100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Let A = \(\begin{pmatrix} 1 & 0 & 1 &0\\ 0 & 1 & 0 &1\\ 1 & 0 & 0 &1 \end{pmatrix}\),B = \(\begin{pmatrix} 0 & 1 & 0 &1\\ 1 & 0 & 1 &0\\ 1 & 0 & 0 &1 \end{pmatrix}\), C = \(\begin{pmatrix} 1 & 1 & 0 &1\\ 0 & 1& 1 &0\\ 1 & 1 & 1 &1 \end{pmatrix}\) be any three boolean matrices of the same type . Find A ^ B
Answer» A ^ B = \(\begin{pmatrix} 1 & 0 & 1 &0\\ 0 & 1 & 0 &1\\ 1 & 0 & 0 &1 \end{pmatrix}\)^ \(\begin{pmatrix} 0 & 1 & 0 &1\\ 1 & 0 & 1 &0\\ 1 & 0 & 0 &1 \end{pmatrix}\) = \(\begin{pmatrix} 0 & 0 & 0 &0\\ 0 &0 & 0 &0\\ 0 & 0 & 0 &0 \end{pmatrix}\)

1.

Let A = \(\begin{pmatrix} 1 & 0 & 1 &0\\ 0 & 1 & 0 &1\\ 1 & 0 & 0 &1 \end{pmatrix}\),B = \(\begin{pmatrix} 0 & 1 & 0 &1\\ 1 & 0 & 1 &0\\ 1 & 0 & 0 &1 \end{pmatrix}\), C = \(\begin{pmatrix} 1 & 1 & 0 &1\\ 0 & 1& 1 &0\\ 1 & 1 & 1 &1 \end{pmatrix}\) be any three boolean matrices of the same type . Find A ^ B

Answer»

A ^ B = \(\begin{pmatrix} 1 & 0 & 1 &0\\ 0 & 1 & 0 &1\\ 1 & 0 & 0 &1 \end{pmatrix}\)^ \(\begin{pmatrix} 0 & 1 & 0 &1\\ 1 & 0 & 1 &0\\ 1 & 0 & 0 &1 \end{pmatrix}\)

= \(\begin{pmatrix} 0 & 0 & 0 &0\\ 0 &0 & 0 &0\\ 0 & 0 & 0 &0 \end{pmatrix}\)

Discussion

2.	A random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is ______. (a) 6 (b) 4 (c) 3 (d) 2
Answer» The correct answer is : (d) 2

Discussion

3.	If X is a binomial random variable with expected value 6 and variance 2.4, Then P {X = 5} is _______.(a) \(\begin{pmatrix} 10\\5\end{pmatrix}\)\(\left(\frac{3}{5}\right)^6\)\(\left(\frac{2}{5}\right)^4\)(b) \(\begin{pmatrix} 10\\5\end{pmatrix}\)\(\left(\frac{3}{5}\right)^5\)(c) \(\begin{pmatrix} 10\\5\end{pmatrix}\)\(\left(\frac{3}{5}\right)^4\)\(\left(\frac{2}{5}\right)^6\)(d) \(\begin{pmatrix} 10\\5\end{pmatrix}\)\(\left(\frac{3}{5}\right)^5\)\(\left(\frac{2}{5}\right)^5\)
Answer» (d) \(\begin{pmatrix} 10\\5\end{pmatrix}\)\(\left(\frac{3}{5}\right)^5\)\(\left(\frac{2}{5}\right)^5\)

Discussion

4.	The operation * defined by a*b = ab/7 is not a binary operation on ______.(a) Q+ (b) Z (c) R (c) C
Answer» The correct answer is : (b) Z

Discussion

5.	The random variable X has the probability density function f(x) = \(\begin{cases} ax\,+\,b, & 0<x<1\\ 0,&otherwise \end{cases}\)and E(X) = 7/12 , then a and b are respectively _______.(a) 1 and 1/2(b) 1/2 and 1(c) 2 and 1(d) 1 and 2
Answer» The correct answer is : (a) 1 and 1/2

Discussion

6.	A binary operation on a set S is a function from ________. (a) S → S (b)(S x S) → S (c) S → (S x S) (d) (S x S) → (S x S)
Answer» (b)( S x S) \(\rightarrow\)S

Discussion

7.	Is cos-1 (-x) = π – cos-1 x true? Justify your answer.
Answer» Let θ = cos (-x) ⇒ cos θ = -x ⇒ -cosθ = x i.e. cos(π – θ) = x ⇒ π – θ = cos^-1 x ⇒ π – cos x = θ i.e. π – cos^-1 x = cos^-1 (-x)

7.

Is cos-1 (-x) = π – cos-1 x true? Justify your answer.

