100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	`If a line is drawn to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.
Answer» For correct, Given, To prove, construction and Figure For correct proof

Discussion

2.	The first term of an A.P. is 5, the last term is 45 and sum is 400. Find the number of terms and the common difference.
Answer» `" "a = 5` `" "a_n = 45` `" "S_n = 400` `rArr (n)/(2) (5+ 45) = 400` `50 n = 800` `n = 16` also `a_n = 45` `5 + 15 d = 45` ` 15 d = 40` `d= 8//3`

Discussion

3.	If the sum of first 14 terms of an A.P. is 1050 and its first term is 10, find the `20^(th)` term.
Answer» Let common difference be `d` `" " rArr (14)/(2) [ 2(10) + (n-1)d] = 1050` `" "rArr d = 10` `" " a_(20) = a + 19 d` `" " = 10 + 19(10) = 200`

Discussion

4.	Find two consecutive positive integers sum of whose squares is 365.
Answer» Let two consecutive positive integers be `x and x+1` `" " therefore x^(2) + (x+1)^(2) = 365` `rArr x^(2) + x- 182=0` `" " (x + 14) (x - 13)=0` `" " therefore x = 13` Hence two consecutive positive integers are 13 and 14

Discussion

5.	Which of the following disproportionate (s) in heating with sodium hydroxidede?A. `P_4`B. `S_8`C. `Cl_2`D. `B`
Answer» Correct Answer - A,B,C (A)`P_4+3NaOH+3H_2O to PH_3+3NaH_2PO_2` (B)`4S+6NaOH overset(Delta)to Na_2S_2O_3 + 2Na_2S +3 H_2O` ( C)`3Cl_2 + 6NaOH to 5 NaCl + NaClO_3 + 3H_2O` (D)`2B+6NaOH to 2Na_3BO_3 + 3H_2`

Discussion

6.	Find the two positive integers whose sum of the squares is 365.
Answer» Let the first integer be x. The next consecutive positive integer will be x + 1. According to the given question, x² + ( x + 1)² = 365 x² +( x + 1)² = 365 x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²] 2x² + 2x + 1 = 365 2x² + 2x + 1- 365 = 0 2x² + 2x - 364 = 0 2(x² + x - 182) = 0 x² + x - 182 = 0 x² + 14x - 13x - 182 = 0 x (x + 14) - 13 (x + 14) = 0 (x - 13) (x + 14) = 0 x - 13 = 0 and x + 14 = 0 x = 13 and x = - 14 The value of x cannot be negative (because it is given that the integers are positive). ∴ x = 13 and x + 1 = 14

6.

Find the two positive integers whose sum of the squares is 365.

Answer»

Let the first integer be x.

The next consecutive positive integer will be x + 1.

According to the given question,

x² + ( x + 1)² = 365

x² +( x + 1)² = 365

x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²]

2x² + 2x + 1 = 365

2x² + 2x + 1- 365 = 0

2x² + 2x - 364 = 0

2(x² + x - 182) = 0

x² + x - 182 = 0

x² + 14x - 13x - 182 = 0

x (x + 14) - 13 (x + 14) = 0

(x - 13) (x + 14) = 0

x - 13 = 0 and x + 14 = 0

x = 13 and x = - 14

The value of x cannot be negative (because it is given that the integers are positive).

∴ x = 13 and x + 1 = 14

Discussion

7.	Use binomial theorem to evaluate up to 4 decmials place (102)6
Answer» (102)⁶ = (100 + 2)⁶ = (100)⁶ + ⁶C₁(1 00)⁵ . 2 + ⁶C₂(100)⁴ .2²+ ⁶C₃(100)³ .2³ + ⁶C₄(100)² 2⁴ + ⁶C₅.100.2⁵ + ⁶C₆2⁶ = 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64 = 1,126,162,419,264

7.

