datavorous/entrance-exam-dataset huggingface dataset:
If the frequency of incident light falling on a metallic surface is doubled, maximum kinetic energy of emitted photoelectrons

Options:
A increases to slightly more than double.
B remains the same.
C becomes twice.
D decreases to slightly more than double.

The correct answer is:
increases to slightly more than double.

Solution:
No Solution

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2020 (12 Oct Shift 2)
If the frequency of incident radiation is kept constant and the experiment is
repeated by using incident light of different intensities, then stopping potential
$\left(\mathrm{V}_{\mathrm{s}}\right)$

Options:
A increases with increase in intensity.
B decreases with increase in intensity.
C depends upon current.
D remains same.

The correct answer is:
remains same.

Solution:
No Solution required

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2020 (15 Oct Shift 1)
If the fringe width obtained in Young's double slit experiment is \( 1.33 \mathrm{~mm} \), then on immersing the whole set up in water with refractive index \( 1.33 \), new fringe width will be

Options:
A Does not change
B \( 1.0 \mathrm{~mm} \)
C \( 2.0 \mathrm{~mm} \)
D \( 2.69 \mathrm{~mm} \)

The correct answer is:
\( 1.0 \mathrm{~mm} \)

Solution:
Î²′=Î²Î¼=0.1331.33=0.1Â cm
=1Â mm

Topic: Physics, Wave Optics, JEE Main
If the function $f:[1, \infty) \rightarrow[1, \infty)$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is

Options:
A $\left(\frac{1}{2}\right)^{x(x-1)}$
B $\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)$
C $\frac{1}{2}\left(1-\sqrt{1+4 \log _2 x}\right)$
D Not defined

The correct answer is:
$\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)$

Solution:
$\begin{aligned} & \text { Given } f(x)=2^{x(x-1)} \Rightarrow x(x-1)=\log _2 f(x) \\ & \Rightarrow x^2-x-\log _2 f(x)=0 \Rightarrow x=\frac{1 \pm \sqrt{1+4 \log _2 f(x)}}{2} \\ & \text { Only } x=\frac{1+\sqrt{1+4 \log _2 f(x)}}{2} \text { lies in the domain } \\ & \therefore f^{-1}(x)=\frac{1}{2}\left[1+\sqrt{1+4 \log _2 x}\right]\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If the function f:−∞,−1→a,b defined by fx=ex3−3x+1 is one-one and onto, then the distance of the point P2b+4, a+2 from the line x+e−3y=4 is:

Options:
A 21+e6
B 41+e6
C 31+e6
D 1+e6

The correct answer is:
21+e6

Solution:
Given: fx=ex3−3x+1

⇒f'x=ex3−3x+1×3x2-3

⇒f'x=3×ex3−3x+1×x-1×x+1

So, for x∈(-∞,-1] f'x>0, hence fx is increasing function in given domain,

Now, finding a and b we get,

⇒a=limx→-∞ex3-3x+1

⇒a=0

And b=f-1

⇒b=e-1+3+1

⇒b=e3

So, P2b+4, a+2

⇒P≡2e3+4, 2

Now, finding the distance from the given line we get,

⇒d=2e3+4+2e−3−41+e−6

⇒d=2e3+e−31+e−6

⇒d=1e-321+e−61+e−6

⇒d=21+e−6e-3

⇒d=21e-6+1

⇒d=21+e6

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (31 Jan Shift 2)
If the function $f:[a, b] \rightarrow\left[-\frac{\sqrt{3}}{4}, \frac{1}{2}\right]$ defined by
$$
f(x)=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin x & 1 \\
1+\cos x & 1 & 1
\end{array}\right|
$$
is one-one and onto, then

Options:
A $a=\frac{-\pi}{4}, b=\frac{\pi}{6}$
B $a=\frac{-\pi}{2}, b=\frac{\pi}{2}$
C $a=\frac{-\pi}{6}, b=\frac{\pi}{4}$
D $a=-\pi, b=\pi$

