datavorous/entrance-exam-dataset huggingface dataset:
Let fx=a-10axa-1ax2axa,a∈R. Then the sum of the squares of all the values of a for 2f'10-f'5+100=0 is

Options:
A 117
B 106
C 125
D 136

The correct answer is:
125

Solution:
Given, fx=a-10axa-1ax2axa

fx=a1-10xa-1x2axa

=a1a2+ax+1ax+x2

⇒fx=ax+a2

Now differentiating the function w.r.t x

We get, f'x=2ax+a

Given, 2f'10-f'5+100=0

⇒2×2a10+a-2a5+a+100=0

⇒40a+4a2-10a-2a2+100=0

2a2+30a+100=0

⇒a2+15a+50=0

a+10a+5=0

a=-10 or a=-5

Required =-102+-52=125

Topic: Mathematics, Determinants, JEE Main
Let fx=a-10axa-1ax2axa,a∈R. Then the sum of the squares of all the values of a for 2f'10-f'5+100=0 is

Options:
A 117
B 106
C 125
D 136

The correct answer is:
125

Solution:
Given, fx=a-10axa-1ax2axa

fx=a1-10xa-1x2axa

=a1a2+ax+1ax+x2

⇒fx=ax+a2

Now differentiating the function w.r.t x

We get, f'x=2ax+a

Given, 2f'10-f'5+100=0

⇒2×2a10+a-2a5+a+100=0

⇒40a+4a2-10a-2a2+100=0

2a2+30a+100=0

⇒a2+15a+50=0

a+10a+5=0

a=-10 or a=-5

Required =-102+-52=125

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let $f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, x \in \mathbb{R}$ wherea, b

and d are non-zero real constants. Then :

Options:
A $f$ is an increasing function of $x$
B $f$ is a decreasing function of $x$
C $f^{\prime}$ is not a continuous function of $x$
D $f$ is neither increasing nor decreasing function of $x$

The correct answer is:
$f$ is an increasing function of $x$

Solution:
$f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{(d-x)}{\sqrt{b^{2}+(d-x)^{2}}}$

$=\frac{x}{\sqrt{a^{2}+x^{2}}}+\frac{(x-d)}{\sqrt{b^{2}+(x-d)^{2}}}$

$\begin{array}{l}

f^{\prime}(x)=\frac{\sqrt{a^{2}+x^{2}}-\frac{x(2 x)}{2 \sqrt{a^{2}+x^{2}}}}{\left(a^{2}+x^{2}\right)} \\

=\frac{a^{2}+x^{2}-x^{2}}{\left(a^{2}+x^{2}\right)^{3 / 2}}+\frac{b^{2}+(x-d)^{2}-(x-d)^{2}}{\left(b^{2}+(x-d)^{2}\right)^{3 / 2}} \\

+\frac{\sqrt{b^{2}+(x-d)^{2}}-\frac{(x-d) 2(x-d)}{2 \sqrt{b^{2}+(x-d)^{2}}}}{\left(a^{2}+(x-d)^{2}\right)} \\

\Rightarrow f^{\prime}(x)>0, \square x \in R

\end{array}$

$\Rightarrow f(x)$ is increasing function.

Hence, $f(x)$ is increasing function

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Let fx=a; x=Ï22x-Ï9+2x-Ï-b; x>Ï2. If fx is continuous at x=Ï2, then the value of a25b is

Options:

The correct answer is:
2.40

Solution:
limxâÏ2+fx=fÏ2
âlimxâÏ2+2x-Ï9+2x-Ï-b=a
For the denominator to be zero, b=3
Now, on rationalization
limxâÏ2+2x-Ï9+2x-Ï+39+2x-Ï-9=a
â´a=6

Topic: Mathematics, Continuity and Differentiability, JEE Main
Let fx=ax2+bx+c be such that f1=3, f-2=λ and f3=4. If f0+f1+f-2+f3=14, then λ is equal to

Options:
A -4
B 132
C 232
D 4

The correct answer is:
4

Solution:
Given fx=ax2+bx+c be such that f1=3, f-2=λ and f3=4. If f0+f1+f-2+f3=14, then λ is equal to