Use binomial theorem to evaluate up to 4 decmials place (102)6

Answer»

(102)⁶ = (100 + 2)⁶ = (100)⁶ + ⁶C₁(1 00)⁵ . 2 + ⁶C₂(100)⁴ .2²+ ⁶C₃(100)³ .2³ + ⁶C₄(100)² 2⁴ + ⁶C₅.100.2⁵ + ⁶C₆2⁶

= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64

= 1,126,162,419,264

Discussion

8.	Which of the following statements is/are true for the solutions of alkali metals and alkaline earth metals in ammonia (l) ?A. Concentrated solutions of alkali metals in ammonia are copper bronzed coloured and have a metallic lustureB. Dilute solutions of alkaline earth metals are deep blue-black in colour due to the spectrum from the solvated electronC. Concentrated solutions of the alkaline earth metals in ammonia are bronze colouredD. Evaporation of the ammonia from solutions of alkali metals yields the metal , but with alkaline earth metals evaporation of ammonia gives hexammoniaes of the metals.
Answer» Correct Answer - A,B,C,D (A)Due to the formation of metal ion clusters (B)`M+(x+y)NH_3toM^(+)(NH_3)_x + e^(-)(NH_3)_y` (C ) due to the formation of metal clusters. (D)`M(NH_3)_6 to` true statement

Discussion

9.	What must be added to each term in the ratio 5 : 6 so that it becomes 8 : 9?
Answer» Let x be added ∴ (5 + x)/(6 + x) = 8/9 Solving to get x = 3

9.

What must be added to each term in the ratio 5 : 6 so that it becomes 8 : 9?

Answer»

Let x be added

∴ (5 + x)/(6 + x) = 8/9

Solving to get x = 3

Discussion

10.	Select the correct statement with respect to sodium peroxide ?A. It decolourises the acidifed `KMnO_4` solutionB. On heating with oxygen at `450^@C` and 300 atm pressure, it becomes paramagneticC. It is obtained along with sodium metal, when sodium oxide is heated to a temperature more than `400^@C`D. It gives both hydrogen peroxide and oxygen gas with water as well as with sulphuric acid.
Answer» Correct Answer - A,B,C (A)`2MnO_4^(-)("pink")+16H^(+)+5O_2^(2-)to2Mn^(2+) + 8 H_2O +5O_2` (B)`Na_2O_2 + O_2 underset(300^@ "atm")overset(gt 450^@C)to2NaO_2` (`O_2^(-)` contains one unpaired electron in `pi^(**)MO`). (C )`2Na_2O overset(400^@C)toNa_2O_2 + 2Na` (D) gives only `H_2O_2`

Discussion

11.	Property of the alkali metals that increases with their atomic number is :A. Ionic molility of their ion in waterB. Solubility of their sulphatesC. Solubility of their carbonatesD. Solubility of their hydroxides
Answer» Correct Answer - A,C,D As size of cation increases, the size of hydrated ions in water decreases and thus ionic mobility increases.

Discussion

12.	Find the middle term in the expansion of (2x2/3 - 3/2x)10
Answer» Identifying middle term as T₆ T_{r + 1} = ¹⁰C_r(2x²/3)^10-r(-3/2x)^r Substituting r= 5 and simplifying to get T₆ = -252x⁵ .

12.

Find the middle term in the expansion of (2x2/3 - 3/2x)10

Answer»

Identifying middle term as T₆

T_{r + 1} = ¹⁰C_r(2x²/3)^10-r(-3/2x)^r

Substituting r= 5 and simplifying to get T₆ = -252x⁵ .

Discussion

13.	Wwhich of the following statements are true about the alkali metals?A. All alkali-metal salts impart a characteristic colour to the Bunsen flameB. Among LiOH, CsOH, KOH and RbOH, CsOH has the highest solubility in waterC. Among the alkali metals, cesium is the most reactiveD. The reducing character of the alkali metal hydrides follow the order : LiHgtNaHgtKHgtRbHgtCsH
Answer» Correct Answer - A,B,C (A)This is because the heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is emission of radiation in the visible region as given below : `{:("Metal","Li","Na","K","Rb","Cs"),("Colour","Crimson red","Yellow","Violet/Lilac","Red violet","Blue"):}` (B)Down the group the change in lattice energy is more than that of hydration energy. (C )Because of low ionization energy and melting point. (D)Reducing nature increases down the group as their stability decreases down the group CsH gt RbH gt KH gt NaH gt LiH .