The correct answer is:
$a=\frac{-\pi}{4}, b=\frac{\pi}{6}$

Solution:
We have, $f:(a, b) \rightarrow\left[\frac{-\sqrt{3}}{4}, \frac{1}{2}\right]$
$$
\begin{aligned}
& f(x)=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin x & 1 \\
1+\cos x & 1 & 1
\end{array}\right| \\
& f(x)=\left|\begin{array}{ccc}
1 & 0 & 0 \\
1 & \sin x & 0 \\
1+\cos x & -\cos x & -\cos x
\end{array}\right| \\
& \quad\left[C_3 \rightarrow C_3-C_1, C_2 \rightarrow C_2-C_1\right] \\
& f(x)=-\sin x \cos x \\
& f(x)=-\frac{1}{2} \sin 2 x \\
& f(x) \in\left[\frac{\sqrt{3}}{4}, \frac{1}{2}\right]
\end{aligned}
$$
$$
\begin{array}{rlrl}
& \therefore & -\frac{\sqrt{3}}{4} & \leq \frac{-1}{2} \sin 2 x \leq \frac{1}{2} \\
\Rightarrow & & -1 & \leq \sin 2 x \leq \frac{\sqrt{3}}{2} \\
\Rightarrow & -\frac{\pi}{4} & \leq x \leq \frac{\pi}{6} \\
& \therefore & a & =-\frac{\pi}{4}, b=\frac{\pi}{6}
\end{array}
$$

Topic: Mathematics, Functions, JEE Main
If the function f defined as fx=1x-k-1e2x-1, x≠0 is continuous at x=0, then ordered pair k,f0 is equal to

Options:
A 2,1
B 3,1
C 3,2
D 13,2

The correct answer is:
3,1

Solution:
Given fx=1x-k-1e2x-1, x≠0

and fx is continuous at x=0.

⇒f0=limx→01x-k-1e2x-1

=limx→01+2x+12!2x2+…-1-xk-12x2e2x-12x

=limx→01+2x+12!2x2+…-1-xk-12x2

Clearly, k=3 and f0=1. ( ∵Equating coefficient of x=0)

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2018 (16 Apr Online)
If the function f defined on -13,1/3 by fx=1xloge1+3x1-2x,when x≠0k,when x=0, is continuous, then k is equal to.

Options:

The correct answer is:
5

Solution:
limx→0⁡fx=limx→0⁡ln⁡1+3xx-ln⁡1-2xx
=limx→0⁡3ln⁡1+3x3x+2ln1-2x-2x=3+2=5

limfx→0ln1+fxfx=1

∴fx, will be continuous if f0=limx→0⁡fx

Thus, k=5

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If the function f defined on π6,π3 by fx=2cosx-1cotx-1, x≠π4k, x=π4 is continuous, then k is equal to

Options:
A 12
B 1
C 2
D 12

The correct answer is:
12

Solution:
∵ fx is continuous at x=π4

∴ fπ4=k=limx→π42cosx-1cotx-1→00 form

Applying L' Hospital Rule

k=limx→π42-sinx-cosec2x

⇒k=2sinπ4cosec2π4=2×1222=12.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the function f given by fx=x3-3a-2x2+3ax+7, for some a∈R is increasing in 0, 1 and decreasing in 1, 5, then a root of the equation, fx-14x-12=0, x≠1 is :

Options:
A 7
B -7
C 6
D 5

The correct answer is:
7

Solution:
At x=1, f′x=0

∴ f′x=3x2-6a-2x+3a

Now, f'1=0

⇒3-6 a-2+3a=0

⇒a=5

∴ fx-14x-12=0

⇒x3-9x2+15x-7x-12=0

⇒x-7x2-2x+1x-12=0

⇒x-7=0⇒x=7

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If the function f:R-1, -1→A defined by fx=x21-x2, is surjective, then A is equal to

Options:
A 0, ∞
B R-{-1}
C R-−1, 0
D R--1, 0

The correct answer is:
R-−1, 0

Solution:
A function is surjective, when the range of the function is same as the co-domain.

Now, y=x21-x2

⇒y-yx2=x2

⇒y=x2y+1

⇒x2=yy+1

∵ x2≥0 ⇒yy+1≥0

⇒y ∈-∞, -1∪0, ∞ or R-−1, 0.