Now given f0+f1+f-2+f3=14

⇒f0+3+λ+4=14

⇒f0=7-λ=c

Now f1=a+b+c=3 ...i

f3=9a+3b+c=4 ...ii

f-2=4a-2b+c=λ ...iii

Now subtracting equation ii-iii

We get, a+b=4-λ5, now putting in equation i

We get, 4-λ5+7-λ=3

⇒6λ=24

⇒λ=4

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let $f(x)=a x^3+b x^2+c x+41$ be such that $f(1)=40, f^{\prime}(1)=2$ and $f^{\prime}(1)=4$. Then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to:

Options:
A 73
B 62
C 51
D 54

The correct answer is:
51

Solution:
$\begin{aligned} & f(x)=a x^3+b x^2+c x+41 \\ & f^{\prime}(x)=3 a x^2+2 b x+c x \\ & \Rightarrow f^{\prime}(1)=3 a+2 b+c=2 \ldots(1)\end{aligned}$
$\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{n})=6 \mathrm{ax}+2 \mathrm{~b} \\ & \Rightarrow \mathrm{f}^{\prime \prime}(1)=6 \mathrm{a}+2 \mathrm{~b}=4 \\ & 3 \mathrm{a}+\mathrm{b}=2 \ldots \ldots \ldots .(2)\end{aligned}$
$\begin{aligned} & (1)-(2) \\ & b+c=0...(3)\end{aligned}$
$\begin{aligned}
& \mathrm{f}(1)=40 \\
& \mathrm{a}+\mathrm{b}+\mathrm{c}+41=40
\end{aligned}$
use (3)
$a+41=40$
by (2)
$\begin{aligned}
& -3+b=2 \Rightarrow b=5 \& c=-5 \\
& a^2+b^2+c^2=1+25+25=51
\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Let fx=ax (a>0) be written as fx=f1x+f2x, where f1(x) is an even function and f2(x) is an odd function. Then f1x+y+f1(x-y) equals:

Options:
A 2f1xf1y
B 2f1x+yf1x-y
C 2f1xf2y
D 2f1x+yf2x-y

The correct answer is:
2f1xf1y

Solution:
fx=ax, a>0

fx=ax+a-x+ax-a-x2
⇒f1x=ax+a-x2
f2x=ax-a-x2

⇒f1x+y+f1x-y

=ax+y+a-x-y2+ax-y+a-x+y2
=ax+a-x2ay+a-y

=f1x×2f1y

=2f1xf1y

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Let fx and gx be two real polynomials of degree 2 and 1 respectively. If fgx=8x2-2x, and gfx=4x2+6x+1, then the value of f2+g2 is ______.

Options:

The correct answer is:
18

Solution:
Given, fgx=8x2-2xand gfx=4x2+6x+1

Let fx=cx2+dx+e and gx=ax+b

So, fgx=cax+b2+dax+b+e=8x2-2x

and gfx=acx2+dx+e+b=4x2+6x+1

On comparing both side we get, a=2, b=-1, c=2, d=3, e=1

So, gx=2x-1⇒g2=3

& fx=2x2+3x+1

f2=8+6+1=15.

So, f2+g2=18

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let fx be a continuous and positive function and the area bounded by y=fx, the x-axis and the lines x=0 and x=a is 4a3+sin⁡a sq. units a>0. If fπ2=kπ2, then the value of k is

Options:

The correct answer is:
3

Solution:
Given area is ∫0afxdx
∴∫0afxdx=4a3+sin⁡a
Differentiating w.r.t. a, we get
fa=12a2+cos⁡a
⇒fπ2=3π2
⇒k=3

Topic: Mathematics, Area Under Curves, JEE Main
Let fx be a continuous and positive function, such that the area bounded by y=fx, x-axis and the lines x2=2ax is 6a2+sin⁡a ∀a>0 sq. units. If fπ=kπ, then the value of k is equal to

Options:

The correct answer is:
3

Solution:
x2-2ax=0⇒x=0,2a

So, the given area is ∫02afxdx
∴∫02afxdx=6a2+sin⁡a
Differentiating with respect to a, we get,