Discussion

14.	Find the value of (1.2)55 using Binomial theorem upto 4 places of decimals.
Answer» (1.2)⁵5 = (1 + 0.2)⁵5 = 1⁵5 + ⁵C₁(0.2) + ⁵C₂(0.2)² + ⁵C₃(0.2)³ + ⁵C₄(0.2)⁴ + (0.2)⁵ = 1 + 5(0.2) + 10 (0.04) + 10(0.008) + 5(00004) + (0.00032) = 2.4883

14.

Find the value of (1.2)55 using Binomial theorem upto 4 places of decimals.

Answer»

(1.2)⁵5 = (1 + 0.2)⁵5

= 1⁵5 + ⁵C₁(0.2) + ⁵C₂(0.2)² + ⁵C₃(0.2)³ + ⁵C₄(0.2)⁴ + (0.2)⁵

= 1 + 5(0.2) + 10 (0.04) + 10(0.008) + 5(00004) + (0.00032)

= 2.4883

Discussion

15.	Evaluate (2 + √3)5 + (2 – 13)5
Answer» (2 + √3)⁵ + (2 – √3)⁵ = 2 [⁵C₀ 2⁵ + ⁵C₂2³ (√3)² + ⁵C₄2¹ (√3)⁴] = 2[32 + 240 + 90] = 724

15.

Evaluate (2 + √3)5 + (2 – 13)5

Answer»

(2 + √3)⁵ + (2 – √3)⁵ = 2 [⁵C₀ 2⁵ + ⁵C₂2³ (√3)² + ⁵C₄2¹ (√3)⁴]

= 2[32 + 240 + 90]

= 724

Discussion

16.	Solve: x6 - 6x5 + 10x4 -9x2 + 6x - 2 = 0 given that (2 + √3) and (1 + i) are roots.
Answer» (2 + √3) and (1 + i) are roots of equation. x⁶ - 6x⁵ + 10x⁴ -9x² + 6x - 2 = 0-------(i) Let p(x) = x⁶ - 6x⁵ + 10x⁴ - 9x² + 6x - 2 \(\because\) (2 + √3) and (1 + i) are roots of equation (i) \(\therefore\) (2 + √3) and (1 - i) are roots of equation (i) ⇒ (x - (2 + √3))(x - (2 - √3)) is a factor of p(x) ⇒ ((x - 2) - √3) ((x - 2) + √3) is a factor of p(x) ⇒ ((x - 2)² - 3) is a factor of p(x) ⇒ (x²- 4x + 1) is a factor of p(x) Now, \(\frac{P(x)}{x^2-4x + 1}=\frac{x^6-6x^5+10x^4-9x^2+6x-2}{x^2-4x+1}\) =(x⁴- 2x³ + x² + 6x + 14) (x²- 4x + 1) + \(\frac{56x-16}{x^2-4x+1}\) \(\therefore\) x² - 4x + 1 is not a divisior of p(x). \(\therefore\) (2 +√3) is not a root or given equation. \(\therefore\) (1 + i) is a root of equation (i) \(\therefore\) (x - (1 + i))(x - (1 - i)) is a factor of P(x) ⇒ ((x - 1)² - i²) is a factor of P(x) ⇒ (x²- 2x + 2) is a factor of P(x) Now, \(\frac{P(x)}{x^2-2x + 2}\) = \(\frac{x^6-6x^5+10x^4-3x^2+6x-2}{x^2-2x+2}\) = x⁴ - 4x³ + 8x + 7 + \(\frac{4x-16}{x^2-2x+2}\) Hence, x² - 2x + 2 is not a divisior of P(x). \(\therefore\) (1 + i) is not a root given equation.

16.

Solve: x6 - 6x5 + 10x4 -9x2 + 6x - 2 = 0 given that (2 + √3) and (1 + i) are roots.

