2+1+x-1

⇒f-x=-logex2+1+x

⇒f-x=-fx

Hence, fx is an odd function.

Now,

gt=∫-π/2π/2cosπ4t+fxdx

⇒gt=cosπ4t∫-π/2π/21 dx+∫-π/2π/2fxdx

⇒gt=πcosπ4t+∫-π/2π/2fxdx

⇒gt=πcosπ4t+0

∵fx is an odd function.

∴g1=π2⇒2g1=π

g0=π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Let gx=∫0xftdt, where f is continuous function in [0,3] such that 13≤ft≤1 for all t∈[0,1] and 0≤ft≤12 for all t∈(1,3].

The largest possible interval in which g(3) lies is :

Options:
A -1,-12
B -32,-1
C 13,2
D [1,3]

The correct answer is:
13,2

Solution:
13≤ft≤1∀t∈0,1

0≤ft≤12∀t∈(1,3]

Now, g3=∫03ftdt=∫01ftdt+∫13ftdt

∵∫0113dt≤∫01ftdt≤∫011.dt ...1

and ∫130dt≤∫13f1dt≤∫1312dt ...2

Adding, we get

13+0≤g3≤1+123-1

13≤g3≤2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let g(x)=3fx3+f(3-x) and f"(x)>0 for all x∈(0,3). If g is decreasing in (0,α) and increasing in (α,3), then 8α is

Options:
A 24
B 0
C 18
D 20

The correct answer is:
18

Solution:
Given: g(x)=3fx3+f(3-x) and f"(x)>0∀x∈(0,3)

⇒f'(x) is increasing function

⇒g'(x)=3×13·f'x3-f'(3-x)

⇒g'(x)=f'x3-f'(3-x)

It is given that g is decreasing in 0, α then g'(x)<0.

⇒f'x3-f'(3-x)<0

⇒f'x3<f'(3-x)

⇒x3<3-x

⇒x<9-3x

⇒x<94

⇒α=94

⇒8α=8×94

⇒8α=18

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Let $g(x)$ be the anti-derivative of $f(x)$. Then the function for which $\log _e\left(1+(g(x))^2\right)+c$ is an anti-derivative is]

Options:
A $\left(1+(g(x))^2\right) g^{\prime}(x) f(x)$
B $\frac{-2 f(x) g(x)}{1+g(x)}$
C $\frac{2 f(x) g(x)}{1+(g(x))^2}$
D $\frac{2 g(x)}{1+(g(x))^2}$

The correct answer is:
$\frac{2 f(x) g(x)}{1+(g(x))^2}$

Solution:
We have a function $\log _{\mathrm{e}}\left(1+(\mathrm{g}(\mathrm{x}))^2\right)+\mathrm{c}$ as antderivative of some function $\mathrm{f}(\mathrm{x})$
$$
\Rightarrow \int \mathrm{f}(\mathrm{x}) \mathrm{dx}=\mathrm{g}(\mathrm{x})+\mathrm{c}
$$
Let $g(x)+c=\log _e\left(1+(g(x))^2\right)+c$
$$
\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{1+(\mathrm{g}(\mathrm{x}))^2} \cdot 2 \mathrm{~g}(\mathrm{x}) \cdot \mathrm{g}^{\prime}(\mathrm{x})
$$
or $f(x)=\frac{2 g(x) f(x)}{1+(g(x))^2} \quad from(i)$
or $\frac{2 \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})}{1+(\mathrm{g}(\mathrm{x}))^2}$

Topic: Mathematics, Differentiation, JEE Main
Let gx=cosx2, fx=x, and α,βα<β be the roots of the quadratic equation 18x2-9πx+π2=0. Then the area (in sq. units) bounded by the curve y=gofx and the lines x=α,x=β and y=0, is

Options:
A 122-1
B 123-1
C 123+1
D 123-2

The correct answer is:
123-1

Solution:
Given, gx=cosx2, fx=x
Given quadratic is 18x2-9πx+π2=0
The roots are x=9π±81π2-72π22×18
⇒x=9π±9π236
⇒x=9±3π36
⇒x=π3, π6
y=gofx=cosx2=cosx