Answer»

Let θ = cos (-x)

⇒ cos θ = -x ⇒ -cosθ = x

i.e. cos(π – θ) = x

⇒ π – θ = cos^-1 x

⇒ π – cos x = θ

i.e. π – cos^-1 x

= cos^-1 (-x)

Discussion

8.	If `b^(2) -ac lt 0 and a gt 0` then the value of the determinant isA. positiveB. negativeC. zeroD. `b^(2)+ae`
Answer» Correct Answer - 2 We have `\|{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}\|` `\|{:(" "a," "b," "0),(" "b," "c," "0),(ax+by,bx+cy,-(ax^(2)+2bxy+cy^(2))):}\|` [Applying `C_(3)toC_(3)-xC_(1)-yC_(2)`] `=-(ax^(2)+2bx+cy^(2))(ac-b^(2))` `(1)/(a)(b^(2)-ac)[(ax+by)^(2)+Y^(2)(ac-b^(2))]lt0` `[therefore b^(2)-aclt 0 and agt 0]`

Discussion

9.	If f(x) is continuous and increasing functin such that the domain of `g(x) = sqrt(f(x)-x)` be R and `h(x) =(1)/(1-x)` then the domain of `phi(x)=sqrt(f(f(f(x)))-h(h(h(x))))`is -A. RB. {0,1}C. R-{0,1}D. `R^(+)-{1}`
Answer» Correct Answer - 3 `h(x) = (1)/(1-x), x ne 1 rArr h(h(x)) = (x-1)/(x),x ne 0,1` `"also",g(x)ge f(x)0 x epsilion R` `rArr f(f(f(x)))ge1(x)gex` `rArr f(f(f(x)))-xge0` `therefore phi (x) "is defined for all" x epsilon R- {0,1}.`

Discussion

10.	Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.
Answer» Given roots is (3 + 2i), the other root is (3 – 2i); Since imaginary roots occur in with real co-efficient occurring conjugate pairs. x² – x(S.O.R) + P.O.R = 0 ⇒ x² – x(6) + (9 + 4) = 0 x² – 6x + 13 = 0

10.

Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.

Answer»

Given roots is (3 + 2i), the other root is (3 – 2i); Since imaginary roots occur in with real co-efficient occurring conjugate pairs.

x² – x(S.O.R) + P.O.R = 0

⇒ x² – x(6) + (9 + 4) = 0

x² – 6x + 13 = 0

Discussion

11.	If `f(x)={alpha+sin[x]/x , x > 0 and 2 ,x=0 and beta+[(sin x-x)/x^3] ,x < 0` (whlenotes the greatest integer function) if `f(x)` is continuous at `x = 0`. then `beta` is equal toA. `alpha-1`B. `alpha+1`C. `alpha+2`D. `alpha-2`
Answer» Correct Answer - 2 `f(0)=2` `f(0+)=underset(4rarr0)lim alpha + (sin[h])/(h) = alpha` `f(0^(-))=beta+underset(hto0)(lim)[((h-(h^(3))/(31)....)-h)/(h^(3))]=beta-1` `alpha=2, beta-1 = 2 rArr beta =3`

Discussion

12.	The polynomial `f(x)=x^4+a x^3+b x^3+c x+d`has real coefficients and `f(2i)=f(2+i)=0.`Find the value of `(a+b+c+d)dot`A. 1B. 4C. 9D. 10
Answer» Correct Answer - 3 If a polynomial has real coefficients then roots occur in complex conjugate and roots are 2i -2i, 2 + I,2-I Hence ,f(x)=(x+2i)(x-2e)(x-2-i)(x-2+I)` `f(1)=(1+xi)(1-2i)(1-2-i)(1-2+i)` `f(1)=5xx2=10` Also, f(1)= 1+a+b+c+d` `therefore 1+a+b+c+d=10` `rArr a+b+c+d=9`

Discussion

13.	Find the greatest value of the term independent of `x`in the expansion of `(xsinalpha+(cosalpha)/x)^(10)`, where `alpha in Rdot`A. `2^(5)`B. `(10!)/(5!)^(2)`C. `(1)/(2^(5))(10!)/(5!)^(2)`D. None of these
Answer» Correct Answer - 3 `Cr(x sin alpha)^(10-r),x^(-1 cos alpha^(r))` `Cr(sin alpha)^(10-r) (cos alpha)^(r ).(x)^(10-2r)` "for independent of x" `10-2r=0 rArr r= 5 ` `rArr T_(5+1)=C_(5)(sin alpha )^(5).(cos alpha)^(5) = C_(5)(sin 2 alpha)^(5)/(2)^(5)` `rArr "greatest value" = (C_(5))/(2)^(5)`

Discussion

14.	In ∆ABC and ∆DEF, ∠F = ∠C, ∠B = ∠E and AB = \(\frac12\)DE. Then the two triangle are(a) Congruent, but not similar.(b) Similar, but not congruent(c) Neither congruent nor similar(d) Congruent as well as similar
Answer» Correct option is: (b) Similar, but not congruent.