Answer»

(2 + √3) and (1 + i) are roots of equation.

x⁶ - 6x⁵ + 10x⁴ -9x² + 6x - 2 = 0-------(i)

Let p(x) = x⁶ - 6x⁵ + 10x⁴ - 9x² + 6x - 2

\(\because\) (2 + √3) and (1 + i) are roots of equation (i)

\(\therefore\) (2 + √3) and (1 - i) are roots of equation (i)

⇒ (x - (2 + √3))(x - (2 - √3)) is a factor of p(x)

⇒ ((x - 2) - √3) ((x - 2) + √3) is a factor of p(x)

⇒ ((x - 2)² - 3) is a factor of p(x)

⇒ (x²- 4x + 1) is a factor of p(x)

Now,

\(\frac{P(x)}{x^2-4x + 1}=\frac{x^6-6x^5+10x^4-9x^2+6x-2}{x^2-4x+1}\)

=(x⁴- 2x³ + x² + 6x + 14) (x²- 4x + 1) + \(\frac{56x-16}{x^2-4x+1}\)

\(\therefore\) x² - 4x + 1 is not a divisior of p(x).

\(\therefore\) (2 +√3) is not a root or given equation.

\(\therefore\) (1 + i) is a root of equation (i)

\(\therefore\) (x - (1 + i))(x - (1 - i)) is a factor of P(x)

⇒ ((x - 1)² - i²) is a factor of P(x)

⇒ (x²- 2x + 2) is a factor of P(x)

Now,

\(\frac{P(x)}{x^2-2x + 2}\) = \(\frac{x^6-6x^5+10x^4-3x^2+6x-2}{x^2-2x+2}\)

= x⁴ - 4x³ + 8x + 7 + \(\frac{4x-16}{x^2-2x+2}\)

Hence, x² - 2x + 2 is not a divisior of P(x).

\(\therefore\) (1 + i) is not a root given equation.

Discussion

17.	Find the value of (1,01)5 using Bonomial upto 4 decimal places.
Answer» (1.01)⁵ = (1 + 0.01)⁵ = 1 + ⁵C₁(0.01) + ⁵C₂(0.01)² + ⁵C₃ (0.01)³ + ⁵C₄ (0.01)⁴ + ⁵C₅ (0.01)⁵ = 1 + 5(0.01) + 10(0.001) +10(0.000001) + ignoring the further terms = 1.0510[correct upto 4 decimals]

17.

Find the value of (1,01)5 using Bonomial upto 4 decimal places.

Answer»

(1.01)⁵ = (1 + 0.01)⁵

= 1 + ⁵C₁(0.01) + ⁵C₂(0.01)² + ⁵C₃ (0.01)³ + ⁵C₄ (0.01)⁴ + ⁵C₅ (0.01)⁵

= 1 + 5(0.01) + 10(0.001) +10(0.000001) + ignoring the further terms

= 1.0510[correct upto 4 decimals]

Discussion

18.	∫(xex/(1 + x)2)dx is equal to which of the following ?
Answer» (A) (e^x/1+ x) + c

Discussion

19.	Which the following is value of sin-1[sin(2π/3)](A) 2π/3(B) π/6(C) π/3(D) -3π/4
Answer» The value of π/3.

Discussion

20.	The direction ratio of a line are 2,3,7 then its direction cosines are which of the following ?(A) 1/6,1/4,7/12(B) √2/62, √3/62, √7/62(C) 2/√62, 3/√62, 7/√62(D) 2/12, 3/12, 7/12
Answer» (C) 2/√62, 3/√62, 7/√62

Discussion

21.	A binary composection * is defined on R × R by (a,b) (c,d) = (ac, bc ≠ d), where a,b, c, d ∈ R then (2,3) * (1) is equal to which of the following(A) (1,2) (B) (2,1) (C) (1,1)(D) 2,2
Answer» option: (B) (2,1)

Discussion

22.	If sin-1x + sin-1 y = π/3 then the value of cos–1x + cos–1y is equal to which of the following :(A) π/6(B) π/3(C) 2π/3(D) π
Answer» Option (C) is correct Explanation sin^-1x+sin^-1y=π/3 =>π/2-cos-^1x+π/2-cos^-1y=π/3 =>cos^-1x+cos^-1y=π-π/3 =>cos^-x+cos^-1y=(2π)/3

Discussion

23.	Which of the following will be value of x and y if (2x, x + y) = (6,2)(A) x = 3, y = –1 (B) x = 1, y = 5 (C) x = –1, y = 3 (D) x = 5, y = 1
Answer» (A) x = 3, y = –1

Discussion

24.	If tan-1x + tan-1y + tan-1 z = π/2 then the value of xy + yz + zx is equal to which of the following.(A) –1 (B) 1 (C) 0 (D) None of these
Answer» Correct option : (B) 1

24.

