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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 21751. |
You conducted a successful job search, and now have three offers from which to choose. What can you do to most thoroughly investigate your potential employers? a. check out their websites b. watch the news to see if the companies are mentioned c. research their financial situations d. speak with people who work for them already |
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Answer» The Correct option is (a, c, d) |
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| 21752. |
Fill in the blanks. Distance of P(x,-y) from the origin will be ……………….. |
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Answer» Distance of P(x,-y) from the origin will be √(x2 +y2). |
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| 21753. |
4 cosec2θ -4cot2 θ is equal to(a) 1 (b) 0 (c) 2 (d) 4 |
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Answer» The correct option is: (d) 4 |
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| 21754. |
Fill in the blanks. If 2 sin θ = 1, then θ = ...... |
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Answer» If 2 sin θ = 1, then θ = 30°. |
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| 21755. |
The difference of maximum value and minimum value of a data is called(a) Range (b) Class Inverval (c) Mid Point (d) None of these |
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Answer» The correct option is:(a) Range |
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| 21756. |
Most power FETS are a. Used in high-current applications b. Digital computers c. RF stages d. Integrated circuits |
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Answer» (a) Used in high-current applications |
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| 21757. |
With CMOS, the upper MOSFET is a. A passive load b. An active load c. Nonconducting d. Complementary |
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Answer» (d) Complementary |
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| 21758. |
Fill in the blanks. The Co-ordinate of any point an y-axis is……… |
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Answer» The Co-ordinate of any point an y-axis is (o,y). |
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| 21759. |
An E-MOSFET with its gate connected to its drain is an example of a. A three-terminal device b. An active load c. A passive load d. A switching device |
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Answer» (b) An active load |
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| 21760. |
An ordinary resistor is an example of a. A three-terminal device b. An active load c. A passive load d. A switching device |
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Answer» (c) A passive load |
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| 21761. |
With active-load switching, the upper E-MOSFET is a a. Two-terminal device b. Three-terminal device c. Switch d. Small resistance |
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Answer» (a) Two-terminal device |
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| 21762. |
Find the 10th term from the end of the AP 4, 9, 14, … … … . , 254. |
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Answer» Given, AP is 4, 9, 17……...,254. First term of AP is a = 4. Common difference of AP is d = a2 – a1 = 9 – 4 = 5. nth term of AP is an = 254. ∵ an = a + (n – 1) d ∴ a + (n – 1) d = 254 ⇒ 4 + (n − 1)5 = 254 (a = 4 & d = 5 and an = 254) ⇒ 5(n − 1) = 254 − 4 = 250 ⇒ n − 1 = \(\frac{250}{5}\) = 50 ⇒ n = 50 + 1 = 51. 10th term from the end of the AP is 51 – (10 – 1) = 51– 9 = 42th term of AP. ∴ a42 = a + (42 – 1)d = 4 + 41 × 5 = 4 + 205 = 209. (∵a = 4,d = 5) Hence, The 10th term from the end of the given AP is 209. |