Required Area =∫π6π3cos⁡x dx=sinxπ6π3
⇒ 32-12

Topic: Mathematics, Area Under Curves, JEE Main
Let gx=cosx2, fx=x, and α,βα<β be the roots of the quadratic equation 18x2-9πx+π2=0. Then the area (in sq. units) bounded by the curve y=gofx and the lines x=α,x=β and y=0, is

Options:
A 122-1
B 123-1
C 123+1
D 123-2

The correct answer is:
123-1

Solution:
Given, gx=cosx2, fx=x
Given quadratic is 18x2-9πx+π2=0
The roots are x=9π±81π2-72π22×18
⇒x=9π±9π236
⇒x=9±3π36
⇒x=π3, π6
y=gofx=cosx2=cosx

Required Area =∫π6π3cos⁡x dx=sinxπ6π3
⇒ 32-12

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2018 (08 Apr)
Let gx=fx+f1-x and f"x>0,x∈0,1. If g is decreasing in the interval 0,α and increasing in the interval α,1, then tan-12α+tan-11α+tan-1α+1α is equal to

Options:
A π
B 5π4
C 3π4
D 3π2

The correct answer is:
π

Solution:
Given,

gx=fx+f1-x and f"x>0,x∈0,1. If g is decreasing in the interval 0,α and increasing in the interval α,1,

Now solving,

gx=fx+f1-x

Now differentiating both side we get,

g'x=f'x-f'1-x

Differentiating again we get,

g"x=f"x+f"1-x>0

So, g'x is increasing as given f"x>0

⇒g'0<g'1

⇒f'0-f'1<f'1-f'0

⇒f'0<f'1

Now finding g'x=0 we get,

⇒f'x=f'1-x

⇒x=1-x

⇒x=12

Now g'x is positive for x∈0,12

And g'x is negative for x∈12,1

∴ α=12

So,

tan-12α+tan-11α+tan-1α+1α

=tan-11+tan-12+tan-13

=π4+tan-12+31-6

=π4+tan-1-1

=π4+3π4=π

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m$ and $n$ are integers, $m \neq 0, n>0$ and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim _{x \rightarrow 1^{+}} g(x)=p$, then

Options:
A
$n=1, m=1$

B
$n=1, m=-1$

C
$n=2, m=2$

D
$n>2, m=n$

The correct answer is:
$n=2, m=2$

Solution:
Given, $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m \neq 0, n$ are integers and $|x-1|=\left\{\begin{array}{l}x-1 ; x \geq 1 \\ 1-x ; x < 1\end{array}\right.$
The left hand derivative of $|x-1|$ at $x=1$ is $p=-1$.
Also,
$$
\begin{aligned}
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(1+h-1)^n}{\log \cos ^m(1+h-1)}=-1 \Rightarrow \lim _{h \rightarrow 0} \frac{h^n}{\log \cos ^m h}=-1 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{h^n}{m \log \cos h}=-1
\end{aligned}
$$
[Using L' Hospital rule]
$$
\begin{array}{ll}
\Rightarrow & \quad \lim _{h \rightarrow 0} \frac{n \cdot h^{n-1}}{m \frac{1}{\cos h}(-\sin h)}=-1 \\
\Rightarrow \quad & \quad \lim _{h \rightarrow 0}\left(-\frac{n}{m}\right) \cdot \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=-1 \\
\Rightarrow \quad & \quad\left(\frac{n}{m}\right) \lim _{h \rightarrow 0} \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=1 \\
\Rightarrow & n=2 \text { and } \frac{n}{m}=1 \\
\therefore & \quad m=n=2
\end{array}
$$

Topic: Mathematics, Limits, JEE Main
Let $g(x)=\log f(x)$, where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3, \ldots . g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)$ is equal to