Discussion

15.	Show that the relation R on the set N x N defined by(a, b) R (c, d) if a2 + d2 = b2 + c2 ∀ a, b, c, d ∈ N, is an equivalence relation.
Answer» We have relation defined on the set N x N defined by (a, b) R(c, d) if a² + d² = b² + c² ∀ a, b, c, d ∈ N. Reflexivity : Since, a² + b² = a² + b² , where a, b ∈ N. ⇒ a² + b² = b² + a² ( \(\because\) Addition on natural numbers is commutative.) ⇒ (a, b) R (a, b) ∀ a, b ∈ N. Hence, relation R is reflexive relation. Symmetricity : Let a, b, c, d ∈ N and (a, b) R (c, d) ⇒ a² + d² = b² + c² ⇒ b² + c² = a² + d² ⇒ c² + b² = d² + d² ( \(\because\) Addition on natural numbers is commutative. ) ⇒ (c, d) R (a, b) ∀ a, b, c, d ∈ N. Hence, relation R is symmetric relation. Transitivity : Let a, b, c, d, e, f ∈ N, (a, b) R (c, d) and (c, d) R(e, f). Since, (a, b) R (c, d) and (c, d) R (e, f) ⇒ a² + d² = b² + c² and c² + f² = d² + e² ⇒ a² + d² + c² + f² = b² + c² + d² + e² (Adding two equations ) ⇒ a² + f² = b² + e² (Cancellation law of addition on N) ⇒ (a, b) R (e, f) ∀ a, b, c, d, e, f ∈ N. Hence, relation R is transitive relation. Since relation R is reflexive, symmetric and transitive relation. Therefore, relation R is an equivalence relation.

15.

Show that the relation R on the set N x N defined by(a, b) R (c, d) if a2 + d2 = b2 + c2 ∀ a, b, c, d ∈ N, is an equivalence relation.

Answer»

We have relation defined on the set N x N defined by

(a, b) R(c, d) if a² + d² = b² + c² ∀ a, b, c, d ∈ N.

Reflexivity : Since, a² + b² = a² + b² , where a, b ∈ N.

⇒ a² + b² = b² + a² ( \(\because\) Addition on natural numbers is commutative.)

⇒ (a, b) R (a, b) ∀ a, b ∈ N.

Hence, relation R is reflexive relation.

Symmetricity : Let a, b, c, d ∈ N and (a, b) R (c, d)

⇒ a² + d² = b² + c²

⇒ b² + c² = a² + d²

⇒ c² + b² = d² + d² ( \(\because\) Addition on natural numbers is commutative. )

⇒ (c, d) R (a, b) ∀ a, b, c, d ∈ N.

Hence, relation R is symmetric relation.

Transitivity : Let a, b, c, d, e, f ∈ N, (a, b) R (c, d) and (c, d) R(e, f).

Since, (a, b) R (c, d) and (c, d) R (e, f)

⇒ a² + d² = b² + c² and c² + f² = d² + e²

⇒ a² + d² + c² + f² = b² + c² + d² + e² (Adding two equations )

⇒ a² + f² = b² + e² (Cancellation law of addition on N)

⇒ (a, b) R (e, f) ∀ a, b, c, d, e, f ∈ N.

Hence, relation R is transitive relation.

Since relation R is reflexive, symmetric and transitive relation.

Therefore, relation R is an equivalence relation.

Discussion

16.	Write the value of (1–sin2θ)sec2θ
Answer» (1– sin²θ) sec²θ = cos²θ. sec²θ (∵ sin²θ + cos²θ = 1) =\(\cfrac{1}{sec^2 \theta}\). sec²θ = 1. (\(\because\) cosθ = \(\cfrac{1}{sec \theta}\)) Hence, the value of (1– sin²θ) sec²θ is 1.

16.

Write the value of (1–sin2θ)sec2θ

Answer»

(1– sin²θ) sec²θ = cos²θ. sec²θ (∵ sin²θ + cos²θ = 1)

=\(\cfrac{1}{sec^2 \theta}\). sec²θ = 1. (\(\because\) cosθ = \(\cfrac{1}{sec \theta}\))

Hence, the value of (1– sin²θ) sec²θ is 1.

Discussion

17.	Which term of the AP 5, 9, 13, 17, …….. is 81?
Answer» Given AP is 5, 9, 13, 17, ........ Hence, the first term of AP is = 5. And common difference of AP is d = a₂ − a₁ = 9 − 5 = 4. Let n^th term of given AP is 81. i.e., a_n= 81 ⇒ a + ( n − 1)d = 81. [∵a_n = a + ( n − 1)d] ⇒ 5 + (n – 1)4 = 81 ⇒ 4(n –1) = 81 – 5 = 76 ⇒ n – 1 = \(\frac{76}{4}\) = 19 ⇒ n = 19 +1 = 20. Hence, 81 is 20^th term of AP.

17.

Which term of the AP 5, 9, 13, 17, …….. is 81?

Answer»

Given AP is 5, 9, 13, 17, ........

Hence, the first term of AP is = 5.

And common difference of AP is d = a₂ − a₁ = 9 − 5 = 4.

Let n^th term of given AP is 81. i.e., a_n= 81

⇒ a + ( n − 1)d = 81. [∵a_n = a + ( n − 1)d]

⇒ 5 + (n – 1)4 = 81

⇒ 4(n –1) = 81 – 5 = 76

⇒ n – 1 = \(\frac{76}{4}\) = 19

⇒ n = 19 +1 = 20.

Hence, 81 is 20^th term of AP.