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| 21763. |
Ram and Rohit can run at a speed of 45 m/min. and 54 m/ min. respectively. They start from a point on a circular track of circumference 1980 meter in the opposite direction and meet at a point. Find the distance traveled by Rohit when they meet.1. 1080 metre2. 980 metre3. 900 metre4. 1030 metre |
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Answer» Correct Answer - Option 1 : 1080 metre Given: Speed of Ram = 45 m/min Speed of Rohit = 54 m/min Calculations: Let time be t Hence, 45 t + 54 t = 1980 ⇒ 99 t = 1980 ⇒ t = 20 min. In the First meeting Distance covered by Ram = 45 × 20 = 900 metre Distance covered by Rohit = 54 × 20 = 1080 m When they meet distance covered by Rohit 1080 m |
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| 21764. |
The sum of two numbers is 18 and the sum of their reciprocals is 1/4.Find the numbers. |
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Answer» Let the numbers are a and b. Given that, The sum of these numbers is 18 and sum of their reciprocals is 1/4. Hence, a + b = 18 … (1) And, \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{4}\) ⇒ \(\frac{a+b}{ab}\) = \(\frac{1}{4}\) ⇒ ab = 4(a + b) = 4 × 18 = 72. (∵ a + b = 18.) Now, (a − b)2 = (a + b)2 − 4ab = 182 − 4 × 72 = 324– 288 = 36. ∴ a − b = \(\sqrt{36}\) = 6 … (2) Now, Adding equations (1) and (2), we get (a + b) + (a – b) = 18 + 6 = 24 ⇒ 2a = 24 ⇒ a = 12. Now, Putting a = 12 in equation (1), we get 12 + b = 18 ⇒ b = 18 − 12 = 6. Hence, The numbers are 6 and 12. |
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| 21765. |
What number must be added to each of the numbers 5, 9, 17, 27 to make new numbers in proportion? |
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Answer» Let the number to be added be x. Then the numbers 5, 9, 17, 27 will be 5 + x, 9 + x, 17 + x, 27 + x. Given that new numbers are in proportional, i.e., 5 + x : 9 + x :: 17 + x : 27 + x. ⇒ \(\frac{5+x}{9+x}\) = \(\frac{17+x}{27+x}\) ⇒ (5 + x)(27 + x) = (9 + x)(17 + x) ⇒ x2 + 32x + 135 = x2 + 26x + 153 ⇒ 32x − 26x = 153 − 135 ⇒ 6 = 18 ⇒ x = \(\frac{18}{6}\) = 3. |
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| 21766. |
The degree of a quadratic equation will be(a) 2 (b) 0 (c) 1 (d) None of these |
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Answer» The correct option is: (a) 2 |
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| 21767. |
An E-MOSFET that operates at cutoff or in the ohmic region is an example of a. A current source b. An active load c. A passive load d. A switching device |
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Answer» (d) A switching device |
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| 21768. |
VGS(on) is always a. Less than VGS(th) b. Equal to VDS(on) c. Greater than VGS(th) d. Negative |
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Answer» (c) Greater than VGS(th) |
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| 21769. |
CMOS stands for a. Common MOS b. Active-load switching c. p-channel and n-channel devices d. Complementary MOS |
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Answer» (d) Complementary MOS |
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| 21770. |
If sinθ = cosθ then the value of θ will be(a) 30° (b) 45° (c) 60° (d) 90° |
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Answer» The correct option is: (b) 45° |
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| 21771. |
Power FETs are a. Integrated circuits b. Small-signal devices c. Used mostly with analog signals d. Used to switch large currents |
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Answer» (d) Used to switch large currents |