2f2a=12a+cos⁡a

⇒fx=3x+12cosx2
⇒fπ=3π

⇒k=3

Topic: Mathematics, Definite Integration, JEE Main
Let fx be a cubic function such that f â² 1 = f â³ 2 =0. If x=1 is a point of local maxima of fx, then the local minimum value of fx occurs at

Options:
A x=0
B x=2
C x=4
D x=3

The correct answer is:
x=3

Solution:
Let f ″ x =6λ x−2
⇒f'x=3λx2−4x+c
As f ′ 1 =0⇒c=9λ

i.e. f'x=3λx2−4x+9λ
Also, f'3=0
Hence, x=3 is the other point of extrema.

Topic: Mathematics, Application of Derivatives, JEE Main
Let fx be a cubic polynomial with f1=-10, f-1=6, and has a local minima at x=1, and f'x has a local minima at x=-1. Then f3 is equal to

.

Options:

The correct answer is:
22

Solution:
Let, fx=ax3+bx2+cx+d

Diff w. r. t. 'x'

f'x=3ax2+2bx+c

Again diff. w. r. t. 'x'

f"x=6ax+2b

f"-1=0

-6a+2b=0

b=3a

f'1=0

3a+6a+c=0

c=-9a

f1=-10

-5a+d=-10 …i

f(-1)=6

11a+d=6 …ii

From equations i-ii

We get a=1, d=-5, b=3, c=-9

Then, fx=x3+3x2-9x-5

Hence, f3=27+27-27-5

f3=22

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let $f(x)$ be a differentiable function and $f^{\prime}(4)=5 .$ Then, $\lim _{x \rightarrow 2} \frac{f(4) - f\left(x^{2}\right)}{x-2}$ equals

Options:
A 0
B 5
C 20
D -20

The correct answer is:
-20

Solution:
Given, $f^{\prime}(4)=5$
$$
\begin{array}{ll}
\text { Now, } \lim _{x \rightarrow 2} \frac{f(4)-f\left(x^{2}\right)}{x-2} & {\left[\frac{0}{0} \text { form }\right]} \\
=\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) \cdot 2 x}{1} \\
=\frac{-f^{\prime}(4) \cdot 2 \times 2}{1} \\
=-(5) \times 4=-20
\end{array}
$$

Topic: Mathematics, Limits, JEE Main
Let fx be a differentiable function at x=a with f'a=2 and fa=4. Then limx→axfa-afxx-a equals:

Options:
A a+4
B 2a-4
C 4-2a
D 2a+4

The correct answer is:
4-2a

Solution:
f'a=2,fa=4

limx→axfa-afxx-a

⇒limx→afa-af'x1 (L'Hospitals rule)

=fa-af'a

=4-2a

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let fx be a differentiable function defined on 0, 2 such that f'x=f'2-x for all x∈0,2,f0=1 and f2=e2. Then the value of ∫02fxdx is

Options:
A 21+e2
B 1+e2
C 1-e2
D 21-e2

The correct answer is:
1+e2

Solution:
f'x=f'2-x

fx=-f2-x+c

put x=0

f0=-f2+c

c=f0+f2=1+e2

so, fx+f2-x=1+e2

I=∫02fxdx

I=∫02f2-xdx

2I=∫02fx+f2-xdx

2I=1+e2∫02dx

I=1+e2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let fx be a differentiable function on xâR such that fx+y=fxâfy for all x,y. If f0â 0, f5=12 and f â² 0 =16 , then f â² 5 is equal to

Options:
A 190
B 186
C 196
D 192

The correct answer is:
192

Solution:
f ′ 5 = lim h→0 f 5+h −f 5 h
=limh→0f5+h-f5+0h
=limh→0f5·fh-f5·f0h
∵fx+y=fx.fyfor all x,y
=limh→0fh-f0h·f5
f ′ 0 ×f 5 =16×12=192

Topic: Mathematics, Differentiation, JEE Main
Let fx be a differentiable function such that f'x=7-34fxx, x>0 and f1≠4. Then limx→0+⁡x f1x

Options:
A does not exist.
B exists and equals 4.
C exists and equals 47.
D exists and equals 0.