Discussion

18.	A book store shopkeeper gives books on rent for reading. He has variety of books in his store related to fiction, stories and quizzes etc. He takes a fixed charge for the first two days and an additional charge for subsequent day. Amruta paid Rs 22 for a book and kept for 6 days; while Radhika paid Rs 16 for keeping the book for 4 days.Assume that the fixed charge be Rs x and additional charge (per day) be Rs y.1. The situation of amount paid by Radhika, is algebraically represented by(a) x - 4y = 16(b) x + 4y = 16(c) x - 2y = 16(d) x + 2y = 162. The situation of amount paid by Amruta, is algebraically represented by(a) x - 2y = 11(b) x - 2y = 22(c) x + 4y = 22(d) x - 4y = 113. What are the fixed charges for a book ?(a) Rs 9(b) Rs 10(c) Rs 13(d) Rs 154. What are the additional charges for each subsequent day for a book ?(a) Rs 6(b) Rs 5(c) Rs 4(d) Rs 35. What is the total amount paid by both, if both of them have kept the book for 2 more days ?(a) Rs 35(b) Rs 52(c) Rs 50(d) Rs 58
Answer» Correct answer is 1. (d) x + 2y = 16 2. (c) x + 4y = 22 3. (b) Rs 10 4. (d) Rs 3 5. (c) Rs 50

18.

A book store shopkeeper gives books on rent for reading. He has variety of books in his store related to fiction, stories and quizzes etc. He takes a fixed charge for the first two days and an additional charge for subsequent day. Amruta paid Rs 22 for a book and kept for 6 days; while Radhika paid Rs 16 for keeping the book for 4 days.Assume that the fixed charge be Rs x and additional charge (per day) be Rs y.1. The situation of amount paid by Radhika, is algebraically represented by(a) x - 4y = 16(b) x + 4y = 16(c) x - 2y = 16(d) x + 2y = 162. The situation of amount paid by Amruta, is algebraically represented by(a) x - 2y = 11(b) x - 2y = 22(c) x + 4y = 22(d) x - 4y = 113. What are the fixed charges for a book ?(a) Rs 9(b) Rs 10(c) Rs 13(d) Rs 154. What are the additional charges for each subsequent day for a book ?(a) Rs 6(b) Rs 5(c) Rs 4(d) Rs 35. What is the total amount paid by both, if both of them have kept the book for 2 more days ?(a) Rs 35(b) Rs 52(c) Rs 50(d) Rs 58

Answer»

Correct answer is

1. (d) x + 2y = 16

2. (c) x + 4y = 22

3. (b) Rs 10

4. (d) Rs 3

5. (c) Rs 50

Discussion

19.	If tan θ = 2/3, then the value of sec θ is(a) \(\frac{\sqrt{13}}{3}\)(b) \(\frac{\sqrt{5}}{3}\)(c) \(\sqrt{\frac{13}{3}}\)(d) \(\frac{3}{\sqrt{13}}\)
Answer» Correct answer is (a) \(\frac{\sqrt{13}}{3}\)

Discussion

20.	The base BC of an equilateral ΔABC lies on the y-axis. The co-ordinates of C are (0, -3). If the origin is the mid-point of the base BC, what are the co-ordinates of A and B ?(a) A(√3, 0), B(0, 3)(b) A\((\pm 3\sqrt{3}, 0)\), B(3, 0)(c) A\((\pm 3\sqrt{3}, 0)\), B(0, 3)(d) A(-√3, 0), B(3, 0)
Answer» Correct answer is (c) A\((\pm 3\sqrt{3}, 0)\), B(0, 3)

Discussion

21.	If sec θ + tan θ = p, then tan θ is(a) \(\frac{p^2 + 1}{2p}\)(b) \(\frac{p^2 - 1}{2p}\)(c) \(\frac{p^2 - 1}{p^2 + 1}\)(d) \(\frac{p^2 + 1}{p^2 - 1}\)
Answer» Correct answer is (b) \(\frac{p^2 - 1}{2p}\)

Discussion

22.	In a ΔABC, ∠A = x°, ∠B = (3x - 2)°, ∠C = y°. Also ∠C - ∠B = 9°. The sum of the greatest and the smallest angles of this triangle is(a) 107°(b) 135°(c) 155°(d) 145°
Answer» Correct answer is (a) 107°

Discussion

23.	(1 + sin θ/ 1 - sin θ) + (1 - sin θ/1 + sin θ)\(\frac{1+sin\theta}{1-sin\theta}+\frac{1-sin\theta}{1+sin\theta}\)
Answer» \(\frac{1+sin\theta}{1-sin\theta}+\frac{1-sin\theta}{1+sin\theta}\) = \(\frac{(1+sin\theta)^2+(1-sin\theta)^2}{1-sin^2\theta}\) = \(\frac{1+sin^2\theta+2sin\theta + 1+sin^2\theta-2sin\theta}{cos^2\theta}\) \(=\frac{2(1+sin^2\theta)}{cos^2\theta}\) = 2(sec² θ + tan²θ) = 2(1 + 2 tan²θ) = 2 + 4 tan²θ

23.