If tan-1x + tan-1y + tan-1 z = π/2 then the value of xy + yz + zx is equal to which of the following.(A) –1 (B) 1 (C) 0 (D) None of these

Answer»

Correct option :

(B) 1

Discussion

25.	The principal value of cosec–1(2) is(A) π/3(B) π/6(C) 2π/3(D) 5π/6
Answer» Correct option: (B) π/6

25.

The principal value of cosec–1(2) is(A) π/3(B) π/6(C) 2π/3(D) 5π/6

Answer»

Correct option:

(B) π/6

Discussion

26.	The equation of plane parallel to the plane 2x + 5y – 6z + 3 = 0 will be(A) 3x + 5y – 6z + 3 = 0 (B) 2x – 5y – 6z + 3 = 0(C) 2x + 5y – 6z + k = 0 (D) None of these
Answer» (C) 2x + 5y – 6z + k = 0

Discussion

27.	If y = tan2x , then dy/dx = ......(A) x3 · cos (x3) (B) sec2x (C) 2tanx · sec2x (D) None of these
Answer» (C) 2tanx · sec²x

Discussion

28.	∫(1 + log x)/x dx , x ∈ [1, e] = ........(A) 3/2(B) 1/2(C) e(D) 1/e
Answer» Correct option: (A) 3/2

28.

∫(1 + log x)/x dx , x ∈ [1, e] = ........(A) 3/2(B) 1/2(C) e(D) 1/e

Answer»

Correct option:

(A) 3/2

Discussion

29.	The plane x = 0 and y = 0 is(A) parallel (B) perpendicular to each other(C) Intersect in x -axis (D) None of these
Answer» (B) perpendicular to each other

Discussion

30.	The direction ratio of the line joining the points A(2, –4,5) & B(1,–1,3) are(A) (1, –3,2) (B) (–3,1,2) (C) (2,1,–3) (D) None of these
Answer» Correct option: (A) (1, –3, 2)

30.

The direction ratio of the line joining the points A(2, –4,5) & B(1,–1,3) are(A) (1, –3,2) (B) (–3,1,2) (C) (2,1,–3) (D) None of these

Answer»

Correct option:

(A) (1, –3, 2)

Discussion

31.	The line (x - 2)/3 = (y - 3)/4 = (z - 4)/5 is parallel to which of the following plane.(A) 3x + 4y + 5z = 7 (B) 2x + 3y + 4z = 0(C) x + y – z = 0 (D) 2x + y – 2z = 0
Answer» (D) 2x + y – 2z = 0

Discussion

32.	\( 28) \) If \( \sin \alpha, \cos \alpha \) are the roots of the equation \( a x^{2}+b x+c-0(c \neq 0) \), then prove that \( (n+c)^{2}-b^{2}+c^{2} \) 29) Find value of a for which the sum of the squares of the equation \( x^{2}-(a-2) x-a-1=0 \) assumes the least value.
Answer» 28} Let the roots of the equation ax2+bx+c= 0(where c≠0), be p and q So by the given condition p = sinα and q = cosα and p+q = -b/a and pq = c/a again p² +q² = (p+q)2 -2pq => sin²α + cos²α = (-b/a)2 - 2c/a => 1= (b²-2ca)/a² =>a²=(b²-2ca) =>a²+2ca +c2=b² +c2 =>(a +c)²=b² +c2 proved 29) Let the roots of the quadratic equation x²−(a−2)x−(a+1)=0 be p and q then p+q =a-2 and pq = - (a+1) So p²+q² = (p+q)2-2pq= ( a-2)²+2(a+1)=a²-4a +4+2a+2=a²-2a+6 = (a-1)²+5 It is obvious that the value of p²+q²will be least only when a-1 = 0 or a =1

32.