∴ Range of f=A=R-−1, 0.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the function f:R→A defined as fx=sin-1⁡x1+x2 is a surjective function, then the set A is

Options:
A –π6,π6
B –π2,π2
C –π3,π6
D 0,π3

The correct answer is:
–π6,π6

Solution:
Let y=x1+x2

⇒yx2-x+y=0
∵x∈R, hence D≥0
⇒-12-4y2≥0⇒y∈-12,12
i.e. sin-1⁡y∈sin-1⁡-12,sin-1⁡12

Hence, the range of fx∈-π6,π6≡A

Topic: Mathematics, Functions, JEE Main
If the function f:R→A defined as fx=tan-1⁡2x31+x6 is a surjective function, then the set A is equal to

Options:
A –π2,π2
B –π4,π4
C –π2,π2
D 0,π4

The correct answer is:
–π4,π4

Solution:
Let, x3=t∈R
Let, y=2t1+t2

⇒yt2-2t+y=0
∵t∈R, hence D≥0
⇒-22-4y2≥0⇒y∈-1,1
tan-1⁡y∈tan-1⁡-1,tan-1⁡1

Hence, the range of fx∈-π4,π4≡A

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If the function $f: R \rightarrow R$ defined by $f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1) x+\sin x}{x}, & x < 0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}}, & x=0 \text { is }\end{array}\right.$ continuous on $[R$, then $a+b=$

Options:
A -1
B 2
C 1
D 3

The correct answer is:
-1

Solution:
Given function
$$
f(x)=\left\{\begin{array}{cc}
\frac{\sin (a+1) x+\sin x}{x} & , x < 0 \\
b & , x=0 \\
\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} & , x>0
\end{array}\right.
$$
Function is continuous so,
$$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)
$$
Now, $\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x+\sin x}{x}\right) \\ = & \lim _{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x \times(a+1)}{(a+1) x}+\frac{\sin x}{x}\right) \\ = & (a+1)+1 \\ & {\left[\because \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x}=1\right] }\end{aligned}$
Now, $\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) & =\lim _{x \rightarrow 0^{+}}+\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} \\ = & \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{1+x}-1) \times \sqrt{1+x}+1}{x \sqrt{x} \times(\sqrt{1+x}+1)}\end{aligned}$
$$
b=\frac{1}{2}
$$
put in Eq. (iii)
$$
\begin{aligned}
a+2 & =\frac{1}{2} \\
a & =\frac{-3}{2}
\end{aligned}
$$
Now,
$$
\begin{aligned}
& a+b=\frac{-3}{2}+\frac{1}{2} \\
& a+b=-1
\end{aligned}
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function $f: R \rightarrow R$ is defined by $f(x)=\left\{\begin{array}{l}2 x-3, \text { if } x < -2 \\ x^2-1, \text { if }-2 \leq x \leq 2 \\ 3 x+2, \text { if } x>2\end{array}\right.$ then $\mathrm{f}$ is

Options:
A an injection but not a surjection
B a surjection but not an injection
C a bijection
D neither injection nor surjection

The correct answer is:
neither injection nor surjection

Solution:
$f(x)=\left\{\begin{array}{lc}2 x-3, & x < -2 \\ x^2-1, & -2 \leq x \leq 2 \\ 3 x+2, & x>2\end{array}\right.$
Clearly for $x \in[-2,2]$
$$
f(-2)=f(2)
$$
$\therefore f(x)$ is not Injective.
Also $y \notin(-7,-1) \cup(3,8)$
i.e. Range $\neq$ Codomain $\Rightarrow$ not surjective
$\therefore f(x)$ is neither injective nor surjective.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
If the function $f: R \rightarrow R$ is defined by $f(x)=x|x|$, then .