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| 21772. |
The probability for a contractor to get a road contract is 2/3 and to get a building contract is 5/9. The probability to get at least one contract is 4/5. Find the probability that he gets both the contracts. |
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Answer» Suppose A is the event of getting a road contract. B is the event of getting a building contract Given P(A) = 2/3; P(B) = 5/9; P(A ∪ B) = 4/5 P(A ∪ B) =P(A) + P(B) – P(A ∩ B) P(A ∩ B) =P(A) + P(B) – P(A ∪ B) = 2/3 + 5/9 - 4/5 = (30 + 25 - 36)/45 = 19/45 Probability to get both contracts = 19/45 |
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| 21773. |
In trangle ABC a4+b4+c4-2c2(a2+b2)=0 proove c=45 or135 |
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Answer» Given condition can be written as a^4+b^4+c^4-2c^2(a^2+b^2)+2a^2b^2=2a^2b^2 =>[(a^2+b^2-c^2)/(2ab)]^2==1/2 =>cos^2C=1/2 =>cosC=+1/√2 or -1/√2 Then C= 45° or135° |
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| 21774. |
A pair of dice are rolled. The probability of obtaining an even prime number on each die is(a) 1/36(b) 1/12(c) 1/6(d) 0 |
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Answer» Answer is (a) 1/36 |
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| 21775. |
SinA+sinB+sinC=S/R |
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Answer» We know a/sinA=b/sinB=c/sinC=2R So a/(2R)=sinA b/(2R)=sinB c/(2R)=sinC Now sinA+sinB+sinC =1/(2R)[a+b+c) =2s/(2R)=s/R Proved |
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| 21776. |
Choose the correct antonyms for the underlined words from the options given.It is important that even the mediocre are given due recognition. (a) middling (b) ordinary (c) exceptional (d) middleman |
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Answer» (c) exceptional |
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| 21777. |
If vector a = vector(4 i + 3 j + 4 k) and vector b = vector(3 i + 2 k) then |vector(b x 2a)| is(A) 18 (B) 20 (C) 22 (D) 25 |
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Answer» Correct option: (C) 22 vec(b×2a)=2vec(3i+2k)×vec(4i+3j+4k) =2vec(12i×i+8k×i+9i×j+6k×j+12i×k+8k×k) =2vec(12.0+8j+9k-6i-12j+8*0) =2vec(-6i-4j+9k) Now |(b×2a)| =2√(6^2+4^2+9^2)=2√133 ~23 |
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| 21778. |
The selling price of one article after allowing a discount of 15% on its cost price, is the same as the selling price of another article after allowing a discount of 25% on its cost price. If the sum of the cost prices of both the articles is Rs. 640, then find the selling price of each article.1. Rs. 2502. Rs. 2553. Rs. 2804. Rs. 340 |
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Answer» Correct Answer - Option 2 : Rs. 255 Given: Selling prices of two articles after allowing a discount of 15% and 25% on their cost price are same. Sum of cost prices of both the articles = Rs. 640 Formula used: (100 - Discount %) of cost price of 1st article = (100 - Discount %) of cost price of 2nd article Calculations: Let the cost price of 1st article and 2nd article be 'x' and 'y' respectively. According to question, (100 - 15)% × x = (100 - 25)% × y ⇒ 85% × x = 75% × y ⇒ x/y = 15/17 Let the value of x and y be 15t and 17t respectively. According to question, ⇒ 15t + 17t = 640 ⇒ 32t = 640 ⇒ t = 20 ⇒ 15t = 300 ⇒ The cost price of 1st article = Rs. 300 ⇒ The selling price of the 1st article = 85% of 300 ⇒ Rs. 255 ∴ The selling price of each article is Rs. 255.
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| 21779. |
Choose the correct antonyms for the underlined words from the options given.I managed to solve the complex problem. (a) intricate (b) compound(c) simple (d) complicated |