The correct answer is:
exists and equals 4.

Solution:
We have, dydx+3y4x=7 ( where y=f(x))
This is a linear differential equation.
Integrating factor, I.F. =e∫3 dx4x=e34lnx=x34
So, integrating the given differential equation using I.F.
⇒y.x34=∫7.x34dx
⇒y.x34=7.4x747+C
⇒yx34=4x74+C
⇒f(x)=4x+Cx-34
⇒limx→0+xf1x⇒limx→0+x4x+Cx34=4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let fx be a differentiable function such that ∫tt2xfxdx=43t3-4t3 ∀t≥0, then f1 is equal to

Options:
A 4
B 43
C 3
D 83

The correct answer is:
83

Solution:
Differentiating both side using Leibnitz rule, we get,
t 2 f t 2 â2tâtf t =4 t 2 â 4 3
Substituting t=1, we get,
f1=83

Topic: Mathematics, Definite Integration, JEE Main
Let fx be a function satisfying fx+fπ-x=π2, ∀x∈ℝ. Then ∫0πfxsinx dx is equal to

Options:
A π24
B 2π2
C π2
D π22

The correct answer is:
π2

Solution:
Let

I=∫0πfxsinxdx ...1

Now using the property ∫abfxdx=∫abfa+b-xdx we get,

⇒I=∫0πfπ-xsinπ-xdx

⇒I=∫0πfπ-xsinxdx ...2

Adding 1 & 2, we get

2I=∫0πfx+fπ-xsinxdx

⇒2I=π2∫0πsinxdx {as given fx+fπ-x=π2}

⇒2I=π2-cosx0π

⇒2I=2π2

⇒I=π2

Hence this is the correct option.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Let $f(x)$ be a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ be a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$, is

Options:
A
$\mathrm{e}+\frac{\mathrm{e}^2}{2}+\frac{5}{2}$

B
$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{5}{2}$

C
$\mathrm{e}+\frac{\mathrm{e}^2}{2}-\frac{3}{2}$

D
$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$

The correct answer is:
$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$

Solution:
Let $f(x)=e^x$
$\therefore \int_0^1 f(x) g(x) d x=\int_0^1 e^x\left(x^2-e^x\right) d x$
$=\int_0^1 x^2 e^x d x-\int_0^1 e^{2 x} d x$
$=\left[\mathrm{x}^2 \mathrm{e}^{\mathrm{x}}\right]_0^1-2\left[\mathrm{xe}^{\mathrm{x}}-\mathrm{e}^{\mathrm{x}}\right]_0^1-\frac{1}{2}\left[\mathrm{e}^{2 \mathrm{x}}\right]$
$=\mathrm{e}-\left[\frac{\mathrm{e}^2}{2}-\frac{1}{2}\right]-2[\mathrm{e}-\mathrm{e}+1]=\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2003
Let fx be a function such that fx+y=fx·fy for all x,y∈N, If f1=3 and ∑k=1nfk=3279, then the value of n is

Options:
A 6
B 8
C 7
D 9

The correct answer is:
7

Solution:
Given,

fx+y=fx·fy and f1=3

Now taking x=1 & y=1 we get,

f1+1=f1·f1⇒f2=32

Similarly f2+1=f2·f1⇒f3=32×3=33

And so on fn=3n

So, ∑r=1nfr=3279

⇒f1+f2+f3.......fn=3279

⇒3+32+33.......+3n=3279

⇒3×3n-13-1=3279

⇒3n-12=1093

⇒3n-1=2186

⇒3n=2187

⇒3n=37

⇒n=7

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let fx be a non-constant polynomial such that fa=fb=fc=2. Then the minimum number of roots of the equation f"x=0 in x∈a,c is/are

Options:
A 2
B 1
C 0
D 3

The correct answer is:
1

Solution:
As fa=fb=fc, then by Rolleâs theorem, we get,
fc1=fc2=0 â for some c1âa,b & c2âb,c
Again, applying Rolleâs theorem to function y=f'x,

we have f"c=0 for some c3âc1,c2
Hence, the equation f"x=0 has at least one root.