Options:
A $f$ is one-one but not onto
B $f$ is onto but not one-one
C $f$ is both one-one and onto
D $f$ is neither one-one nor onto

The correct answer is:
$f$ is both one-one and onto

Solution:
Given, $f(x)=x|x|=\left\{\begin{array}{l}x(x), x \geq 0 \\ x(-x), x < 0\end{array}\right.$
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2, x \geq 0 \\ -x^2, x < 0\end{array}\right.$, Also, $f: R \rightarrow R$
According to the graph,
domain of $f(x)=$ Range of $f(x)=R$
$\therefore$ Range $=$ Codomain $=R$
$\Rightarrow f(x)$ is an onto function.
If a line parallel to $X$-axis is drawn,
it will intersect at only one point of $f(x)$.
Hence, $f(x)$ is one-one also.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
If the function fx=1+|cos x|λ|cos x|,0<x<π2μ,x=π2ecot 6xcot 4x,π2<x<πis continuous at x=π2, then 9λ+6 logcμ+μ6-e6λ is equal to

Options:
A 11
B 8
C 2e4+8
D 10

The correct answer is:
10

Solution:
Given,

fx is continuous at x=π2

Now, solving L.H.L. at x=π2 we get,

limx→π+2ecot 6xcot 4x=limx→π+2esin 4x·cos 6xsin 6x·cos 4x=e2/3

Similarly, on simplification for R.H.L. we get

limx→π2-1+|cosx|λ|cosx|=elimx→π2-λ=eλ

∵ fπ2=μ

For continuous function,

fπ2-=fπ2=fπ2+

⇒e23=eλ=μ

⇒λ=23, μ=e23

Now,

9λ+6 logeμ+μ6-e6λ

=9(23)+6(23)+e236-e6(23)

=6+4+e4-e4

=10

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2023 (25 Jan Shift 2)
If the function fx=1-x2tan⁡πx2 is continuous at x=1, then f1 is equal to

Options:
A 1π
B π2
C 0
D π

The correct answer is:
1π

Solution:
f1=limx→1fx=limx→11-x2tan⁡πx2
Put x=1+h
=limh→0⁡1-1+h2tan⁡π21+h
=limh→0⁡-h2tan⁡π2+πh2
=limh→0⁡-h2-cot⁡πh2
=limh→0π2⋅htan⁡πh2⋅1π=1π

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function fx=1xloge1+xa1-x b , x<0k , x=0cos2x-sin2x-1 x2+1-1 , x>0

is continuous at x=0, then 1a+1 b+4k is equal to :

Options:
A 4
B 5
C -4
D -5

The correct answer is:
-5

Solution:
Given that

fx=1xloge1+xa1-x b , x<0k , x=0cos2x-sin2x-1 x2+1-1 , x>0

fx is continuous at x = 0

L.H.L = R.H.L = f0

LHL=limx→0-log1+xa1-xbx

=limx→0-log1+xa-log1-xbx

=limx→0-log1+xaxa×1a-log1-xb-xb×-1b=1b+1a

RHL=limx→0+cos2x-sin2x-1x2+1-1×x2+1+1x2+1+1

=limx→0+-2sin2xx2x2+1+1=-4

So, 1b+1a=k=-4 ..

k = -4 ⇒4k = -1.

1a + 1b = -4.

∴1a + 1b + 4k = -5.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If the function fx=1|x|,|x|≥2ax2+2b,|x|<2 is differentiable on R, then 48(a+b) is equal to _______.

Options:

The correct answer is:
15

Solution:
Given,

fx=1x, x≥2ax2+2 b, -2<x<2-1x, x≤-2

Also given fx is differentiable on R

So, finding LHD and RHD at x=2 we get,

-1x2=2ax

⇒-14=4a

⇒a=-116

Now, for function to be continuous at x=2 we get,

12=4a+2b

⇒12=-416+2b

⇒b=38

Hence, 48a+b=4838-116

⇒48a+b=48516=15

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If the function fx=2+cosx-1π-x2,x≠πk,x=π is continuous at x=π, then k equals

Options:
A 14
B 0
C 2
D 12

The correct answer is:
14

Solution:
Given fx=2+cosx-1π-x2,x≠πk,x=π is continuous at x=π

Hence, fπ=limx→πfx

⇒ k=limx→π2+cosx-1π-x2

Consider L.H.L.=limx→π-2+cosx-1π-x2

limh→02+cos(π-h)-1π-π-h2
Also, cosπ-h=-cos h
⇒ k=limh→02-cosh-1h2

⇒ k=limh→02-cosh-1h2×2-cosh+12-cosh+1

⇒ k=limh→02-cosh-1h22-cosh+1

⇒ k=limh→01-coshh2limh→012-cosh+1

Now, using cos2x=1-2sin2x, ⇒1-cos2x=2sin2x,

⇒ k=limh→02sin2h24h22×12-1+1

⇒ k=12×12

⇒ k=14.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2014 (19 Apr Online)
If the function $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2-9 a \mathrm{x}^2+12 \mathrm{a}^2 \mathrm{x}+1$, where $\mathrm{a}>0$, attains its maximum and minimum at $\mathrm{p}$ and $\mathrm{q}$ respectively such that $\mathrm{p}^2=\mathrm{q}$, then a equals