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Answer» Correct Answer is : (c) simple |
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| 21780. |
Choose the correct antonyms for the underlined words from the options given.It is compulsory to donate one day salary for the flood victims. (a) obligation (b) optional (c) obeisance (d) obstruction |
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Answer» (b) optional |
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| 21781. |
Le t `f(x) ={{:((sinpix)/(5x)",",xne0),(" "k",",x=0):}` if (x) is continous at `x=0`, then k is equal toA. `(pi)/(5)`B. `(5)/(pi)`C. 1D. 0 |
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Answer» Correct Answer - A Since , f(x) is continuous at `x=0`, `therefore lim_(xto0)f(x)=f(0)` `rArr lim_(xto0)(sinpi x)/(5x)=k` `rArr 1((pi)/(5))=krArr k=(pi)/(5)` |
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| 21782. |
The value of `alpha` so that `sin^(-1)((2)/(sqrt(5)))`, `sin^(-1)((3)/(sqrt(10)))` and `sin^(-1)alpha` are the angles of a Delta isA. `(1)/(2)`B. `(sqrt(3))/(2)`C. `(1)/(sqrt(2))`D. `(1)/(sqrt(3))` |
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Answer» Correct Answer - C `becausesin^(-1)((2)/(sqrt(5)))+sin^(-1)((3)/(sqrt(10)))+sin^(-1)alpha=pi` `impliestan^(-1)(2)+tan^(-1)(3)+sin^(-1)alpha=pi` `impliespi-tan^(-1)(1)+sin^(-1)(alpha)=piimpliesalpha=(1)/(sqrt(2))` |
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| 21783. |
Negation of the statement `pto(q^^r)` isA. `pto~(q^^r)`B. `~pvv(q^^r)`C. `p^^(~p^^~r)`D. `(q^^r)top` |
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Answer» Correct Answer - C `p^^(-(q^^r))=p^^(~qvv-r)` |
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| 21784. |
Let `f(x)= {(((xsinx+2cos2x)/(2))^((1)/(x^(2))),,,,xne0),(e^(-K//2),,,,x=0):}` if `f(x)` is continous at `x=0` then the value of k isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `becauseunderset(xto0)limf(x)=e^(underset(xtoo)(lim)((xsinx+2cos2x)/(2)-1)(1)/(x^(2))=e^(-3//2))` and `f(0)=e^(-k//2)` `thereforek=3` |
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| 21785. |
Mean of 5 observation is 7. if four of these observation are 6,7,8,10 and one is missing, then te variance of all the five observations isA. 2B. 6C. 8D. 4 |
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Answer» Correct Answer - D `because7=(6+7+8+10+x_(5))/(5)impliesx_(5)=4` `becausesigma=(sum(x_(i)-overline(x))^(2))/(5)=(1)/(5)[1+0+1+9+9]=4` |
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| 21786. |
Let `hata,hatb` and `hatc` be unit vectors and `alpha, beta`, and `gamma` the angles between the vectors `hata,hatb,hatb,hatc` and `hatc, hata` respectively . If `hata + hatb+hatc` is also a unit vectors, then `cosalpha + cosbeta +cosgamma` is equal to -A. `-1`B. `3`C. `-3`D. `1` |
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Answer» Correct Answer - 1 `|hata|= |hatb|=|hatc|=1` &`|hata +hatb+ hatc|=1" ".....(1)` By (1) `rArr |hata + hatb+ hatc|^(2)=1` `|hata|^(2) +|hatb|^(2) + |hatc|^(2) + 2|hata||hatb| cos alpha + 2|hatb||hatc|cos beta +2 |hatc||hata| cos gamma=1` `rArr 3+2 (cos alpha + cos beta + cos gamma)=-1` `rArr (cos alpha + cos beta + cos gamma )=-1` |
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| 21787. |
`(1 + sec 2 theta)(1 + sec 2^(2) theta)(1 + sec 2^(3)` is equal isA. `cot 8 theta tan theta`B. `tan 4theta cot theta`C. `tan 8theta cot theta`D. none of these |
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Answer» Correct Answer - 3 `((1 + cos 2 theta)/(cos 2 theta))((1+ cos 4 theta)/(cos 4theta))((1+ cos 8 theta)/(cos 8 theta))` `= (2cos^(2) theta)/(cos 2 theta)(2 cos^(2) 2theta)/(cos 2 theta)(2cos^(2)4theta)/(cos8 theta)` `=(cos theta cos 2theta(sin 4theta)(2cos^(2) 4 theta))/(sin theta cos 2 theta cos2^(2)theta cos 8 theta)` `=(cos theta cos 2 theta(sin 4 theta)(2 cos^(2) 4 theta))/(sin theta cos 2theta cos 2^(2) thetacos 8 theta)` `=(cos theta cos 2theta cos 4theta cos 8 theta)/(sin theta cos 2theta cos2^(2) theta cos 8theta)` `=tan 8 theta cot theta` |