Topic: Mathematics, Application of Derivatives, JEE Main
Let $\mathrm{f}(\mathrm{x})$ be a non-constant polynomial with real coefficients such that $\mathrm{f}\left(\frac{1}{2}\right)=100$ and $\mathrm{f}(\mathrm{x}) \leq 100$ for all real $\mathrm{x}$. Which of the following statements is NOT necessarily true?

Options:
A The coefficient of the highest degree term in $\mathrm{f}(\mathrm{x})$ is negative
B $f(x)$ has at least two real roots
C If $x \neq 1 / 2$ then $f(x) < 100$
D At least one of the coefficients of $\mathrm{f}(\mathrm{x})$ is bigger than 50.

The correct answer is:
If $x \neq 1 / 2$ then $f(x) < 100$

Solution:
Coefficient of highest degree term must be negative becuase ifit is positive, then $\mathrm{x} \rightarrow \infty, \mathrm{y} \rightarrow \infty$ and it is not possible, since $f(x) \leq 100$.
Now, graph will be like
at least two real roots will be there, \& if $x \neq \frac{1}{2}$, then $f(x) < 100$, it is not always true, as the graph can be like. this also
Now, let the highest coefficients, it can have is 49
then, $f\left(\frac{1}{2}\right)=49+\frac{49}{2}+\frac{49}{2^{2}}+\ldots$
But the sum cannot be equal to 100 .

Topic: Mathematics, Quadratic Equation, JEE Main
Let $f(x)$ be a non-negative continous function such that the area bounded by the curve $y=f(x), x$-axis and the ordinates $x=\frac{\pi}{4}, \quad x=\beta\gt\frac{\pi}{4}$ is $\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)$. Then $f\left(\frac{\pi}{2}\right)$ is

Options:
A $\left(1-\frac{\pi}{4}-\sqrt{2}\right)$
B $\left(1-\frac{\pi}{4}+\sqrt{2}\right)$
C $\left(\frac{\pi}{4}+\sqrt{2}-1\right)$
D $\left(\frac{\pi}{4}-\sqrt{2}+1\right)$

The correct answer is:
$\left(1-\frac{\pi}{4}+\sqrt{2}\right)$

Solution:
Given that, $\int_{\pi / 4}^\beta f(x) d x=\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta$
Differentiating w.r.t. $\beta$, we get
$\therefore f(\beta)=\sin \beta+\beta \cos \beta-\frac{\pi}{4} \sin \beta+\sqrt{2}$
Hence,
$f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}+\sqrt{2}\right)$

Topic: Mathematics, Area Under Curves, JEE Main
Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y=f(x), x$-axis and the ordinates $x=\frac{\pi}{4}$ and $x=\beta>\frac{\pi}{4}$ is $\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)$. Then $f\left(\frac{\pi}{2}\right)$ is

Options:
A
$\left(\frac{\pi}{4}+\sqrt{2}-1\right)$

B
$\left(\frac{\pi}{4}-\sqrt{2}+1\right)$

C
$\left(1-\frac{\pi}{4}-\sqrt{2}\right)$

D
$\left(1-\frac{\pi}{4}+\sqrt{2}\right)$

The correct answer is:
$\left(1-\frac{\pi}{4}+\sqrt{2}\right)$

Solution:
Given that $\int_{\pi / 4}^\beta f(x) d x=\beta \sin \beta+\frac{\pi}{4} \cos \beta+2 \beta$
Differentiating w. r. t $\beta$
$f(\beta)=\beta \cos \beta+\sin \beta-\frac{\pi}{4} \sin \beta+\sqrt{2}$
$f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right) \sin \frac{\pi}{2}+\sqrt{2}=1-\frac{\pi}{2}+\sqrt{2}$.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2005
Let $f(x)$ be a polynomial function of second degree. If $f(1)=f(-1)$ and $a, b, c$ are in A.P, then $f^{\prime}(a), f^{\prime}(c)$ are in

Options:
A
Arithmetic-Geometric Progression

B
A.P.