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

B
3

C
1

D
2

The correct answer is:
2

Solution:
$f(x)=2 x^3-9 a x^2+12 a^2 x+1$
$f^{\prime}(x)=6 x^2-18 a x+12 a^2 ; f^{\prime \prime}(x)=12 x-18 a$
For max. or min. $6 x^2-18 a x+12 a^2=0 \Rightarrow x^2-3 a x+2 a^2=0$
$\mathrm{x}=\mathrm{a}$ or $\mathrm{x}=2 \mathrm{a}$, at $\mathrm{x}=\mathrm{a}$ max.and at $\mathrm{x}=2 \mathrm{a} \min$.
$p^2=q$
$a^2=2 a \Rightarrow a=2$ or $a=0$
but $\mathrm{a}>0$, therefore, $\mathrm{a}=2$.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2003
If the function $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2}=q$, then $a$ equals

Options:
A 0
B 1
C 2
D None of these

The correct answer is:
2

Solution:
Given, $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ attains maximum and minimum at $p$ and $q$ respectively. $\therefore$
$$
f^{\prime}(p)=0, f^{\prime}(q)=0
$$
Now,
$$
\begin{array}{c}
f^{\prime \prime}(p) < 0 \text { and } f^{\prime \prime}(q)>0 \\
f^{\prime}(p)=0
\end{array}
$$
and
$$
\begin{array}{r}
f^{\prime}(q)=0 \\
6 p^{2}-18 a p+12 a^{2}=0 \\
6 q^{2}-18 a q+12 a^{2}=0
\end{array}
$$
and
$$
\begin{array}{ll}
\Rightarrow & p^{2}-3 a p+2 a^{2}=0 \\
\text { and } & q^{2}-3 a q+2 a^{2}=0
\end{array}
$$
$p=a, 2 a, q=a, 2 a$
$\begin{array}{lrl}\Rightarrow & p=a, 2 a, q=a, \\ \text { Now, } & f^{\prime \prime}(p) < 0\end{array}$
$\Rightarrow \quad 12 p-18 a < 0$
$\Rightarrow \quad p < \frac{3}{2} a$
and $f^{\prime \prime}(q)>0 \Rightarrow 12 q-18 a>0$
$\Rightarrow \quad q>\frac{3}{2} a$
From Eqs. (i), (ii) and (iii), we get $\begin{aligned} p &=a, q=2 a \\ \text { Now, } \quad p^{2} &=q \end{aligned}$
$$
\begin{array}{lc}
\Rightarrow & a^{2}=2 a \\
\Rightarrow & a=0,2
\end{array}
$$
But for $a=0, f(x)=2 x^{3}+1$ which does not
attains a maximum or minimum for any value of $x$. Hence, $a=2$.
$$

Topic: Mathematics, Application of Derivatives, JEE Main
If the function $f(x)=2 x^3-9 x^2+12 \mathrm{a}^2 x+1, \mathrm{a}>0$ has a local maximum at $x=\alpha$ and a local minimum at $x=\alpha^2$, then $\alpha$ and $\alpha^2$ are the roots of the equation :

