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Given v = (0,0) and axis is y-axis 

The parabola is of the form x² = 4ay or x² = -4ay 

But this equation passes through (-1,-0) which is in III qhadrant. 

∴ The curve turn down words and is of the form x² = – 4ay, 

But it passes through (-1,-3) 

∴ (-1)² = -4a(-3) ⇒ 1 = 12a ⇒ a = \(\frac{1}{12}\)

∴ The required equation is x² = -4 . \(\frac{1}{12}\) y 

⇒ x² = \(\frac{-1}{3}\) or 3x² + y = 0

The given family of curves represents the equation of straight lines. 

∴ y = mx ………. (1) 

Differentiate w.r.t x, we get dy

dy/dx = m

∴ substitute m = dy/dx, we have

y = mx

⇒ y = dy/dx x

dy/dx = y/x

dy/dx - y/x = 0

which is the required differential equation.

(c) Rayleigh 

The Probability density function of the envelope of narrowband Gaussian noise is Rayleigh.

(c) Gaussian 

The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain

(d) a triangular pulse of duration 2T

Autocorrelation function of a rectangular pulse of duration T is a triangular pulse of duration 2T The autocorrelation function is an even function of

Correct answer is: (b) 84000

Expected number of votes = np 

= \(\cfrac{70}{100}\) × 120000 = 84000

y = 3x 

R – {(1,3) (2,6) (3, 9) (4,12)} 

R is not reflexive as (1, 1) ∉ R 

R is not symmetric as (1,3) ∈ R but 

(3,1) ∉ R R is not transitive as 

(1,3) ∈ R, (3,9) ∈ R but (1,9)∉ R

We need to evaluate ∫ sin^-1x dx

Let u = sin^-1x, dv = dx

⇒ du = \(\frac{1}{\sqrt{1-x^2}}\)dx, v = x

∫ sin^-1x dx = x sin^-1x - ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx + c ....(1)

where c is integration constant.

Consider ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx

Let w = 1 - x²

⇒ dw = -2xdx

⇒ xdx = -dw/2

∫ \(\frac{x}{\sqrt{1-x^2}}\)dx = ∫ \(-\frac{dw}{2\sqrt w}\)

= -1/2 ∫ dw/√w

= -1/2 (2√w)

= -√w

substitute back w = 1 - x² we get

∴ ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx = \(-\sqrt{1-x^2}\)

Therefore (1) becomes

∫ sin^-1x dx = x sin^-1 x + \(\sqrt{1-x^2}\) + c

Correct answer is: (b) 2, 2

For Poisson distribution

Mean = variance = np

= 20000 × \(\cfrac{1}{10000}\) = 2

ΣP(x) = 1

1 = kx + 5k = kx

k = 1/5

Correct option is: (C) Vorticella

Let I = \(\int\limits_0^{3/2}\)|x cos π x|dx

\(\because\) |x cos x| = \(\begin{cases}xcos\pi x&;0\leq x\leq\frac12\\-xcos \pi x&;\frac12\leq x\leq \frac32\end{cases}\)

\(\therefore\int\limits_0^{3/2}\)|x cos π x| dx = \(\int\limits_0^{1/2}xcos\pi x-\int\limits_{1/2}^{3/2}xcos\pi xdx\) 

\(=[\frac{xsin\pi x}{\pi}+\frac{cos\pi x}{\pi^2}]_0^{1/2}\)\(-[\frac{xsin\pi x}{\pi}+\frac{cos\pi x}{\pi^2}]_{1/2}^{3/2}\)

\((\because\int xcos\pi x= \frac{xsin\pi x}{\pi}+\frac{cos\pi x}{\pi^2})\) 

\(=\frac1{2\pi}sin\frac{\pi}2+\frac1{\pi^2}cos\frac{\pi}2-\frac1{\pi^2}cos0\) \(-(\frac3{2\pi}sin\frac{3\pi}2+\frac1{\pi^2}cos\frac{3\pi}2-\frac1{2\pi}sin\frac{\pi}2-\frac1{\pi^2}cos\frac{\pi}2)\)

\(=\frac1{2\pi}-\frac1{\pi^2}-(-\frac{3}{2\pi}-\frac1{2\pi})\)

(\(\because\) sin \(\frac{\pi}2=1,\) sin \(\frac{3\pi}2=-1\), cos 0 = 1 and cos \(\frac{\pi}2=cos\frac{3\pi}2=0\))

\(=\frac1{2\pi}-\frac{1}{\pi^2}+\frac4{2\pi}\) 

\(=\frac{5}{2\pi}-\frac1{\pi^2}=\frac{5\pi-2}{2\pi^2}\)

\(\int\limits^2_1 \frac{(x + 1)^3}{x^2}dx\)

\(= \int\limits^2_1 \left(x + 3 + \frac3x + \frac1{x^2}\right)dx\)

\(= \left(2 + 6 +3log2 - \frac12\right)- \left(\frac12 + 3+ 3log1-1\right)\)

\(= 8 + 3log2 - \frac12 - \frac12 - 2\)       \((\because log 1 = 0)\)

\(= 5 + 3log2\)

Correct option (A) 1    

Explanation:

sec²θ – tan² θ  = 1

Correct option  (D) 1  

Explanation:

sin 83° = cos 7°

∴ the given expression is 1 – 1 + 1 = 1.

