Explore topic-wise InterviewSolutions in .

Correct option is  (2) √3/4

Correct option is (4) 0

\(\lim\limits_{x\to 0^+} (x\, cos\, x + x\,log\, x)\)

\(= \lim\limits_{x\to 0^+} x\, cosx + \lim\limits _{x \to 0^+} x\, log x\)

\(= \lim\limits_{h\to 0}(0 + h)\, cos(0 + h) + \lim\limits_{x \to 0^+} \frac{log x}{\frac1x}\)   \(\left(\frac{\infty}{\infty}- case\right)\) 

\(= 0 \times cos0 + \lim\limits_{x\to 0^+} \cfrac{\frac1x}{\frac{-1}{x^2}}\)     (By using D.L.H. Rule)

\(= 0 + \lim\limits_{x\to 0 ^+}\)

\(= \lim\limits_{h\to 0}- (0 + h)\)

\(= -0\)

\(= 0\)

\(\lim\limits_{x\to 0} \frac{ln(cos\, x)}{(1+ x^2)^{\frac14} - 1}\)                      (\(\frac00\) - case)

\(= \lim\limits_{x\to 0} \frac{-sin\,x}{cos\, x\left(\frac14 (1 + x^2) ^{-\frac34}.2x\right)}\)    (By using D.L.H. Rule)

\(=\frac12 \lim\limits _{x \to 0} \frac{-sin\, x}{x} . \frac{1}{cos\,x}.(1 + x^2)^{\frac34}\)

\(= \frac{-1} 2 \left(\lim\limits_{x\to 0} \frac{sin\, x} x\right) \left( \lim\limits_{x\to 0} \frac1{cos\,x}\right) \left(\lim\limits_{x\to 0} (1 + x^2)^{\frac34}\right)\)

\( = \frac{-1}{2} \times 1 \times 1\times1\)          \(\left(\because \lim\limits_{x \to 0} \frac{sin\, x}{x} = 1\right)\)

\(= \frac{-1}{2}\)

Global Challenges

Let I = \(\int\cfrac{2x(1+2x^2)}{1+x^2+x^4}dx\)

Let 1 + x² + x⁴ = t

= (2x + 4x³)dx = dt

= 2x(1 + 2x²) dx = dt

\(\therefore\) I  = \(\int\cfrac{dt}t\) = log t + c

= log|1 + x² + x⁴ | + c

\(\int_c \vec F. d \vec r =\int_c (x^2 \hat i + xy\hat j) (dx\,\hat i + dy \hat j)\)

= \(\int _c (x^2dx+xydy)\)

= \(\int_cx^2dx+\int_cy^2dy\) (∵ y = x)

= \(\int\limits_0^1 x^2dx + \int\limits_0^1y^2dy\)

= \((\frac {x^3}{3})_0^1 + (\frac {y^3}{3})_0^1\)

= \(\frac {1}{3} + \frac {1}{3}= \frac 23\)

Given \(x = \frac \pi3\), \(y = \frac \pi 6\)

L.H.S. = \(cos(x + y) = cos(\frac\pi3 + \frac\pi 6) = cos(\frac \pi 2) = 0\)

R.H.S. = \(cos x + cos y = cos(\frac \pi3) + cos(\frac\pi 6) = \frac 12 +\frac{\sqrt 3}2 \ne 0\)

Hence, \(L.H.S. \ne R.H.S.\)

Hence, at \(x = \frac\pi3 \) & \(y = \frac \pi 6, cos (x + y) \ne cos x + cos y\).

f(x₁) = f(x₂) 

X₁² + 4 = X₂² + 4 

x₁ = x₂ 

∴ f (x) is one-one 

Let y = f(x) y = 4x + 3 

x = y - 3/4 ∈R₊ 

∴ f (x) is onto 

Let f^-1 (x) = y 

f(y) = x 

4y +3 = x

y = x - 3/4 

∴ f^-1 : [4,8) → R+ is defined by 

f^-1 (x) = x - 3/4

Given

\(\vec A = 3\)

\(\vec B = 4\)

\(\vec C = 5\)

\(\vec A + \vec B = \vec C\)

Squaring both sides

\(|\vec A|^2 + |\vec B|^2 + 2\vec A.\vec B = \vec C .\vec C\)

\(|\vec A|^2 + |\vec B|^2 + 2|\vec A||\vec B| \,cos \theta = |\vec C|^2\)

\((3)^2 + (4)^2 + 2\times 3 \times 4 \, cos \theta = (5)^2\)

\(cos\theta = 0\)

\(\theta = 90° \)

Correct option:

(C) I 

Correct option:

(D) π/3

Correct option:

(B) –2 

Correct option:

(B) - 1/3

Parametric equation of parabola x² = 36y is (18t, 9t²)

∵ \(t = \frac13\) given

∴ Point on parabola is \(\left(\frac{18}3,\frac99\right) = (6,1)\).

Hence, required point is (6, 1).

In the word ASSISTANT there are 9 letters, of which S appears 3 times, A appears 2 times, T appears 2 times and the rest all are different. Therefore the total number of ways is
\(\frac{9 !}{3! \times 2 ! \times 2 !}\) = 15120.

log 5040 = log(2⁴ x 3² x 5 x 7)

= log2⁴ + log3² + log5 + log7

= 4log2 + 2log3 + log5 + log7

Let S = 1.6 + 1.66 + 1.666 + ..... + n terms

= (1 + 0.6) + (1 + 0.6 + 0.06) + (1 + 0.6 + 0.06 + 0.006) + ......+ n  terms

= (1 + 1 + ..... + 1)(n times) + (\(\frac{6}{10}\) + \(\frac{6}{10}\) + \(\frac{6}{10}\) + .....+\(\frac{6}{10}\)) (n times)

+ (\(\frac{6}{100}\) + \(\frac{6}{100}\) + \(\frac{6}{100}\) + .... + \(\frac{6}{100}\)) ((n-1) times)

+ (\(\frac{6}{10^3}\) + \(\frac{6}{10^3}\) + \(\frac{6}{10^3}\) + .... + \(\frac{6}{10^3}\))((n-2) times)

+ .... + \(\frac{6}{10^n}\)

= n + \(\frac{6n}{10}\) + \(\frac{6(n-1)}{100}\) + .... + \(\frac{6}{10^n}\)

Probability of Sangeeta winning the match P(S) = 0.62.

∴  Probability of Reshma winning the match

P(R) = 1 – P(S) = 1 – 0.62 = 0.38

Given:

Number of men = 10

Number of women = 8

Concept used:

Number of ways in which one man and woman can be chosen from a group of ‘n’ men and m woman = ⁿC₁ × ^mC₁

ⁿC_r = n!/[r! × (n – r)!]

Calculation:

Number of ways in which one man and woman can be chosen from a group of 10 men and 8 women = ¹⁰C₁ × ⁸C₁ = 10 × 8

Number of ways in which one man and woman can be chosen from a group of 10 men and 8 women = 80

∴ The number of ways in which one man and woman can be chosen from a group of 10 men and 8 women is 80

Statement 1:

We get:

Y = 3X

⇒ X/Y = 1/3

Now, we can obtain the value of X/4Y, by simply multiplying the denominator Y by 4

⇒ X/(4 × Y) = 1/(4 × 3) = 1/12

Hence, Statement 1 can be used alone.

Statement 2:

We get X = 5.

But there is nothing mentioned, using which we can get the value of Y, and hence, the value of X/4Y.

Hence, Statement 2 cannot be used alone.

∴ Statement 1 alone is sufficient, while Statement 2 alone is not sufficient.

Correct option is C) when he had graduated from the University.