Options:
A $x^2-6 x+8=0$
B $x^2+6 x+8=0$
C $8 x^2+6 x-1=0$
D $8 x^2-6 x+1=0$

The correct answer is:
$x^2-6 x+8=0$

Solution:
$\begin{aligned} & \alpha+\alpha^2=3 \mathrm{a} \& \alpha \times \alpha^2=2 \mathrm{a}^2 \\ & \quad \downarrow \\ & \left(\alpha+\alpha^2\right)^3=27 \mathrm{a}^3 \\ & \Rightarrow 2 \mathrm{a}^2+4 \mathrm{a}^4+3(3 \mathrm{a})\left(2 \mathrm{a}^2\right)=27 \mathrm{a}^3 \\ & \Rightarrow 2+4 \mathrm{a}^2+18 \mathrm{a}=27 \mathrm{a} \\ & \Rightarrow 4 \mathrm{a}^2-9 \mathrm{a}+2=0 \\ & \Rightarrow 4 \mathrm{a}^2-8 \mathrm{a}-\mathrm{a}+2=0 \\ & \Rightarrow(4 \mathrm{a}-1)(\mathrm{a}-2)=0 \Rightarrow \mathrm{a}=2 \\ & \text { so } 6 \mathrm{x}^2-36 \mathrm{x}+48=0\end{aligned}$
$\Rightarrow x^2-6 x+8=0$ ...(1)
If we take $\mathrm{a}=\frac{1}{4}$ then $\alpha=\frac{1}{2}$ which is not possible

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If the function $f(x)=\left\{\begin{array}{l}\frac{k \cos x}{\pi-2 x}, \text { when } x \neq \frac{\pi}{2} \\ 3, \quad \text { when } x=\frac{\pi}{2}\end{array}\right.$ be continuous at $x=\frac{\pi}{2}$, then k=

Options:
A 3
B 6
C 12
D None of these

The correct answer is:
6

Solution:
$f(\pi / 2)=3$. Since $f(x)$ is continuous at $x=\pi / 2$
$\Rightarrow \lim _{x \rightarrow \pi / 2}\left(\frac{k \cos x}{\pi-2 x}\right)=f\left(\frac{\pi}{2}\right) \Rightarrow \frac{k}{2}=3 \Rightarrow k=6$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function $\mathrm{f}(\mathrm{x})=\frac{2 \mathrm{x}-\sin ^{-1} \mathrm{x}}{2 \mathrm{x}+\tan ^{-1} \mathrm{x}}$ is continuous at each
point in its domain, then what is the value of $\mathrm{f}(0)$ ?

Options:
A $-\frac{1}{3}$
B $\frac{1}{3}$
C $\frac{2}{3}$
D 2

The correct answer is:
$\frac{1}{3}$

Solution:
$f(0)=\lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$
$=\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=\frac{2-1}{2+1}=\frac{1}{3}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2018 (Phase 2)
If the function fx=aπ-x+1,  x≤5bx-π+3,  x>5 is continuous at x=5, then the value of a-b is:

Options:
A 25-π
B -2Ï+5
C 2Ï+5
D 2Ï-5

The correct answer is:
25-π

Solution:
For continuity at x=5

limx→5+fx=limx→5-fx=f5

⇒limx→5+bx-π+3=limx→5-aπ-x+1=a5-π+1

We know that x= x,x≥0-x,x<0

⇒b5-π+3=a5-π+1=a5-π+1

⇒b5-π+3=a5-π+1

⇒3-1=a5-π-b5-π

⇒a-b5-π=2

⇒a-b=25-π.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If the function $f(x)=a x^3+b x^2+26 x-24$ satisfies the conditions of Rolle's theorem in $[2,4]$ and $f^{\prime}\left(3+\frac{1}{\sqrt{3}}\right)=0$, then the value of $a b$ is equal to

