Explore topic-wise InterviewSolutions in .

Correct option is B) to know

to know 

Is a right answer

225 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

This can easy be find that given x² ,we can analyse that first x odd numbers sum is 225 .

\(|x^2 +x |-5 < 0\)

⇒ \(|x (x + 1)| - 5 < 0\)

Case I: \(x \le 0\)

Then

\(x(x + 1)> 0\)

⇒ \(|x (x + 1)| = x(x +1)\)

⇒ \(|x^2 + x| - 5 < 0\)

⇒ \(x^2 + x - 5 < 0\)

⇒ \(\frac{-1-\sqrt{21}}{2}< x< \frac{-1 + \sqrt{21}}{2}\)

But \(x\le -1\)

\(\therefore x \in \left(\frac{-1-\sqrt{21}}{2}, -1\right]\)    .....(1)

Case II: \(- 1 < x \le 0\)

Then

\(x(x + 1)< 0\)

⇒ \(|x^2 + x| = -(x^2 +1)\)

⇒ \(|x^2 + x|-5 < 0\)

⇒ \(- (x^2 + x) - 5 < 0\)

⇒ \(x^2 + x + 5 > 0\)

⇒ \(x\in R\)

Hence, \(x \in ( -1 , 0]\)     .....(2)

Case III: x > 0

Then

 \(x(x + 1)> 0\)

⇒ \(|x^2 + x| = x^2 +x\)

\(\therefore |x^2 + x| - 5 < 0\)

⇒ \(x^2 +x - 5 < 0\)

⇒ \(\frac{-1 - \sqrt{21}}{2} < x < \frac{-1 + \sqrt{21}}{2}\)

But \(x > 0\)

\(\therefore x \in \left(0,\frac{-1+\sqrt{21}}{2}\right)\)    ....(3)

From (1), (2) & (3), we get 

\(x \in \left(\frac{-1-\sqrt{21}}{2},\frac{-1 + \sqrt{21}}{2}\right)\)

\(\int sin^2x \;sin\,x \,dx\)

\(= \int (1 -cos^2x)sin\,x\, dx\)

\(= \int sin \,x\,dx - \int cos^2 x\; sin x \,dx\)

\(= -cos\, x + \frac{cos^3x}{3} +C\)    (By taking cos x = t ⇒ -sin x dx = dt)

(√1+x²) - tan-¹x +c

\(\int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int\frac{x}{\sqrt{x^2 + 1}}dx - \int\frac{1}{\sqrt{x^2 + 1}}dx\)

Let \(x^2 + 1 = t^2\)

Then 

\(2x \,dx = 2t \,dt\)

⇒ \(x\,dx = t\,dt\)

\(\therefore \int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int \frac{t\,dt}t - log\left|x + \sqrt{x^2 +1}\right|+ C\)

\(= t - log \left|x + \sqrt{x^2 + 1}\right| + C\)

\(= \sqrt{x^2 + 1} - log \left|x + \sqrt{x^2 + 1}\right| + C\)

 \(\int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int\frac{x}{\sqrt{x^2 + 1}}dx - \int\frac{1}{\sqrt{x^2 + 1}}dx\)

Let \(x^2 + 1 = t^2\)

Then 

\(2x \,dx = 2t \,dt\)

⇒ \(x\,dx = t\,dt\)

\(\therefore \int \frac{x - 1}{\sqrt{x^2 + 1}}dx = \int \frac{t\,dt}t - log\left|x + \sqrt{x^2 +1}\right|+ C\)

\(= t - log \left|x + \sqrt{x^2 + 1}\right| + C\)

\(= \sqrt{x^2 + 1} - log \left|x + \sqrt{x^2 + 1}\right| + C\)

Let \(1 + xe^x = t\)

Then

\((xe^x + e^x)dx = dt\)

⇒ \(e^x (x + 1)dx = dt\)

⇒ \((x + 1) dx = \frac{dt}{e^x}\)

\(\therefore \frac{(x + 1)dx}{x} = \frac{dt}{xe^x} = \frac{dt}{t -1}\)