We need to find the smallest number of flowers that can be put in the vase in groups of 8 or 12.

For this first find the LCM of 8 and 12.

8 = 2 × 2 × 2

12 = 2 × 2 × 3

The LCM of 8 and 12 = 2 × 2 × 2× 3 = 24

24 is the smallest number which is divisible by 8 and 12.

Hence, 24 flowers can be put in the vase in groups of 8 or 12.

She will be above forty.

2NaOH  +  Cl₂  →  NaCl  +   NaClO  +  H₂O 

WHAT MILK IS BEST FOR MAGIC MILK EXPERIMENT?
HOW DOES WATER AFFECT SEED GERMINATION?
HOW CAN YOU MAKE A LEGO RUBBER BAND CAR TRAVEL FURTHER?
WHY DO LEAVES CHANGE COLOR IN THE FALL?
WHY DO SKITTLES DISSOLVE IN WATER?
WHAT MAKES ICE MELT FASTER?
WHY DO APPLES TURN BROWN?

Correct answer is a. Cell Style 

Correct answer is b. Back-ends 

  Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged.
Side band frequencies: The AM wave contains three frequencies f_c, (f_c + f_m) and (f_c -f_m),fc is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c +f_m): Upper side band (USB) frequency
(f_c -f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
According to the problem, frequency of carrier signal is f_c = 1 MHz and frequency of speech signal = 3 kHz
= 3 x 10^-3 MHz
= 0.003 MHz
We know that, Frequencies of side bands = (f_c ± f_m) = (1 + 0.003) and (1 – 0.003)
So, side band frequencies are 1.003 MHz and 0.997 MHz.  

Correct Option(d) zero

Explanation:

ln an L-C-R series AC circuit, the voltage across inductor L leads the current by 90° and the voltage across capacitor C lags behind the current by 90°. [ V = V_L ~ V_C = 50 - 50 = 0]

Hence, the voltage across L-C combination will be zero.

Answer is (1) Three

F = dp/dt = mνn 

n = F/mν = 144/(0.04 x 1200) = 3 

Answer is (2) 3.6 J

W = mgh

where h is the distance moved by centre of gravity

W = 0.6 x (4/2) x 9.8 x 0.3 = 3.6 J 

1. +ve

2. -ve

3. -ve

4. +ve

5. -ve

Transfer of pollen grains from the anther to the stigma of another flower of the same plant.

The correct answer is pilgrimage to the tombs of Sufi saints.

• During the Medieval period in India Sufism emerged as a powerful movement.
• Sufis were so-called because of the purity (safa) of their hearts. 
• Unity in God, complete self-surrender, charity, Ibadat, love for mankind, etc. are the main teaching of Sufism.
• Ziyarat means the pilgrimage to the tomb on the death anniversary of the Sufi saint. 
• Its main objective was to seek spiritual grace from the Sufi.
• Music and dance are an important part of Ziyarat.

• Some of the prominent Sufi Shaikhs in India:
  • Data Gunj Bakhsh
  •  Khwaja Muinuddin Chishti
  • Shaikh Qutbuddin
  • Bakhtiyar Kaki
  • Fariduddin Gunj-i Shakar
  • Shaikh Nizamuddin Auliya 

The correct answer is both 1 and 2.

• The uprising began at Supa village in the district of Poona when moneylenders refused the farmer's loans.
• In 1875, farmers attacked a market place where many moneylenders lived.
• They burnt account books and looted grain shops.
• They also burnt the houses of sahukars.
• Sahukars were people who were both traders and moneylenders.
• When the revolt spread in the Deccan, the Government of India, worried by the memory of 1857, pressurised the Government of Bombay to set up a commission of enquiry to investigate into the causes of the riots.
• The commission produced a report that was presented to the British Parliament in 1878.
• The commission held enquiries in the districts where the riots spread, not in unaffected districts. Hence statement 1 is correct.
• They recorded statements of ryots, sahukars and eyewitnesses, compiled statistical data on revenue rates, prices and interest rates in different regions, and collated the reports sent by district collectors. Hence statement 2 is correct.

Correct option is A) both of whom live in Turkey

Correct option is C) mistakes