Options:
A $-9$
B $9$
C $-3$
D $3$

The correct answer is:
$-9$

Solution:
$f(x)=a x^3+b x^2+26 x-24$ ...(i) on $[2,4]$
$\because f(x)$ satisfies the Role's theorem
$\begin{array}{ll}\Rightarrow & f(2)=f(4) \\ \Rightarrow & a(2)^3+b\left(2^2\right)+26(2)-24\end{array}$
$=a(4)^3+b(4)^2+26(4)-24$
$\Rightarrow \quad 8 a+4 b+28=64 a+16 b+80$
$\Rightarrow \quad 56 a+12 b+52=0$
$14 a+3 b+13=0$ ...(ii)
On differentiating Eq. (i) w.r.t. $x$
$\begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left(a x^3+b x^2+26 x-24\right) \\ & f^{\prime}(x)=3 a x^2+2 b x+26\end{aligned}$
At, $x=3+\frac{1}{\sqrt{3}}$
$f^{\prime}\left(3+\frac{1}{\sqrt{3}}\right)=3 a\left(3+\frac{1}{\sqrt{3}}\right)^2+2 b\left(3+\frac{1}{\sqrt{3}}\right)+26$
$\begin{aligned} & 0=3 a\left(9+\frac{1}{3}+\frac{6}{\sqrt{3}}\right)+6 b+\frac{2 b}{\sqrt{3}}+26 \\ & 0=28 a+6 b+6 \sqrt{3} a+\frac{2 b}{\sqrt{3}}+26\end{aligned}$
$\Rightarrow \quad(28+6 \sqrt{3}) a+\left(\frac{6 \sqrt{3}+2}{\sqrt{3}}\right) b+26=0$ ...(iii)
$(28 a+6 b+26)+(18 a+2 b) \frac{1}{\sqrt{3}}=0+0 \frac{1}{\sqrt{3}}$
On comparing rational and irrational part,
$28 a+6 b+26=0$ and $18 a+2 b=0$
and $54 a+6 b=0$
$\therefore \quad 28 a-54 a+26=0$ and $6 b=-54 a$
$\begin{aligned} \Rightarrow & & -26 a+26 & =0 \\ \Rightarrow & & a & =1\end{aligned}$
$b=\frac{-54 \times 1}{6}$
$\begin{aligned} \Rightarrow & b & =-9 \\ \therefore & a b & =-9\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
If the function fx=ax+7;0≤x<2bx+5;x≥2 is differentiable ∀x≥0, then 2a+4b is equal to

Options:
A 24016
B 5
C 8516
D 25016

The correct answer is:
5

Solution:
fx must be continuous and differentiable at x=2
For continuity at x=2
f2-=f2+=f2
a2+7=b2+5⇒3a=2b+5…i
For differentiability at x=2
f ′ x = a 2 x+7 ; 0≤x<2 b ; x≥2
f ′ 2 − = f ′ 2 +
a22+7=b⇒a=6b…ii
From i⇒18b=2b+5⇒b=516
So, 2a+4b=12b+4b=16b=5

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function fx=ax+7:0â¤x<9bx+5:xâ¥9 is differentiable for xâ¥0, then the value of 5a+6b is equal to

Options:
A 24023
B 10
C 8023
D 25023

The correct answer is:
10

Solution:
fx must be continuous and differentiable at x=9
For continuity at x=9
f9-=f9+=f9
⇒a9+7=b9+5⇒4a=9b+5 …. 1
For differentiability at x=9
f'x=a2x+7:0≤x<9b:x≥9
f ′ 9 − = f ′ 9 +
a29+7=b⇒a=8b … 2
From (1), 32b=9b+5⇒b=523
Hence, 5a+6b=40b+6b=46b=10

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function fx=cos-1â¡x32-1-x-x2+x3 (where, â0<x<1), then the value of 3f'12 is equal to (take 3=1.73)

Options:

The correct answer is:
3.73

Solution:
fx=cos-1â¡xx-1-x1-x2
=cos-1â¡xx-1-x21-x2
=cos-1â¡x+cos-1â¡x
Now, f â² x = â1 1â x 2 â 1 1âx â 1 2 x
f'12=â11â14â112Ã1212=â23â1

3f'12=-3â23â1=3.73

Topic: Mathematics, Differentiation, JEE Main
If the function fx=cos(sinx)-cosxx4 is continuous at each point in its domain and f0=1k, then k is _________.

Options:

The correct answer is:
6

Solution:
Given limx→0cos(sinx)-cosxx4=f0

⇒limx→02sinsinx+x2sinx-sinx2x4=1k

⇒limx→02sinsinx+x2sinx-sinx2sinx+x2x-sinx2x4×sinx+x2x-sinx2=1k

⇒limx→02sinx+x2xx-sinx2x3=1k, ∵limx→0sinxx=1

⇒2limx→0sinx+x2xlimx→0x-sinx2x3=1k

Apply L'Hospital rule for an indeterminate form of 00, we get,

⇒2×(1+1)2×12×16=1 k

⇒k=6

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If the function fx, defined as fx=a1-xsinâ¡x+bcosâ¡x+5x2: xâ 0 3: x=0 is continuous at x=0, then the value of b4+a5+a is equal to