\(\therefore \int \frac{(x + 1) dx}{x(1 + xe^x)^2} = \int \frac{dt}{(t - 1)t^2}\)

\(= \int \left(\frac{-1}{t} - \frac1{t^2} + \frac1{t -1}\right)dt\)

\( = -log\, t + \frac1 t + log(t - 1) + C\)

\(= log\left|\frac{t-1}{t}\right| + \frac1t + C\)

\(= log\left|\frac{xe^x}{xe^x + 1}\right| + \frac1{xe^x + 1} + C\)

Concept:

For a second-order differential function, the characteristic equation will have two roots D1 and D2. The roots can have three possible forms i.e.

• Real, distinct roots, D1 ≠ D2
• Complex roots, D1, D2 = a ± ib
• Double roots D1 = D2 = 0

The solution for the second-order differential equation with equal roots of the characteristic equation is given by:

\(y = \left( {{c_1} + {c_2}x} \right){e^{{D_1x}}}\)

If two roots are real and distinct (i.e. D1 ≠ D2), the general solution will be:

\(y = {C_1}{e^{{D_1}x}} + {C_2}{e^{{D_2}x}}\)

If the roots of the characteristic equation are complex (i.e. D1, 2 = a ± ib), the general solution to the differential equation will be:

Calculation:

Given \(\frac{{{d^2}y}}{{d{x^2}}} + p\frac{{dy}}{{dx}} + qy = 0\) 

The given solution is:

\(y = {c_1}{e^{ - x}} + {c_2}{e^{ - 3x}}\)    ---(1)

The general solution of differential equation with unequal roots is given by:

\(y = {c_1}{e^{mx}} + {c_2}{e^{nx}}\)    ---(2)

Here m and n are the roots of ordinary differential equation

Comparing equation (1) and (2), we get:

m = -1 and n = -3

Sum of roots will be:

m + n = -p

-1 – 3 = -p

p = 4

And product of roots is mn = q, i.e.

(-1)(-3) = q

It's auxiliary equation is m² + m + 1 = 0

⇒ \(m = \frac{-1\pm \sqrt{1-4}}{2}= \frac{-1\pm\sqrt3 C}{2}\)

\(C.F. = e^{-\frac x2} \left(C_1 cos\left(\frac{\sqrt3}2x\right)+ C_2sin\left(\frac{\sqrt3}2x\right)\right)\)

\(P.I. = \frac1{D^2 + D + 1}xe^{2x}cos 3x \)

\(= e^{2x} \frac1{(D + 2)^2 + (D + 2) + 1}x \,cos 3x\)

\(= e^{2x} \frac1{D^2 + 5D + 7}x\, cos3x\)

\(= e^{2x} \left(x\frac1{D^2 + 5D + 7}cos 3x + \frac{-(2D + 5)}{(D^2 + 5D + 7)^2}cos3x\right)\)

\(= e^{2x} \left(x\frac{1}{-9 + 5D + 7}cos3x - \frac{2D + 5}{(-9 + 5D + 7)^2}cos 3x\right)\)

\(= e^{2x}\left(x\frac{5D+2}{25D^2 -4}cos3x- \frac{(2D+5)}{25D^2 - 20D + 4}cos3x\right)\)

\(= e^{2x}\left(x\frac{5D \, cos3x + 2cos3x}{-25 \times 9 - 4}+ \frac{(2D + 5)(20D - 221)}{400D^2- 48841}cos 3x\right)\)

\(= e^{2x}\left(\frac{-x}{229}(-15 sin3x + 2cos 3x) - \frac{1}{52441}(40D^2 - 342D - 1105)cos3x\right)\)

\(= e^2x\left(\frac{-x}{229} (-15 sin3x + 2 cos 3x) - \frac1{52441} (-360cos3x + 1026 sin3x - 1105 cos 3x)\right)\)

\(= e^{2x} \left(\frac x{229} (15sin3x - 2 cos 3x) + \frac1{52441} (1465 cos3x - 1026 sin 3x)\right)\)

y = C.F. + P.I is solution of given differential equation.