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| 21788. |
The mean of five observation is 4 and their variance is 5.2 . If three of these observations are 1,2 and 6, then the other two are,A. 2 and 9B. 3 and 8C. 4 and 7D. none of these |
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Answer» Correct Answer - 3 Let two unkown observation are x and y `barx= 4 = (1 + 2 +6 + x+y)/(5)` `x + y = 11" "...(1)` `sigma= 5.2` `(1^(2)+2^(2) +6^(2) +x^(2) + y^(2))/(5) -(barX)^(2) = 5.2` `x^(2)+ y^(2) = 65" "...(2)` by (1) and (2) `x = 4, y= 7` or ` x= 7, y = 4` |
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| 21789. |
If x = - 1 and x =2 are exterme points of the function y=a `log x + bx^(2) + x`, then-A. `a=2,b= 1//2`B. `a=2,b=-1//2`C. `a=-2,b= 1//2`D. `a=- 2, b=- 1//2` |
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Answer» Correct Answer - 2 `(dy)/(dx) = (a)/(x) + 2 bx +1` Since `x=-1` and `x=2` are extreme points so gy/dx at these points must be zero . So `- a - 2b + 1 = 0 and a//2 + 4b + 1 = 0` `rArr a+ 2b -1=0 and a+ 8b + 2=0` `rArr a= 2, b=- 1//2` |
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| 21790. |
adj- `[{:( 1,0,2),(-1,1,-2),(0,2,1):}]=[{:(5,a,-2),(1,1,0),(-2,-2,b):}]`, then [a, b] is equal to-A. `[-4 1]`B. `[-4 -1]`C. `[4 1]`D. `[4-1]` |
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Answer» Correct Answer - 3 `(Adj A)_(12)= C_(21)= "cofactor of" a_(21)=- |{:(0,2),(2,1):}| =4` `(Adj A)_(33)= C_(33)= "cofactor of" a_(33)= + |{:(1,0),(-1,1):}| =1` |
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| 21791. |
If `a =( 4sqrt(6))/(sqrt(2)+sqrt(3))` then the value of `(a+2sqrt(2))/(a-2sqrt(2))+(a+2sqrt(3))/(a-2sqrt(3))` |
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Answer» Correct Answer - 2 If `x= (4sqrt(6))/(sqrt(2)+sqrt(3))` then ………… `x = (4sqrt(6))/(sqrt(2)+sqrt(3))` , `:. (x)/(2sqrt(2)) = (2sqrt(3))/(sqrt(3)+sqrt(3))` using `C-D` rule `(x+2sqrt(2))/(x-2sqrt(2)) = (3sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2))` ……… (1) Also `(x)/(2sqrt(3)) = (2sqrt(2))/(sqrt(2)+sqrt(3))` `implies (x+2sqrt(3))/(x-2sqrt(3)) = (3sqrt(2)+sqrt(3))/(sqrt(2)-sqrt(3))/(sqrt(2)-sqrt(3))` ........ (2) `(x+2sqrt(2))/(x-2sqrt(2))+(x+2sqrt(3))/(x-2sqrt(3)) = 2` |
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| 21792. |
Find the number of different terms in the sum `(1+x)^(2009) +(1+x^(2))^(2008)+(1+x^(3))^(2007)`. |
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Answer» Number of terms in `(1+x)^(2009)=2010`….(1) + additional terms in `(1+x^(2))^(2008)=x^(2010)+x^(2012)+x…..+x^(2016)=1004` …..(2) + additional terms in `(1+x^(3))^(2007)=x^(2010)+x^(2013)+x…..+x^(4014)+….+x^(6021)=1338`…..(3) -(common to 2 and 3)`=x^(2010)+x^(2016)+....+x^(4014)=335` Hence total `=2010+1004+1338-335` `=4352-335=4017` |
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| 21793. |
If f(x) is a polynomial satiysfying `f(x) =(1)/(2) |{:(f(x),f((1)/(x)),-f(x)),(1,,f((1)/(x))):}|` and f(2) = 17, then the value of f(5) is -A. 126B. 626C. `-124`D. 626 |
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Answer» Correct Answer - 2 `f(x) =(1)/(2)[f(x) f((1)/(x)) -f((1)/(x))+f(x)]` `rArr f(x) f((1)/(2)) = f(x)+ f((1)/(2))` then `f(x) = 2x^(n)+ 1 or -x^(n) +1` `f(2) = 2^(n) +1=17 rArr 2^(n) = 16 rArr n=4` `f(x) =x^(4) + 1rArr f(5) =626` |
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| 21794. |
Find the sum of all possible values of x satisfying arc `cos((2)/(pi) arc cos x)=arc sin((2)/(pi) arc sinx)`. |