Options:

The correct answer is:
63.75

Solution:
Since, fx is continuous at x=0 so, at x=0 both left hand and right hand limits must exist and both must be equal to 3.
Now,
limx→0a1-xsin⁡x+bcos⁡x+5x2=f0

⇒limx→0a-axx-x33!+....+b1-x22!+....+5x2=f0
⇒limx→0a+b+5+-a-b2x2+...x2=3
If limx→0fx=3 exists, then a+b+5=0
and -a-b2=3

⇒a=-1 and b=-4

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the function $f(x)$ is continuous on its domain $[-2,2]$ where $f(x)=\frac{\sin a x}{x}+3,-2 \leq x < 0=x+5,0 \leq x \leq 1=\sqrt{x^2+8}-b, 1 < x \leq 2$, then $7 a+b+1$ is equal to

Options:
A 10
B 11
C 14
D 12

The correct answer is:
12

Solution:
for continuity at $x=0, \lim _{x \rightarrow 0}-\frac{\sin a x}{x}+3=5$
$\begin{aligned}
& \Rightarrow a+3=5 \\
& \Rightarrow a=2
\end{aligned}$
for continuity at $x=1$,
$\begin{aligned}
& 6=\lim _{x \rightarrow 1}+\sqrt{x^2+8}-b \\
& \Rightarrow 6=\sqrt{1^2+8}-b \\
& \Rightarrow b=-3
\end{aligned}$
Now $7 a+b+1=7 \times 2-3+1=12$

Topic: Mathematics, Continuity and Differentiability, MHT CET, MHT CET 2022 (11 Aug Shift 1)
If the function fx is symmetric about the line x=3, then the value of the integral I=∫-28fxfx+f6-xdx is

Options:
A 0
B 5
C 10
D 16

The correct answer is:
5

Solution:
As fx is symmetric about x=3, we get,

f3+x=f3-x
Replacing x by 3-x, we get,

f6-x=fx
Now, using a+b-x property in I, and adding, we get,
⇒2I=∫-28fx+f6-xfx+f6-xdx
⇒2I=∫-281dx=x-28
⇒2I=10
⇒I=5

Topic: Mathematics, Definite Integration, JEE Main
If the function fx=k1(x-π)2-1,x≤πk2cosx,x>π is twice differentiable, then the ordered pair k1,k2 is equal to:

Options:
A 12,1
B 1,0
C 12,−1
D 1,1

The correct answer is:
12,1

Solution:
f(x) is differentiable then will also continuous then f(π)=k1π-π2-1=-1

fπ+=limh→0k2cosπ+h=-k2

fπ+=fπ⇒k2=1…1,f is continuous.

Now, f'(x)=ddxk1(x-π)2,x≤πddxk2cosx,x>π

f'(x)=2k1(x-π),x≤π-k2sinx,x>π

then, f'π-=2k1π-π=0

f'π+=-k2sinπ=0

⇒f'π-=f'π+=0

Similarly, we get

f''(x)=2k1,x≤π-k2cosx,x>π

As the function is twice differentiable, the second-order derivatives, LHD=RHD

⇒f''π-=f''π+⇒2k1=-k2cosπ

⇒2k1=k2

⇒k1=12 (from 1)

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If the function fx=loge1-x+x2+loge1+x+x2sec x-cos x,x∈-π2,π2-0k, x=0 is continuous at x=0, then k is equal to:

Options:
A 1
B -1
C e
D 0

The correct answer is:
1

Solution:
Given,

fx=loge1-x+x2+loge1+x+x2sec x-cos x,x∈-π2,π2-0k, x=0

⇒limx→0ln1+x2+x4cosx1-cos2x=k

Taking L.H.S=limx→0ln1+x2+x4cosx1-cos2x

=limx→0ln1+x2+x4x2+x4x21+x2cosxsin2xx2x2

=limx→0ln1+x2+x4x2+x41+x2cosxsin2xx2

=11+011=1

Now equating with R.H.S we get, k=1

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (26 Jul Shift 1)