C
G.P.

D
H.P.

The correct answer is:
A.P.

Solution:
$f(x)=a x^2+b x+c$
$f(1)=f(-1) \Rightarrow a+b+c=a-b+c$ or $b=0$
$\therefore f(x)=a x^2+c$ or $f^{\prime}(x)=2 a x$
Now $f^{\prime}(a) ; f^{\prime}(b) ;$ and $f^{\prime}(c)$ are $2 a(a) ; 2 a(b) ; 2 a(c)$. If $a, b, c$ are in A.P. then $f^{\prime}(a) ; f^{\prime}(b)$ and $f^{\prime}(c)$ are also in A.P.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2003
Let fx be a polynomial function such that fx+f'x+f''x=x5+64. Then, the value of limx→1fxx-1 is equal to

Options:
A -15
B 15
C -60
D 60

The correct answer is:
-15

Solution:
Let fx=x-1x4+ax3+bx2+cx+d

fx+f'x+f''x=x5+a+4x4+b+3a+16x3+

9a+2b+c-12x2+d+c+4b-6ax+c-2b

∵ fx+f'x+f''x=x5+64

⇒ a+4=0, b+3a+16=0, 9a+2b+c-12=0, d+c+4b-6a=0

⇒ a=-4, b=-4, c=56, d=-64

So limx→1fxx−1=limx→1x4−4x3−4x2+56x−64

=1-4-4+56-64=-15

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Let fx, be a polynomial of degree 3, such that f-1=10, f1=-6, fx, has a critical point at x=-1 and f'x, has a critical point at x=1. Then fx, has local minima at x=

Options:

The correct answer is:
3

Solution:
f"x=λx-1

Integrating both the side with respect to x

⇒f'x=λx-122+c

As f'1=0⇒c=-2λ
f'x=λx-122-2λ

Integrating both the side with respect to x

fx=λx-136-2λx+u

f1=-6⟹-2λ+u=-6
f-1=10⟹2λ3+u=10
∴λ=6 and u=6
f'x=0⟹x-1 or x=3

f''3>0
∴x=3, is a point of local minima.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Let $f(x)$ be a polynomial of degree 4 having extreme values at $x=1$ and $x=2$.
If $\lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3$ then $f(-1)$ is equal to

Options:
A
$\frac{1}{2}$

B
$\frac{3}{2}$

C
$\frac{5}{2}$

D
$\frac{9}{2}$

The correct answer is:
$\frac{9}{2}$

Solution:
$\because f(x)$ has extremum values at $x=1$ and $x=2$
$\because f^{\prime}(1)=0$ and $f^{\prime}(2)=0$
As, $f(x)$ is a polynomial of degree 4 .
Suppose $f(x)=A x^4+B x^3+C x^2+D x+E$
$$
\begin{aligned}
&\because \lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3 \\
&\Rightarrow \lim _{x \rightarrow 0}\left(\frac{A x^4+B x^3+C x^2+D x+E}{x^2}+1\right)=3 \\
&\Rightarrow \lim _{x \rightarrow 0}\left(A x^2+B x+C+\frac{D}{x}+\frac{E}{x^2}+1\right)=3
\end{aligned}
$$
As limit has finite value, so $D=0$ and $E=0$
Now $A(0)^2+B(0)+C+0+0+1=3$
$$
\begin{aligned}
&\Rightarrow c+1=3 \Rightarrow c=2 \\
&f^{\prime}(x)=4 A x^3+3 B x^2+2 C x+D \\
&f^{\prime}(1)=0 \Rightarrow 4 A(1)+3 B(1)+2 C(1)+D=0
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 4 A+3 B=-4 \\
f^{\prime}(2)=0 & \Rightarrow 4 A(8)+3 B(4)+2 C(2)+D=0 \\
& \Rightarrow 8 A+3 B=-2
\end{aligned}
$$
From equations (1) and (2), we get $A=\frac{1}{2}$ and $B=-2$
So, $f(x)=\frac{x^4}{2}-2 x^3+2 x^2$
Therefore, $f(-1)=\frac{(-1)^4}{2}-2(-1)^3+2(-1)^2$
$$
=\frac{1}{2}+2+2=\frac{9}{2}
$$
Hence $f(-1)=\frac{9}{2}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Let fx be a polynomial of degree 5 such that x=±1 are its critical points. If limx→02+fxx3=4, then which one of the following is not true?