(1 + sin θ/ 1 - sin θ) + (1 - sin θ/1 + sin θ)\(\frac{1+sin\theta}{1-sin\theta}+\frac{1-sin\theta}{1+sin\theta}\)

Answer»

\(\frac{1+sin\theta}{1-sin\theta}+\frac{1-sin\theta}{1+sin\theta}\)

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

= \(\frac{1+sin^2\theta+2sin\theta + 1+sin^2\theta-2sin\theta}{cos^2\theta}\)

\(=\frac{2(1+sin^2\theta)}{cos^2\theta}\)

= 2(sec² θ + tan²θ)

= 2(1 + 2 tan²θ)

= 2 + 4 tan²θ

Discussion

24.	Find the derivative of y = sec4x - tan4x.
Answer» y = sec⁴x - tan⁴x \(\therefore\) \(\frac{dy}{dx}\) = 4sec³x \(\frac{d}{dx}\)sec x - 4 tan³x \(\frac d{dx}tanx\) = 4 sec⁴x tan x - 4 tan³x sec²x = 4 tan x sec²x (sec²x - tan²x) = 4 tan x sec²x

24.

Find the derivative of y = sec4x - tan4x.

Answer»

y = sec⁴x - tan⁴x

\(\therefore\) \(\frac{dy}{dx}\) = 4sec³x \(\frac{d}{dx}\)sec x - 4 tan³x \(\frac d{dx}tanx\)

= 4 sec⁴x tan x - 4 tan³x sec²x

= 4 tan x sec²x (sec²x - tan²x)

= 4 tan x sec²x

Discussion

25.	Find domain & range of y = sec-1 x + cosec-1 x + tan-1 x.
Answer» y = sec^-1x + cosec^-1x + tan^-1x Domain of sec^-1x is R - (2n + 1)\(\frac{\pi}2\); n \(\in\) Z Domain of cosec^-1x is R - nπ, n \(\in\) Z Domain of tan^-1x is R \(\therefore\)Domain of y is R - {(2n + 1) \(\frac{\pi}2\), nπ}, n \(\in\) Z \(\because\) sec^-1x + cosec^-1x = \(\frac{\pi}2\) \(\therefore\) y = \(\frac{\pi}2\) + tan^-1x Principal Range of tan^-1x is (\(-\frac{\pi}2,\frac{\pi}2\)) \(\therefore\) Range of y is \(\frac{\pi}2+(-\frac{\pi}2,\frac{\pi}2)\) or (0, π).

25.

Find domain & range of y = sec-1 x + cosec-1 x + tan-1 x.

Answer»

y = sec^-1x + cosec^-1x + tan^-1x

Domain of sec^-1x is R - (2n + 1)\(\frac{\pi}2\); n \(\in\) Z

Domain of cosec^-1x is R - nπ, n \(\in\) Z

Domain of tan^-1x is R

\(\therefore\)Domain of y is R - {(2n + 1) \(\frac{\pi}2\), nπ}, n \(\in\) Z

\(\because\) sec^-1x + cosec^-1x = \(\frac{\pi}2\)

\(\therefore\) y = \(\frac{\pi}2\) + tan^-1x

Principal Range of tan^-1x is (\(-\frac{\pi}2,\frac{\pi}2\))

\(\therefore\) Range of y is \(\frac{\pi}2+(-\frac{\pi}2,\frac{\pi}2)\) or (0, π).

Discussion

26.	Prove that:Sin-1 (1/√17) + cos-1 (9/√85) = tan-1 (1/2)
Answer» sin^-1(\(\frac1{(\sqrt{17})}\)) + cos^-1\((\frac9{\sqrt{85}})\) = tan^-1(\(\frac1{\sqrt{17-1}}\)) + tan^-1(\(\frac{\sqrt{85-91}}9\)) = tan^-1(\(\frac14\)) + tan^-1(\(\frac29\)) = tan^-1\(\left(\cfrac{\frac14+\frac29}{1-\frac14\times\frac29}\right)\) = tan^-1\((\frac{9+8}{36-2})\) = tan^-1\((\frac{17}{34})\) = tan^-1(\(\frac12\))

26.

Prove that:Sin-1 (1/√17) + cos-1 (9/√85) = tan-1 (1/2)

Answer»

sin^-1(\(\frac1{(\sqrt{17})}\)) + cos^-1\((\frac9{\sqrt{85}})\)

= tan^-1(\(\frac1{\sqrt{17-1}}\)) + tan^-1(\(\frac{\sqrt{85-91}}9\))

= tan^-1(\(\frac14\)) + tan^-1(\(\frac29\))

= tan^-1\(\left(\cfrac{\frac14+\frac29}{1-\frac14\times\frac29}\right)\)

= tan^-1\((\frac{9+8}{36-2})\)

= tan^-1\((\frac{17}{34})\) = tan^-1(\(\frac12\))

Discussion

27.	A particle is performing U.C.M. along the cirqumference of a circle of diameter 50 cm with frequency 2 Hz. The acceleration of the particle in `m//s^(2)` isA. `2 pi^(2)`B. ` 8 pi^(2)`C. `pi^(2)`D. `4pi^(2)`
Answer» Correct Answer - D (d) Given, diameter of circle ,d=50cm ` " " =50xx10^(-2)m` and frequency ,f= 2Hz The acceleration of particle in a uniform circular motion can be given as ` " "a = omega ^(2)x ` where, `omega =` angular frequency = ` 2pi f ` ` " " x =` distance from centre ` =(d)/(2)` `rArr " "a= 4pi ^(2)f^(2) xx(d)/(2) " "...(i)` Substituting given values in Eq. (i) we get ` " "a= 4pi ^(2)xx 4xx(50 xx10^(-2) )/( 2) =4pi ^(2) `