Options:
A
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$

B
$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$

C
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$

D
$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$

The correct answer is:
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$

Solution:
Since, $\quad f(x)=e^{g(x)}$
$$
\begin{aligned}
\Rightarrow \quad e^{g(x+1)} & =f(x+1) \\
& =x f(x) \\
& =x e^{g(x)}
\end{aligned}
$$
and
$$
g(x+1)=\log x+g(x)
$$
$$
\Rightarrow \quad g(x+1)-g(x)=\log x
$$
Replacing $x$ by $x-\frac{1}{2}$, we get
$$
\begin{aligned}
g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) & =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\
\therefore g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) & =-\frac{4}{(2 x-1)^2}
\end{aligned}
$$
Substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get
$$
g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .
$$

Topic: Mathematics, Functions, JEE Main
Let gx=x-2 and hx=ggx be two functions, then the value of h'-1+h'1+h'3+h'5 is equal to (where, h' denotes the derivative of h)

Options:
A 2
B -1
C 0
D 1

The correct answer is:
0

Solution:
gx= x-2,x>22-x,x<2
hx=gx-2,gx>22-gx,gx<2
hx=x-4,x≥4-x,x≤04-x,x,x∈2,4x∈0,2

Then, h'-1+h'1+h'3+h'5=0

Topic: Mathematics, Differentiation, JEE Main
Let gx=xfx, where fx=x2sin⁡1x:x≠00:x=0. At x=0,

Options:
A g is differentiable but g ′ is not continuous
B g is differentiable while f is not differentiable
C both f and g are non differentiable
D g is differentiable and g â² is continuous

The correct answer is:
g is differentiable and g ′ is continuous

Solution:
We have, gx=x3sin⁡1x:x≠00:x=0
For x≠0,
g ′ x = x 3 cos 1 x − 1 x 2 +3 x 2 sin 1 x
=-xcos⁡1x++3x2sin⁡1x
For x=0,
g'0=limx→0gx−g0x−0=limx→0x3sin1x−0x
=limx→0x2sin⁡1x=0.
∴g'x=3x2sin1x−xcos1x, x≠00, x=0
g ′ is continuous at x=0
Also, g is differentiable at x=0 .

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let \( g(x) \) be a polynomial of degree one and \( f(x) \) be a continuous and differentiable function defined by
\( f(x)=\left\{\begin{array}{cl}g(x), & x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}, & x>0\end{array} .\right. \) If \( f^{\prime}(1)=f^{\prime}(-1) \), then

Options:
A \( f^{\prime}(-1)=-\frac{2}{3}\left(\frac{1}{6}+\ln \frac{3}{2}\right) \)
B \( f^{\prime}(-1)=+\frac{3}{2}\left(\frac{1}{6}+\ln \frac{3}{2}\right) \)
C \( f^{\prime}(-1)=-\frac{2}{3}\left(\frac{6}{1}+\ln \frac{2}{3}\right) \)
D \( f^{\prime}(-1)=-\frac{2}{3}\left(\frac{1}{6}-\ln \frac{3}{2}\right) \)

The correct answer is:
\( f^{\prime}(-1)=-\frac{2}{3}\left(\frac{1}{6}+\ln \frac{3}{2}\right) \)

Solution:
Let gx=ax+b

LHL=lim x→0-fx=lim x→0-gx=b

and RHL=limx→0+fx=1+02+010=12∞=0

As fx is continuous, LHL=RHL=f0⇒b=0

For x>0, fx=1+x2+x1/x

lnfx=1xln1+x-ln2+x

1fxf'x=-1x2log1+x2+x+1x11+x-12+x

f'1=f1-ln23+112-13

=23-ln23+16

f'-1=a

∵ f'1=f'-1⇒a=-2316+ln32

f'-1=-2316+ln32

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let $H_1, H_2, \ldots, H_n$ be mutually exclusive events with $P\left(H_i\right)>0, i=1,2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
Statement I $P\left(H_i / E\right)>P\left(E / H_i\right) P\left(H_i\right)$ for $i=1,2, \ldots, n$.
Statement II $\sum_{i=1}^n P\left(H_i\right)=1$