Options:
A f is an odd function
B f1-4f-1=4 . x=1 is a point of maximum and x=-1
C x=1is a point of local minimum and x=-1 is a point of local maximum
D x=1 is a point of local maxima of f

The correct answer is:
x=1 is a point of local maxima of f

Solution:
fx=ax5+bx4+cx3

Consider, limx→0⁡2+ax5+bx4+cx3x3=4

⇒2+c=4

⇒c=2

Differentiating fx with respect to x

f'x=5ax4+4bx3+6x2

=x25ax2+4bx+6

Critical points of the function are given as ±1.

f'1=0⇒5a+4b+6=0

f'-1=0⇒5a-4b+6=0
Solving both we get, b=0, a=-65
So, fx=-65x5+2x3
And f'x=-6x4+6x2

=-6x2x+1x-1

Hence, local minima at x=-1 and local maxima at x=1.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let $f(x)$ be a polynomial of degree 5 with leading coefficient unity such that $f(1)=5, f(2)=4$, $f(3)=3, f(4)=2, f(5)=1$. Then $f(6)$ is equal to :

Options:
A 0
B 24
C 120
D 720

The correct answer is:
120

Solution:
$\begin{aligned} & f(x)+x-6=(x-1)(x-2)(x-3)(x-4)(x-5) \\ & \Rightarrow \quad f(6)=120\end{aligned}$

Topic: Mathematics, Functions, JEE Main
Let fx be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x=-1 and x=1. If limx→0fxx3=1, then 5·f2 is equal to

Options:

The correct answer is:
144

Solution:
Let fx=x6+ax5+bx4+cx3+dx2+ex+f

as limx→0fxx3=1 non-zero finite

So, d=e=f=0

and fx=x3x3+ax2+bx+c

Hence, limx→0fxx3=c=1

Now, as fx=x6+ax5+bx4+x3

and f'x=0 at x=1 and x=-1

i.e., f'x=6x5+5ax4+4bx3+3x2

f'1=0

⇒6+5a+4b+3=0

⇒5a+4b=-9

and f'-1=0

⇒-6+5a-4b+3=0

⇒5a-4b=3

Solving both we get,

a=-610=-35; b=-32

∴ fx=x6-35x5-32x4+x3

∴ 5f2=564-35·32-32·16+8

=320-96-120+40

=144

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Let fx be a polynomial of degree four and having its extreme values at x=1 and x=2. If limx→0 1+fxx2=3, then f2 is equal to

Options:
A 4
B -8
C -4
D 0

The correct answer is:
0

Solution:
Let fx=ax4+bx3+cx2+dx+e
Then limx→01+fxx2=3
⇒limx→01+ax2+bx+c+dx+ex2=3
This limit exists when d=e=0.
So, limx→01+ax2+bx+c=3
⇒c+1=3

⇒c=2
It is given, x=1 and x=2 are the solution of f'x=0.
⇒f'x=4ax3+3bx2+2cx
⇒x4ax2+3bx+2c=0
Given 1, 2 are the roots of the quadratic equation.
⇒Sum of roots=-3b4a=1+2=3
⇒b=-4a
Product of roots =2c4a=1·2=2
⇒a=c4
⇒a=12, b=-2
∴ fx=12x4-2x3+2x2
⇒f2=8-16+8
⇒f2=0

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2015 (04 Apr)
Let fx be a quadratic polynomial such that f–1+f2=0. If one of the roots of fx=0 is 3, then its other root lies in

Options:
A -1,0
B 1,3
C –3,–1
D 0,1

The correct answer is:
-1,0

Solution:
Let fx=ax2+bx+c be a quadratic equation with roots α, β

f2+f-1=0⇒5a+b+2c=0

Also f3=0⇒9a+3b+c=0

⇒a-5=b13=c6

Product of roots αβ=ca=-65

We know α=3⇒β=-25∈-1,0

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let fx be a quadratic polynomial such that f-2 +f3=0. If one of the roots of fx=0 is -1, then the sum of the roots of fx=0 is equal to