Discussion

28.	Find the wrong statement from the following about the equation of stationary wave given by `Y = 0.04 cos(pix) sin(50 pit)` m where tis in second. Then for the stationary wave.A. Time period= 0.02 sB. Wavelength= 2 mC. Velocity = 50 m/sD. Amplitude= 0.02 m
Answer» Correct Answer - A The given equation of wave is ` " " y= 0.04cos (pix ) sin ( 50 pit ) ` `=0.02sin (50pit + pix ) +0.02sin ( 50pi t -nx )` ` " "[because 2sin A cos B =sin (A+B)_ cos (A-B) ]` Thus the given wave is the combination of two waves, ` y_1 =0.02sin ( 50pi t + pix) " " ` (in -ve x - direction ) and ` y _2 =0.02sin (50 pit -npi ) " " ` (in + ve x- direction ) Comparing them the general equation of wave a ` sin (omega t +kx) ` ,we get Amplitude ` " "a= 0.02m ` time period ` " "T = (2pi)/( 50 pi ) =(1)/(25) =0.04s ` Wavelength `lambda =(2pi)/( pi) =2m ` velocity , `v =( 50 pixxlambda )/(2pi) ` ` " "= ( 100 )/(2) =ms^(-1) ` So, option (a) is wrong

Discussion

29.	An ammeter has resistance `R_(0)` and range l what resistance should be connected in parallel with it to increase its range by nl ?A. a series resistance of `(G)/(n +1) Omega`B. a shut of `(G)/(n +1) Omega`C. a shut of `(G)/(n +1) Omega`D. a series resistance of `(G)/(n -1) Omega`
Answer» Correct Answer - B The range of an ammeter can be increased by connecting a shunt or parallel resistance to it, whose value is given by `R = ((l_(g))/(l-l_(g))) G Omega` To increase the range from l to nl, the current passing through it is l` = nl_(g)`, Using this value in Eq. (i), we get `R = ((l_(g))/(nl-l_(g))) G Omega` ` =(G)/((n-1)) Omega`

Discussion

30.	The first and the last terms of an AP are 8 and 350 respectively. If its commondifference is 9, how many terms are there and what is their sum?
Answer» AP: 8.......350(Given) A1=8 An=350 D. =9 So the AP is, A2: a+d = 8+9=17 A3: a+2d = 8+2(9)=27 So we have the AP as 8,17, 26 Now we have to find number of terms present in the AP, Using An=a+(n-1)d 350=8+(n-1)9 350-8=(n-1)9 342/9=n-1 38=n-1 n=39 Now, we found out the number of terms, that is 39 terms in total. The next step is calculating the sum of all the 39 terms, so Using, Sn=n/2{2a+(n-1)d} Sn = 39/2{2(8)+(39-1)9} =39/2{16+342} =39/2358 =6981* Therefore, we got the sum as 6981 Hope my answer helps you! Thank you. First term, a=8, Common Difference, d=9 Last Term, An= 350 a+(n-1)d= 350 8+(n-1)9=350 (n-1)9=350-8=342 (n-1)9= 342 n-1= 342/9= 38 n-1= 38 n= 38+1= 39, So, there are 39 terms. Sum of n terms, Sn=n/2(a+An) S39= 39/2(8+350) S39= (39/2) * 358 S39= 39* 179= 6981 a=8 An=350 d=9 An=? An=a+(n-1) d 350=8+(n-1) 9 350-8=(n-1) 9 342=(n-1) 9 38=n-1 38+1=n 39=n a=8 d=9 l=350 Sn=? Sn=n/2(a+l) S39=39/2(8+350) S39=39/2(358) S39=39(179) S39=6981 An= a+(n-1)d 350=8+(n-1)9 350-8=(n-1)9 342=(n-1)9 342/9=n-1 38=n-1 38+1=n 39=n Sn=n/2[a+an] Sn=39/2[8+350] Sn=39/2[358] Sn=39/2×358 Sn=39×179 Sn=6981

Discussion

31.	Prove that (3 − √5) is irrational.
Answer» Part I :- We have to prove that √5 is an irrational number. Let us assume contrary that √5 is a rational number. ∴√5 can be written as \(\frac{p}{q}\) form where q ≠ 0 and p & q has no common factor other than 1. ∴ √5 = \(\frac{p}{q}\) ⇒ p = √5q ⇒ p² = 5q² … (1) (By squaring both sides) ⇒ 5 divides p² ⇒ 5 divides p (∵ if a prime number divides a² then that prime number must divides a) ∴ p = 5m where m is an integer. Now, Putting p= 5m in equation (1), we get 25m² = 5q² ⇒ q² = 5m² ⇒ 5 divides q² ⇒ 5 divides q i.e., 5 divides both p and q which implies that 5 is a common factor of both integers p and q which is a contradiction of the fact that p & q have no common factor other than 1. ∴ Our assumption is wrong. ∴ √5 is an irrational number. Part II :- Now, We have to prove that (3 – √5) is an irrational number. Let us assume contrary that 3 – √5 is a rational number 3 – √5 = \(\frac{p}{q}\) where q ≠ 0 and p & q ∈ I. (∵ Every rational number can be written in \(\frac{p}{q}\) form) ⇒ √5 = 3 − \(\frac{p}{q}\) = \(\frac{3q-p}{q}\) … (2) L.H.S = √5 = irrational number (According to result (1)) R.H.S = \(\frac{3q-p}{q}\), ≠ 0 & 3q − p ∈ I is a rational number. (Because every number which can be written in the form p q , q ≠ 0 & p, q ∈ is a rational number) But rational ≠ irrational which is contradiction of equation (2). Hence, Our assumption is wrong. ∴ (3 – √5) is an irrational number.