Let x = e^z

\(\therefore x \frac{dy}{dx} = Dy\), where \(D = \frac d{dz}\)

\(x^2\frac{d^2y}{dx^2} = D(D - 1)y\), where \(D = \frac d{dz}\) 

∴ Given differential equation converts into 

D(D - 1)y + Dy - y = e^2z + 1

It's auxiliary equation is 

m(m - 1) + m - 1 = 0

⇒ m² - 1 = 0

⇒ m = \(\pm \)1

∴ \(C.F. = C_1e^z + C_2e^{-z} = C_1x + \frac{C_2}x\)  (∵ \(e^z = x\))

\(P.I. = \frac1{D^2 - 1} (e^{2z} + 1)\)

\(= \frac1{D^2 - 1}e^{2z} + \frac1{D^2 - 1}e^{0z}\)

\(= \frac{e^{2z}}{4 - 1} + \frac{e^{0z}}{0 - 1}\)

\(= \frac{e^{2z}}{3} + \frac1{-1}\)

\(= \frac{e^{2z}}{3} - 1\)

\(= \frac{x^2}{3} - 1\)  (∵ \(e^z = x\))

∴ \(y = C.F. + P.I.\)

\(= C_1x+ \frac{C_2}x + \frac{x^2}3 - 1\)

which is complete solution of given differential equation.

1-Sinx = sin²x/2 + cos²x/2 -2sinx/2cosx/2 = ( cosx/2 - sinx/2  )²

cosx = cos²x/2 - sin²x/2 = ( cosx/2 - sinx/2 )(cosx/2 + sinx/2 )

then put this values in given integration and substitute  t = cosx/2 + sinx/2  

it reduces in simple integration.

differentiate wrt x 

 f'(x)=(2xf(x)/1+x^2 )+f^2(x)

f(x)=-3(x^2+1)/x^3+3x+c

F(x) = 0→x²∫​​​​​f(√t)dt

F'(x) = f(x)\(d \over dx\)(x²) - f(0)\(d \over dx\)(0)

F'(x) = f(x).2x

f'(x) = f(x).2x          ....[ F'(x) =f'(x) ]

\(f'(x) \over f(x)\) = 2x

\(∫{ f'(x) \over f(x)}\)\(dx\) = \(∫ 2x dx\)

log(f(x)) = x² + C

f(x) = (\(e ^x\))².\(e^c\)

1 = \(e^0\).Q  → Q = 1

f(x) = \((e^x)^2\)

Now,

F(x) = 0 → x² \(∫e^t dt\)

F(x) = \((e^x)²\) - \(e^0\) = \((e^x)² - 1\)

F(2) = \(e^4 -1\)

(a) (i) Given equation of circle is

(x - 2)² + (y + 3)² = 9

⇒ (x - 2)² + y(-(-3))² = 3²

\(\therefore\) Centre of circle is (2, -3) and radius of the circle is 3.(But comparing the standard form of circle whose centre is (h, k) and radius is r; i.e; (x - h)² + (y - k)² = r²)

(ii) Given equation of circle is

2x² + 2y² - 6x + 10y = 1

⇒ x² + y² - 3x + 5y = 1/2 (Dividing both sides by 2)

Compare with x² + y² + 2fx + 2gy + c = 0

we get 2f = -3 ⇒ f = -3/2

2g = 5 ⇒ g = 5/2

& c = -1/2

\(\therefore\) Centre of the circle = (-f, -g) = (3/2, -5/2)

and radius of the circle is r = \(\sqrt{f^2+g^2-c}\) 

⇒ r = \(\sqrt{(\frac{-3}2)^2+(5/2)^2-(-1/2)}\)

 = \(\sqrt{\frac94+\frac{25}4+\frac12}\) 

 = \(\sqrt\frac{9+25+2}4\) = \(\sqrt\frac{36}4=\sqrt9\) = 3

\(\therefore\) Radius of the circle is r = 3

& centre of the circle is(-3/2, 5/2)