Options:
A 113
B 73
C 123
D 143

The correct answer is:
113

Solution:
Let fx=ax2+bx+c

Given, f-2+f3=0

⇒4a-2b+c+9a+3b+c=0

⇒ 13a+b+2c=0 ...i

Since -1 is a root of fx=0. So, f-1=0

⇒a-b+c=0 ...ii

On solving equations i & ii, we get

11a+3b=0 ⇒-ba=113

Now, we know that sum of the roots of ax2+bx+c=0 is -ba.

∴ The sum of the roots of fx=0 is 113.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Let $f(x)$ be an indefinite integral of $\cos ^3 x$.
Statement 1: $f(x)$ is a periodic function of period
$\pi$.
Statement 2: $\cos ^3 x$ is a periodic function.

Options:
A
Statement 1 is true, Statement 2 is false.

B
Both the Statements are true, but Statement 2 is not the correct explanation of Statement 1.

C
Both the Statements are true, and Statement 2 is correct explanation of Statement 1.

D
Statement 1 is false, Statement 2 is true.

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

Solution:
Statement $-2: \cos ^3 x$ is a periodic function. it is a true statement.
Statement-1
Given $f(x)=\int \cos ^3 x d x$
$$
\begin{aligned}
& =\int\left(\frac{\cos 3 x}{4}+\frac{3 \cos x}{4}\right) d x \\
& =\frac{1}{4} \frac{\sin 3 x}{3}+\frac{3}{4} \sin x \\
& =\frac{1}{12} \sin 3 x+\frac{3}{4} \sin x
\end{aligned}
$$
Now, period of $\frac{1}{12} \sin 3 x=\frac{2 \pi}{3}$
Period of $\frac{3}{4} \sin x=2 \pi$
Hence period of $f(x)=\frac{\text { L.C.M. }(2 \pi, 2 \pi}{\text { HCF of }(1,3)}$
$$
=\frac{2 \pi}{1}=2 \pi
$$
Thus, $f(x)$ is a periodic function of period $2 \pi$.
Hence, Statement $-1$ is false.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2012 (07 May Online)
Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$, and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then, $f(x)$ is

Options:
A
$\frac{1}{3 x}+\frac{2 x^2}{3}$

B
$-\frac{1}{3 x}+\frac{4 x^2}{3}$

C
$-\frac{1}{x}+\frac{2}{x^2}$

D
$\frac{1}{x}$

The correct answer is:
$\frac{1}{3 x}+\frac{2 x^2}{3}$

Solution:
$$
\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1
$$
$$
\begin{aligned}
& \Rightarrow & x^2 f^{\prime}(x)-2 x f(x)+1 & =0 \\
& \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} & =0 \\
& \Rightarrow & \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) & =-\frac{1}{x^4} \\
\Rightarrow & & f(x) & =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\
& \text { Hence, } & f(x) & =\frac{2}{3} x^2+\frac{1}{3 x}
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main
Let $\quad f(x)=e^x, \quad g(x)=\sin ^{-1} x \quad$ and $h(x)=f(g(x))$, then $\frac{h^{\prime}(x)}{h(x)}$ is equal to

Options:
A $\sin ^{-1} x$
B $\frac{1}{\sqrt{1-x^2}}$
C $-\frac{1}{\sqrt{1-x^2}}$
D $e^{\sin ^{-1} x}$

The correct answer is:
$\frac{1}{\sqrt{1-x^2}}$

Solution:
We have,
$$
\begin{aligned}
h(x) & =f(g(x))=f\left(\sin ^{-1} x\right) \\
h(x) & =e^{\sin ^{-1} x} \\
\log h(x) & =\sin ^{-1} x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
\frac{h^{\prime}(x)}{h(x)}=\frac{1}{\sqrt{1-x^2}}
$$

Topic: Mathematics, Differentiation, JEE Main