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Answer (A) Nylon-6

Correct option is (2) Neoprene

The Correct option is (b) dacron.

In Black colour does the sky appear for an astronaut.

(a) Both the Statements are true and Statement-II is the correct explantion of Statement -I

(d) Statement I is false, but Statement -II is true.

(c) Statement -I  is true, but Statement -II is false.

Solution: The correct answer is option (A)

Total energy = - Kinetic energy = - E
i.e E energy will have to be supplied to escape the electron to infinity.

Solution:
cos x + sinx = √ 2 cos x , then (√2 -1 ) cos  x = sin x
on multiplying both sides by (√2+1) , we get
(√2+1)(√2-1) cos x  =  (√2+ 1) sin x
⇒ cos x = √2 sin x + sin x
⇒ cos x -sin x = √ 2 sin x

We have

12x² - 20xy + 7y² ≡(2x - y)(6x - 7y)

 Therefore, the sides of the triangle are

2x - y = 0 ......(1)

6x - 7y = 0   .....(2)

2x - 3y = -4   ......(3)

Solving Eqs. (1) and (2), we get (1,2) as the vertex. Solving Eqs. (2) and (3), we obtain (7, 6) as another vertex.By hypotheses, (0, 0) is the third vertex. Hence, the centroid of the triangle is

(0 + 1 + 7/3 , 0 + 2 + 6/3) = (8/3,8/3)

No. of students whose weight in the range of (0 - 38) kg is 0. By converting normal form into less than and more than series.

Answer: Error

Given differential equation is,

\(\frac{d^2y}{dx^2}+\frac{dy}{dx}+x=\sqrt{1+\frac{d^2y}{dx^2}}\)

By Squaring both sides, we get

\((\frac{d^2y}{dx^2}+\frac{dy}{dx}+x)^2=1+\frac{d^2y}{dx^2}\)

\(\Rightarrow\) \((\frac{d^2y}{dx^2})^ 2-\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2+2\,\frac{d^2y}{dx^2}\frac{dy}{dx}+2x\frac{d^2y}{dx^2}+2x\frac{dy}{dx}+x^2=1\)

which is differential equation of order 2 and degree 2.

(A+A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T

∴ A +A^T is a symmetric matrix

Given different equation is 

(D² + 4)y = 4 sec² 2x

It,s auxiliarly equation is

m² + 4 = 0

= m = \(\pm\) 2i

\(\therefore\) C . F = c₁ cos 2x + c₂ sin 2x

Let y₁ = cos 2x and y₂ = sin 2x

R = 4 sec² 2x

W(y₁,y₂) = \(\begin{vmatrix} y_1 & y_2 \\[0.3em] y_1^1 & y_2^1 \\[0.3em] \end{vmatrix}\)

= \(\begin{vmatrix} cos2x & sin2x \\[0.3em] -2sin2x & 2cos2x \\[0.3em] \end{vmatrix}\)

= 2cos² 2x + 2sin² 2x

= 2(sin² 2x + cos² 2x)

= 2

\(\therefore\) u₁ = \(\int \frac{\begin{vmatrix} 0& sin2x \\[0.3em] 4sec^2 2x & 2cos2x \\[0.3em] \end{vmatrix}}{w(y_1,y_2)}dx\) 

= \(\int \frac{\begin{vmatrix} -4\frac{sin2x}{cos2x}.sec2x \\[0.3em] \end{vmatrix}}{2}dx\)

= -2 \(\int\) sec 2x . tan 2x dx 

= - 2 \(\cfrac{sec\,2x}2\) = - sec 2x

u₂ = \(\int \frac{\begin{vmatrix} cos2x&0 \\[0.3em] -2sin2x & 4sec^22x \\[0.3em] \end{vmatrix}}{w(y_1,y_2)}dx\)

= \(\int\cfrac{4cos2x.sec^2\,2x}2 dx\)

= 2 \(\int\) sec 2x dx

= 2 \(\cfrac{log|sec2x+tan2x|}2\)

= log |sec 2x x tan 2x |

P . I = u₁y₁ + u₂y₂

= - sec2x . cos 2x + log |sec 2x + tan 2x| sin 2x

= - 1+ log |sec2x + tan 2x| sin2x

\(\therefore\) complete solution of given differential equation is 

y = C . F . + P . I .