\( 28) \) If \( \sin \alpha, \cos \alpha \) are the roots of the equation \( a x^{2}+b x+c-0(c \neq 0) \), then prove that \( (n+c)^{2}-b^{2}+c^{2} \) 29) Find value of a for which the sum of the squares of the equation \( x^{2}-(a-2) x-a-1=0 \) assumes the least value.

Answer»

28} Let the roots of the equation ax2+bx+c= 0(where c≠0), be p and q

So by the given condition p = sinα and q = cosα

and p+q = -b/a and pq = c/a

again p² +q² = (p+q)2 -2pq

=> sin²α + cos²α = (-b/a)2 - 2c/a

=> 1= (b²-2ca)/a²

=>a²=(b²-2ca)

=>a²+2ca +c2=b² +c2

=>(a +c)²=b² +c2 proved

29) Let the roots of the quadratic equation x²−(a−2)x−(a+1)=0 be p and q

then p+q =a-2 and pq = - (a+1)

So p²+q² = (p+q)2-2pq= ( a-2)²+2(a+1)=a²-4a +4+2a+2=a²-2a+6 = (a-1)²+5

It is obvious that the value of p²+q²will be least only when a-1 = 0 or a =1

Discussion

33.	Two parallel wires P and Q placed at a separation d =6 cm on x-axis carry electric current `i_(1) = 5A and i_(2) = 2 A` in opposite directions as shown in Fig. Find the point on the line PQ where the resultant mgnetic field is zero. A. 4 cm left PB. 4 cm right of QC. middle of PQD. At no position
Answer» Correct Answer - B At the desired point, the magnetic field due to two wires must have equal magnitude but opposite directions. The point should be either to the left of P or to the right of Q. As the wire Q has smaller current, hte point should be closer to Q. Let this point R be at a distance x form Q. The magnetic field at R due to the current `i_(1)` will have magnitude. `(B_1)=(mu_(0)i_(1))/(2 pi (d+x)` and will be directed in the plane of the figure. The field at the same point due to the current `(i_2)` will be `B_(2)=(mu_(0)i_(2))/(2 pi x)` directed upward in the plane of the figure. If the resultant field at R is zero, we should have `B_(1)=B_(2)`, so that `(i_1)/(d+x_=(i_2)/(x)` Hence, `x=(i_(2)d)/(i_(1)-i_(2)) =((2A)(6 cm))/((3A))=4 cm`.

Discussion

34.	Which among the following is(are) double displacement reaction(s)?
Answer» Basically double displacement reactions are occur in ionic compound. CH₄ + 2O₂ → CO₂ + 2H₂O is a combustion reaction.

34.

Which among the following is(are) double displacement reaction(s)?

Answer»

Basically double displacement reactions are occur in ionic compound.

CH₄ + 2O₂ → CO₂ + 2H₂O is a combustion reaction.

Discussion

35.	For an ellipse having major and minor axis along `x` and `y` axes respectivley, the product of semi major and semi minor axis is `20`. Then maximum value of product of abscissa and ordinate of any point on the ellipse is greater than(A) `5`(B) `8`(C) `10`(D) `15`A. `5`B. `8`C. `10`D. `15`
Answer» Correct Answer - A::B Equation of ellipse is `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` For any point `(x_(1),y_(1))` on the curve `((x^(2))/(a^(2))+(y^(2))/(b^(2)))/2gesqrt((x_(1)^(2)y_(1)^(2))/(a^(2)b^(2)))implies1/2ge(x_(1)y_(1))/(ab)impliesx_(1)y_(1)le(ab)/2` `impliesx_(1)y_(1)le10` Which is greater than `5` and `8`

Discussion

36.	For an ellipse having major and minor axis along `x` and `y` axes respectivley, the product of semi major and semi minor axis is `20`. Then maximum value of product of abscissa and ordinate of any point on the ellipse is greater thanA. `5`B. `8`C. `10`D. `15`
Answer» Correct Answer - A::B Equation of ellipse is `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` For any point `(x_(1),y_(1))` on the curve `((x^(2))/(a^(2))+(y^(2))/(b^(2)))/2gesqrt((x_(1)^(2)y_(1)^(2))/(a^(2)b^(2)))implies1/2ge(x_(1)y_(1))/(ab)impliesx_(1)y_(1)le(ab)/2` `impliesx_(1)y_(1)le10` Which is greater than `5` and `8`