(iii) Given equation of circle is 

x² + y² + 8x - 2y + 13 = 0

⇒ x² + 8x + 16 - 16 + y² - 2y + 1 - 1 + 13 = 0

⇒ (x + 4)² + (y - 1)² + 13 - 17 = 0

⇒ (x - (-4))² + (y - 1)² = 2²

\(\therefore\) Centre of the circle is(-4, 1)

and radius of the circle is r = 2

(b)(i) Polar equationo of the curve is

\(\frac2r=1+cos\theta\)----(1)

\(\because\) x = r cos θ and y = r sin θ

\(\therefore\) cos θ = x/r

Then equation(1) converts into

2/r = 1 + x/r

⇒ 2 = x + r

⇒ 2 - x = r

⇒ r² = (2 - x)²

⇒ x² + y² = x² - 4x + 4 (\(\because\) r² = x² + y²)

⇒ y² = -4x + 4 which is a cartesian equation of the given curve.

(ii) Given cartesian equation of the curve is

x² + y² = 9

⇒ r² = 9 (\(\because\) x² + y² = r²)

⇒ r = 3

(c) (i) Equation of line which passes through two points (-2, 3) & (2, -5) is

\(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

⇒ y - 3 = \(\frac{-5-3}{2-(-2)}(x - (-2))\) 

⇒ 2x + y + 1 = 0 which is equation of required line.

(ii) Given equation of curve is

x² + y² + 2x + 4y - 3 = 0----(1)

By diferentiating equation (1) w.r.t. x

we get

2x + 2y dy/dx + 2 + 4 dy/dx = 0

⇒ (2y + 4)dy/dx = -(2x + 2)

⇒ dy/dx = \(\frac{-(2x+2)}{2y+4} = \frac{-(x+1)}{y+2}\)

\(\therefore\) slope of tangent to curve (1) at point (1, -4) is

m = (dy/dx)(1, -4) = \(\frac{-(1+1)}{-(4+2)}=\frac{-2}{-2}=1\)

\(\therefore\) Equation of tangent to curve (1) at point (1, -4) is y-(-4) = m(x - 1)

⇒ y + 4 = 1(x - 1)

⇒ x - y = 5

which is equation of tangent to given curve at point(1, - 4)

The correct option is (ii) B.

Intial Energy = 2 + 0 = 2 ev

Final energy = \(\frac{13.6\times1}1\)

2 = -13.6 + E_p

E_p = 15.6 ev

ɸ = \(\frac{12420}{4600}\)

ɸ = 2.7

then 

E = hv - ɸ

E = 15.6 - 2.7

E = 12.9 ev

Given E = 1.5 ev

(i) Kinetic energy of electron

K.E. = -E

K.E. = -5(-1.5 ev)

K. E. = 1.5 ev

(ii) Potential energy of electron

P.E. = -2 K.E.

P.E. = -2 x 1.5

P.E. = -3.0 ev

(iii) Energy of photon \(\Delta\)E = E₂ - E₁

\(\Delta\)E = -1.5 - (-13.6)

\(\Delta\)E = 12.1 ev

\(\Delta\)E = 12.1 x 1.6 x 10^-19 J

\(\Delta\)E = 19.36 x 10^-19 J

Energy of photon

\(\Delta\)E = \(\frac{hc}\lambda\)

\(\frac{hc}\lambda\) = 19.36 x 10^-19

Wavelength of radiation \(\lambda\) = \(\frac{6.6\times10^{-34}\times3\times10^8}{19.36\times10^{-19}}\)

\(\lambda=1.022\times10^{-7}\) m

Correct option is: (B) Nucleus

Correct option is: (B) 5’ → 3’ direction

Correct option is: (C) C-residue

Correct option is: (B) Morphine

Correct option is: (C) Eichhornia

Correct option is: (D) It reduces the oxygen carrying capacity of haemoglobin

Correct option is: (D) Sewage

Correct option is: (D) 80-99 dB