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Answer» 1 arc `cos((2)/(pi) arc cos x)=arc sin ((2)/(pi) arc sin x) ` `cos^(-1)((2)/(pi) ((pi)/(2)-sin^(-1)x))=sin^(-1)((2)/(pi) sin^(-1)x)` `cos^(-1)(1-(2)/(pi) sin^(-1)x)=sin^(-1)((2)/(pi) sin^(-1)x)` Let `(2)/(pi) sin^(-1)x=alpha` were `alpha in [0,1]` think ! `implies cos^(-1)(1-alpha)=sin^(-1) alpha` `implies sin^(-1) sqrt(2alpha-alpha^(2))=sin^(-1) alpha implies sqrt(2alpha-alpha^(2))=alphaimplies 2 alpha-alpha^(2)=alpha^(2)` `implies 2 alpha=2alpha^(2)` Hence `alpha` is either 0 or 1 If `alpha=0` then x=0 if `alpha=1` then x=1 hence sum of all possible value of x is 1 |
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| 21795. |
`underset("n-digits")((666 . . . .6)^(2))+underset("n-digits")((888 . . . .8))` is equal toA. `(4)/(9) (10^(n)-1)^(2)`B. `(4)/(9) (10^(n)-1)`C. `(4)/(9) (10^(2n)-1)`D. `(4)/(9) (10^(2n)-1)^(2)` |
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Answer» Correct Answer - C `(666............n" times")^(2)+ (888.............n" times ")` `implies (6+60+600+....n" times")^(2)+(8+80+800...n" times")` `implies{6[(10^(n)-1)/(10-1)]}^(2)+8[(10^(n)-1)/(10-1)]` ` implies(4)/(9)(10^(n)-1)^(2)+(8)/(9)(10^(n)-1)` `implies(4)/(9)(10^(n)-1){10^(n)-1+2}` ` implies (4)/(9) (10^(n)-1)(10^(n)+1)` `implies (4)/(9)(10^(2n)-1)` |
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| 21796. |
the total number of ways of selecting two number from the set {1,2,3,4,…..3n} so that their sum divisible by 3 is equal to -A. `(2n^(2)-n)/(2)`B. `(3n^(2)-n)/(2)`C. `2n^(2)-n`D. `3n^(2)-n` |
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Answer» Correct Answer - B Given number can be rearranged as ` 1,4,7,.....,3n-2 to 3lamda-2` `2,5,8,.....3n-1 to 3 lamda-1` ` 3,6,9,....,3nto 3lamda ` that means , we must take two number from last row or one number each from first and second rows , Therefore , the total number of ways is . `""^(a)C_(2)+^(a)C_(1)xx^(a)C_(1)=(n(n-1))/(2)+n^(2)` ` =(3n^(2)-n)/(2)` |
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| 21797. |
The solution the differential equation `"cos x sin y dx" + "sin x cos y dy" =0` isA. `(sinx)/(siny) = c`B. `cos x + cos y = c`C. `sin x + sin y = c`D. `sin x. sin y=c` |
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Answer» Correct Answer - 4 `"cos x sin y dx + sin x cosy dy" =0` `"sin y d(sinx) + sin x d (sin y)=0"` `"d (sin x sin y) =0"` Intergrating `sin x sin y =c` |
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| 21798. |
Total number of ways of selecting two numbers from the set `{1, 2, 3, ..., 90}` so that their sum is divisible by 3, is |
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Answer» Given number can be rerranged as ` 1" " 4" " 7…….88 to 30` `2" " 5" "8…..89 to 30` `3" "6" "9 to 30` That means we must take two numbers from last row or one number each from from first and second law. Total ways `=""^(30)C_(2)+""^(30)C_(1)*""^(30)C_(1)=435+900=1335` |
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| 21799. |
If `f(theta)=(s intheta+cos e ctheta)^2+(costheta+s e ctheta)^2,`then minimum value of `f(theta)`is`7``8``9`4. Noneof these |
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Answer» Correct Answer - 9 If `f(theta) = (sin theta+cossec theta)^(2)`……….. `f(theta) = 1+4+cossec^(2)theta+sec^(2)theta` `implies 5 + (1)/(sin^(2)theta cos^(2)theta) implies 5+ (4)/(sin^(2)2theta)` min. value `implies 9` |
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| 21800. |
If `log_(10)x +log_(10)y le 2` then smallest possible value of `x + y` is `P`. Then `(P)/(10)` is |
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Answer» Correct Answer - 2 If `log_(10)x +log_(10)y ge 2` then …………… `xy ge 100` `(x+y)^(2)ge (x-y)^(2)+4xy` `ge400` `ge 20` |
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