Discussion

37.	If `f(x)=root (3)(8x^(3)+mx^(2))-nx` such that `lim_(xrarroo)f(x)=1` then(A) `m+n=15`(B) `m-n=10`(C) `m-n=12`(D) `m+n=14`A. `m+n=15`B. `m-n=10`C. `m-n=12`D. `m+n=14`
Answer» Correct Answer - B::D `lim_(xrarroo)root(3)(8x^(3)+mx^(2))-nx` `=lim_(xrarroo)((8-n^(3))x^(3)+mx^(2))/(root(3)((8x^(3)+mx^(2))^(2))+nx^(3)sqrt(8x^(3)+mx^(2))+n^(2)x^(2))` for limit to be finite `8-n^(3)=0impliesn=2` Hence `lim_(xrarroo)f(x)=lim_(xrarroo)m/(root(3)((8+m/x)^(2))+2root(3)((8+m/x))+4)` `=m/12=1impliesm=12`

Discussion

38.	If `f(x)=root (3)(8x^(3)+mx^(2))-nx` such that `lim_(xrarroo)f(x)=1` thenA. `m+n=15`B. `m-n=10`C. `m-n=12`D. `m+n=14`
Answer» Correct Answer - B::D `lim_(xrarroo)root(3)(8x^(3)+mx^(2))-nx` `=lim_(xrarroo)((8-n^(3))x^(3)+mx^(2))/(root(3)((8x^(3)+mx^(2))^(2))+nx^(3)sqrt(8x^(3)+mx^(2))+n^(2)x^(2))` for limit to be finite `8-n^(3)=0impliesn=2` Hence `lim_(xrarroo)f(x)=lim_(xrarroo)m/(root(3)((8+m/x)^(2))+2root(3)((8+m/x))+4)` `=m/12=1impliesm=12`

Discussion

39.	The co-ordinates of the points on the curve `y=x^(2)+3x+4` at which the tangent passes through the origin are(A) `(-2,14)`(B) `(2,14)`(C) `(2,-2)`(D) `(-2,2)`A. `(-2,14)`B. `(2,14)`C. `(2,-2)`D. `(-2,2)`
Answer» Correct Answer - B::D `(dy)/(dx)` at `x_(1),y_(1)` must equal to `(y_(1)-0)/(x_(1)-0)` Therefore `2x_(1)+3=(x_(1)^(2)+3x_(1)+4-0)/(x_(1)-0)` Hence `x_(1)^(2)=4impliesx_(1)=+-2impliesy_(1)=14,2` `(2,14),(-2,2)`s

Discussion

40.	For the curve `y=4x^3-2x^5,`find all the points at which the tangent passes through the origin.A. `(-2,14)`B. `(2,14)`C. `(2,-2)`D. `(-2,2)`
Answer» Correct Answer - B::D `(dy)/(dx)` at `x_(1),y_(1)` must equal to `(y_(1)-0)/(x_(1)-0)` Therefore `2x_(1)+3=(x_(1)^(2)+3x_(1)+4-0)/(x_(1)-0)` Hence `x_(1)^(2)=4impliesx_(1)=+-2impliesy_(1)=14,2` `(2,14),(-2,2)`s

Discussion

41.	Minimum value of `(sin^(-1)x)^(2)+(cos^(-1)x)^(2)` is greater thanA. `(pi^(2))/4`B. `(pi^(2))/16`C. `(3pi^(2))/4`D. `(3pi^(2))/32`
Answer» Correct Answer - B::D `(sin^(-1)x)^(2)+((pi)/2-sin^(-1)x)^(2)=t^(2)+((pi)/2-t)^(2)` where `sin^(-1)x=t` `=2t^(2)-pit+(pi^(2))/4` is minimum when `t=(pi)/4` Minimum value `=(pi^(2))/8`