31.

Prove that (3 − √5) is irrational.

Answer»

Part I :- We have to prove that √5 is an irrational number.

Let us assume contrary that √5 is a rational number.

∴√5 can be written as \(\frac{p}{q}\) form where q ≠ 0 and p & q has no common factor other than 1.

∴ √5 = \(\frac{p}{q}\)

⇒ p = √5q

⇒ p² = 5q² … (1)

(By squaring both sides)

⇒ 5 divides p²

⇒ 5 divides p

(∵ if a prime number divides a² then that prime number must divides a)

∴ p = 5m where m is an integer.

Now,

Putting p= 5m in equation (1), we get

25m² = 5q²

⇒ q² = 5m²

⇒ 5 divides q²

⇒ 5 divides q

i.e., 5 divides both p and q which implies that 5 is a common factor of both integers p and q which is a contradiction of the fact that p & q have no common factor other than 1.

∴ Our assumption is wrong.

∴ √5 is an irrational number.

Part II :-

Now,

We have to prove that (3 – √5) is an irrational number.

Let us assume contrary that 3 – √5 is a rational number 3 – √5 = \(\frac{p}{q}\) where q ≠ 0 and p & q ∈ I.

(∵ Every rational number can be written in \(\frac{p}{q}\) form)

⇒ √5 = 3 − \(\frac{p}{q}\) = \(\frac{3q-p}{q}\) … (2)

L.H.S = √5 = irrational number

(According to result (1))

R.H.S = \(\frac{3q-p}{q}\), ≠ 0 & 3q − p ∈ I is a rational number.

(Because every number which can be written in the form p q , q ≠ 0 & p, q ∈ is a rational number)

But rational ≠ irrational which is contradiction of equation (2).

Hence,

Our assumption is wrong.

∴ (3 – √5) is an irrational number.

Discussion

32.	150 is a term of the AP : \( 11,8,5,2 \ldots \)8. An AP consists of 50 terms of which 3 rd term is 12 and the last tem is 736 . tem. zero?
Answer» (i) 11, 8, 5, 2, .... is an A.P. a = 11 d = 8 - 11 = -3 150^th term = a₁₅₀ = a + (150 - 1)d = 11 + 149 x -3 = 11 - 447 = -436 (ii) a₃ = 12, a_n = a₅₀ = 736 \(\therefore\) a + 2d = 12---(1) a + 49d = 736----(2) ⇒ 47d = 724 (By (2) - (1)) ⇒ d = 724/47 ⇒ a = 12 - 2d = 12 - \(\frac{724\times2}{47}\) = \(\frac{564-1448}{47}\) = \(\frac{-448}{47}\) Let n^th term be zero \(\therefore\) a +(n -1)d = 0 ⇒ \(\frac{-884}{47}+(n-1)\frac{724}{47}=0\) ⇒ 724(n - 1) = 884 ⇒ n -1 = 884/724 ⇒ n = 1608/724 which is not a whole number \(\therefore\) No term of A.P. will be zero.

32.

150 is a term of the AP : \( 11,8,5,2 \ldots \)8. An AP consists of 50 terms of which 3 rd term is 12 and the last tem is 736 . tem. zero?

Answer»

(i) 11, 8, 5, 2, .... is an A.P.

a = 11 d = 8 - 11 = -3

150^th term = a₁₅₀ = a + (150 - 1)d

= 11 + 149 x -3

= 11 - 447 = -436

(ii) a₃ = 12, a_n = a₅₀ = 736

\(\therefore\) a + 2d = 12---(1)

a + 49d = 736----(2)

⇒ 47d = 724 (By (2) - (1))

⇒ d = 724/47 ⇒ a = 12 - 2d

= 12 - \(\frac{724\times2}{47}\) = \(\frac{564-1448}{47}\)

= \(\frac{-448}{47}\)

Let n^th term be zero

\(\therefore\) a +(n -1)d = 0

⇒ \(\frac{-884}{47}+(n-1)\frac{724}{47}=0\)

⇒ 724(n - 1) = 884

⇒ n -1 = 884/724

⇒ n = 1608/724

which is not a whole number

\(\therefore\) No term of A.P. will be zero.

Discussion

33.	A computerized system is not complete after the execution and testing phase? What is the next phase to complete the system?
Answer» Documentation is the next phase to complete the system.