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement II

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:
Statement : I If $P\left(H_i \cap E\right)=0$ for some $i$, then
$$
P\left(\frac{H_i}{E}\right)=P\left(\frac{E}{H_i}\right)=0
$$
If $P\left(H_i \cap E\right) \neq 0$ for $\forall i=1,2, \ldots . n$, then
$$
\begin{aligned}
P\left(\frac{H_i}{E}\right) & =\frac{P\left(H_i \cap E\right)}{P\left(H_i\right)} \times \frac{P\left(H_i\right)}{P(E)} \\
& \left.=\frac{P\left(\frac{E}{H_i}\right) \times P\left(H_i\right)}{P(E)}>P\left(\frac{E}{H_i}\right) \cdot P\left(H_i\right) \quad \text { [as } 0 < P(E) < 1\right]
\end{aligned}
$$
Hence, Statement I may not always be true.
Statement II
Clearly, $H_1 \cup H_2 \cup \ldots . \cup H_n=S$ (sample space)
$$
\Rightarrow \quad P\left(H_1\right)+P\left(H_2\right)+\ldots+P\left(H_n\right)=1 .
$$
Hence, Statement II is true.

Topic: Mathematics, Probability, JEE Main
Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4 \sqrt{3}$. Suppose the point $(\alpha, 6), \alpha>0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2+\beta$ is equal to

Options:
A 172
B 171
C 169
D 170

The correct answer is:
171

Solution:
$\begin{aligned} & \mathrm{H}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1, \mathrm{e}=\sqrt{3} \\ & \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{3} \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=2 \\ & \qquad \mathrm{a}^2=2 \mathrm{~b}^2 \\ & \text { length of L.R. }=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=4 \sqrt{3} \\ & \qquad a=\sqrt{6}\end{aligned}$
$P(\alpha, 6)$ lie on $\frac{y^2}{3}-\frac{x^2}{6}=1$
$\begin{aligned}
& 12-\frac{\alpha^2}{6}=1 \Rightarrow \alpha^2=66 \\
& \text { Foci }=(0, \pm \text { be })=(0,3) \&(0,-3)
\end{aligned}$
Let $d_1 \& d_2$ be focal distances of $\mathrm{P}(\alpha, 6)$
$\begin{aligned} & \mathrm{d}_1=\sqrt{\alpha^2+(6+b e)^2}, \mathrm{~d}_2=\sqrt{\alpha^2+(6-b e)^2} \\ & \mathrm{~d}_1=\sqrt{66+81}, \mathrm{~d}_2=\sqrt{66+9} \\ & \beta=\mathrm{d}_1 \mathrm{~d}_2=\sqrt{147 \times 75}=105 \\ & \alpha^2+\beta=66+105=171\end{aligned}$
SECTION - B

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let H be the hyperbola, whose foci are 1±2,0 and eccentricity is 2. Then the length of its latus rectum is:

Options:
A 3
B 52
C 2
D 32

The correct answer is:
2

Solution:
Given the foci points of a hyperbola are 1±2,0

So, S1+2,0,S'1-2,0

SS'=2ae=(1+2)-(1-2)

⇒2ae=22

⇒2×a×2=22, given e=2

⇒a=1

Since b2=a2(e2-1)

⇒b2=1(2-1)=1

So, the latus rectum of a hyperbola is given by,

L.R. =2b2a

L.R. =2

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let H:x2a2-y2 b2=1,a>0, b>0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 422+14. If the eccentricity H is 112, then value of a2+b2 is equal to ______.

Options:

The correct answer is:
88

Solution:
Given x2a2-y2b2=1, 2a+2b=422+14 and e=112

We know e2=1+b2a2⇒114=1+b2a2⇒b2=74a2

∴x2a2-y272a2=1

Now 2a+2·7a2=422+14

a2+7=422+7

a=42⇒a2=32

b2=74×16×2=56

Hence a2+b2=88

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (29 Jun Shift 1)