Discussion

42.	If `y+a=m_(1)(x+3a),y+a=m_(2)(x+3a)` are two tangents to the parabola `y^(2)=4ax`, thenA. `m_(1)+m_(2)+m_(1)m_(2)=0`B. `m_(1)+m_(2)-m_(1)m_(2)=2/3`C. `m_(1)+m_(2)+2m_(1)m_(2)=1/3`D. `m_(1)+m_(2)-2m_(1)m_(2)=1`
Answer» Correct Answer - A::B::D Let equation of tangent is `y=mx+a/m` `implies-a=-3am+a/m` `implies3m^(2)-m-1=0` `m_(1)+m_(2)=1/3,m_(1),m_(2)=-1/3`

Discussion

43.	Give an application of cyclotron.
Answer» (i) It is used to accelerate positively charged particles to very high energies. (ii) It is used to produce radioactive material for medical purpose ex; for the purpose ex; for the purpose of diagnostics and treatment of chronic disease. (iii) It is used to synthesize fresh materials. (iv) It is used to improve the quality of solids by adding ions. (v) It is used to bombard the atomic nuclei with highly accelerated particles to study the nuclear reactions.

43.

Give an application of cyclotron.

Answer»

(i) It is used to accelerate positively charged particles to very high energies.

(ii) It is used to produce radioactive material for medical purpose ex; for the purpose ex; for the purpose of diagnostics and treatment of chronic disease.

(iii) It is used to synthesize fresh materials.

(iv) It is used to improve the quality of solids by adding ions.

(v) It is used to bombard the atomic nuclei with highly accelerated particles to study the nuclear reactions.

Discussion

44.	If `f(x)=lim_(m->oo) lim_(n->oo)cos^(2m) n!pix` then the range of f(x) isA. `f(sqrt(3))=1`B. `f(3)=1`C. `f(sqrt(2))=0`D. `f(1)=2`
Answer» Correct Answer - B::C When `x=` rational then `cosn!pix=+-1` so `f(x)=1` When `x=` irrational then `cosn!pixepsilon(-1,0)uu(0,1)` `impliesf(x)=0`

Discussion

45.	Define ‘drift velocity’ of free electrons.
Answer» It is defined as the average velocity with which free electrons get drifted towards the positive end of a conductor (opposite to the electric field) under the influence of an external electric field.

Discussion

46.	The radius of the circle passing through the point (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is ____. (a) 10 (b) 2√5(c) 6 (d) 4
Answer» The correct answer is : (b) 2√5

Discussion

47.	What is an equipotential surface?
Answer» An equipotential surface is a surface with a constant value of potential at all points on the surface.

Discussion

48.	In the following nuclear reaction identify the particle X.n → P + \(\bar e\) + X
Answer» The given nuclear reaction: ¹₀n → ¹₁p + ⁰_-1e + ^A_ZX Conservation of mass number, 1 = 1 + 0 + A ⇒ A = 0 Conservation of atomic number, 0 = 1 + (-1) + Z ⇒ Z = 0 Thus the particle is⁰₀X which is a gamma photon(γ)

48.

In the following nuclear reaction identify the particle X.n → P + \(\bar e\) + X

Answer»

The given nuclear reaction:

¹₀n → ¹₁p + ⁰_-1e + ^A_ZX

Conservation of mass number,

1 = 1 + 0 + A

⇒ A = 0

Conservation of atomic number,

0 = 1 + (-1) + Z

⇒ Z = 0

Thus the particle is⁰₀X which is a gamma photon(γ)

Discussion

49.	tan-1 (1/4) + tan-1( 2/9) is equal to ________.(a) 1/2 cos-1(3/5)(b) 1/2 sin-1(3/5)(c) 1/2 tan-1(3/5)(d) tan-1 (1/2)
Answer» (d) tan^-1(1/2)

Discussion

50.	The equation of the latus rectum of y2 = 4x is _______. (a) x = 1 (b) y = 1 (c) x = 4 (d) y = -1
Answer» The correct answer is : (a) x = 1

Discussion

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If `f(x)=lim_(m->oo) lim_(n->oo)cos^(2m) n!pix` then the range of f(x) isA. `f(sqrt(3))=1`B. `f(3)=1`C. `f(sqrt(2))=0`D. `f(1)=2`

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In the following nuclear reaction identify the particle X.n → P + \(\bar e\) + X

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