= c₁ cos 2x + c₂ sin 2x - 1 + log |sec 2x + tan 2x| sin 2x

\(\frac{d}{dx}cos^{-1}(\frac{x}{x+1})=\cfrac{-1}{\sqrt{1-(\frac{x}{x+1})^2}}\frac{d}{dx}(\frac{x}{x+1})\)

 = \(\frac{-(x+1)}{\sqrt{(x+1)^2-x^2}}\times\frac{(x+1).1-x(1)}{(x+1)^2}\) 

 = \(\frac{-(x+1)}{\sqrt{2x+1}}\times\frac1{(x+1)^2}\)

 = \(\frac{-1}{\sqrt{2x+1}(x+1)}\)

\(L = \lim\limits_{n \to \infty} (2^n + 3^n + 11^n) \)   \((\infty^0 - case)\)

\(log\, L = \lim\limits_{n \to \infty} \frac1n (log (2^n + 3^n + 11^n) )\)  \(\left(\frac\infty\infty - case\right)\)

\(= \lim\limits_{n \to \infty } \frac{2^n log2 + 3^n log2 + 11^n log11}{2^n + 3^n + 11^n}\)    (By using D.L.H. Rule)

\(=\lim\limits_{n\to\infty} \cfrac{11^n\left(\left(\frac2{11}\right)^n log2 +\left(\frac3{11}\right)^n log2 + log 11 \right)}{11^n \left(\left(\frac2{11}\right)^n + \left(\frac3{11}\right)^n + 1\right)}\)

\(=log11\left(\because\lim\limits_{n\to\infty}\left(\frac2{11}\right)^n =\lim\limits_{n\to\infty}\left(\frac3{11}\right)^n = 0 \right)\)

\(\therefore L = 11\)

⇒ \(\lim\limits_{n\to \infty}(2^n + 3^n + 11^n)^\frac1n = 11\)

 \(\frac{d}{dx}cos^{-1}(\frac{x}{x+1})=\cfrac{-1}{\sqrt{1-(\frac{x}{x+1})^2}}\frac{d}{dx}(\frac{x}{x+1})\)

 = \(\frac{-(x+1)}{\sqrt{(x+1)^2-x^2}}\times\frac{(x+1).1-x(1)}{(x+1)^2}\) 

 = \(\frac{-(x+1)}{\sqrt{2x+1}}\times\frac1{(x+1)^2}\)

 = \(\frac{-1}{\sqrt{2x+1}(x+1)}\)

The son told his father that that day his teacher (a) had asked him to write an essay. His father asked him what topic she (b) had given The son told that it was “Indian Culture’. His father asked if he (c) had attempted the essay. The son replied that he (d) had written only a few lines and (e) requested to tell him something about it.

\(y = tan^{-1}\left(\frac{\sqrt{1+x^2}+1}{x}\right)\)

Let \(x = tan \theta \)

⇒ \(\theta = tan^{-1}x\)

\(\therefore y= tan^{-1}\left(\frac{\sqrt{1+tan^2\theta}+1}{tan\theta}\right)\)

\(= tan^{-1}\left(\frac{sec\theta + 1}{tan\theta}\right)\)

\(= tan^{-1}\left(\frac{1+cos\theta}{sin\theta}\right)\)

\(= tan^{-1} \left(\frac{2cos^2\frac{\theta}{2}}{2sin\frac{\theta}2cos\frac{\theta}{2}}\right)\)

\(= tan^{-1}\left(cot\frac{\theta}{2}\right)\)

\(= tan^{-1}\left(tan\frac{\pi}{2}- \frac{\theta}{2}\right)\)

⇒ \(y = \frac{\pi}{2}- \frac{\theta}{2}\)

\(\therefore \frac{dy}{dx} = \frac{-1}{2} \frac{d\theta}{dx}\)

\(= \frac{-1}{2} \frac{d}{dx} tan^{-1}x\)          \((\because \theta = tan^{-1}x)\)

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

\(\therefore \frac{d}{dx} tan^{-1} (\frac{\sqrt{1 +x^2} + 1}{x})= \frac{-1}{2(1 + x^2)}\)

Correct option is D) It’s Jane.

log_ab x log_ba = log_ab x \(\frac1{log_a^b}=1\)

Sarah called out to Jane and asked her (a) to bring her a sheet of writing paper, as (b) she had to write a letter Jane asked her (c) where she was to find it. Sarah replied that ther (d) was plenty in (e) her mistress’s letter case in the parlour.

Hari proposed Mohan that (a) they should go to the playground. Mohan regretted that (b) he could not He further remarked that (c)he must reach home as early as possible. Hari asked him (d)why he was in such a hurry Mohan replied that (e) he had to take her sister to the doctor.