Discussion

34.	The roots of the equation a2q2x2-p2 = 0 will be(a) a2p2/q2(b) a2/ap(c) ± p/aq(d) aq/p
Answer» The correct option is:(c) ± p/aq

Discussion

35.	Which of the following is a prime number?(a) 6 (b) 9 (c) 15 (d) 11
Answer» 11 is a prime number.

35.

Which of the following is a prime number?(a) 6 (b) 9 (c) 15 (d) 11

Answer»

11 is a prime number.

Discussion

36.	Total surface area of a sphere is(a) 2πr2 Sq unit(b) 3πr2 Sq unit(c) 4πr2 Sq unit(d) πr2 Sq unit
Answer» Total surface area of a sphere is 4πr² Sq unit.

Discussion

37.	The administrator in a organization consider a microcomputer asA. a simple electronic machineB. as important component of information systemC. a powerfull tool of productivityD. a calculating machine
Answer» Correct Answer - C

Discussion

38.	The disk-caching feature improves theA. performance of hard diskB. speed of processorC. performance of monitorD. performance of CD drive
Answer» Correct Answer - A

Discussion

39.	The symbol used for copy right is a ....... (a) @ (b) Copy (c) & (d) ©
Answer» The symbol used for copy right is a ©

Discussion

40.	The symbol used for copy right is a ........ (a) & (b) Copy (c) # (d) ©
Answer» The symbol used for copy right is a ©.

Discussion

41.	Some phases in programming are given below. 1) Source coding 2) Execution 3) Translation 4) Problem study These phases should follow a proper order. Choose the correct order from the following: a) 4 —> 2 —> 3 —> 1 b) 1 —> 3 —> 2 —> 4 c) 1 —> 3 —> 4 —> 2 d) 4 —> 1 —> 3 —> 2
Answer» d) 4 —> 1 —> 3 —> 2

Discussion

42.	निम्न में से कौन-सा लूप स्टेटमेण्ट नहीं है? (a) if(b) do-while(c) while(d) for
Answer» (a) if एक ब्रांचिंग स्टेटमेण्ट है, जो प्रोग्राम में निर्णय लेने के लिए प्रयोग किया जाता है।

Discussion

43.	<TABLE> टैग के निम्न एट्रिब्यूट्स को परिभाषित कीजिए।(i) cellpadding(ii) bordercolor
Answer» (i) cellpadding एट्रिब्यूट का प्रयोग सेल के किनारे व टेक्स्ट के बीच स्पेस निर्दिष्ट करने के लिए किया जाता है। प्रारूप <TABLE cellpadding=”pixel”> (ii) bordercolor एट्रिब्यूट का प्रयोग टेबल में बॉर्डर कलर देने के लिए किया जाता है। प्रारूप <TABLE bordercolor=”color_name/hex_num/rgb_num”>

43.

<TABLE> टैग के निम्न एट्रिब्यूट्स को परिभाषित कीजिए।(i) cellpadding(ii) bordercolor

Answer»

(i) cellpadding एट्रिब्यूट का प्रयोग सेल के किनारे व टेक्स्ट के बीच स्पेस निर्दिष्ट करने के लिए किया जाता है।
प्रारूप <TABLE cellpadding=”pixel”>

(ii) bordercolor एट्रिब्यूट का प्रयोग टेबल में बॉर्डर कलर देने के लिए किया जाता है।
प्रारूप <TABLE bordercolor=”color_name/hex_num/rgb_num”>

Discussion

44.	Some of the components in the phases of programming are given below. Write them in order of their occurrence. a) Translation b) Documentation c) Problem identification d) Coding of a program
Answer» The chronological order is as follows 1) c) Problem Identification 2) d) Coding of a program 3) a) Translation 4) b) Documentation

44.

Some of the components in the phases of programming are given below. Write them in order of their occurrence. a) Translation b) Documentation c) Problem identification d) Coding of a program

Answer»

The chronological order is as follows

1) c) Problem Identification

2) d) Coding of a program

3) a) Translation

4) b) Documentation

Discussion

45.	Program written in HLL is known as .......
Answer» Program written in HLL is known as Source code.

Discussion

46.	Mr. George wants to check a number is greater than zero and to perform an operation while drawing a flow chart which symbol is used for this?
Answer» Rhombus symbol is used.

Discussion

47.	ANSI means ......
Answer» American National Standards Institute

Discussion

48.	Following are the advantages of flowcharts one among them is wrong. Find it. (a) Better communication (b) Effective analysis (c) proper program documentation (d) Modification easy
Answer» (d) Modification easy. It is a disadvantage.

Discussion

49.	State and prove work energy theorem in case of rectilinear motion under constant acceleration.
Answer» work energy theorem state that change in kinetic energy of an object is equal to the net work done on it by the net force W = K_f- k_i where k_f = final kinetic energy k_i= initial kinetic energy for rectilinear motion and constant acceleration. v² - u² = 2as......(1) where v = final speed, u = initial speed. multiplying equation (1) both side by \(\cfrac{m}2\) \(\cfrac{1}2\)mv² - \(\cfrac{1}2\)mu² = \(\cfrac{m}2\) .2 as \(\because\) F = ma \(\cfrac{1}2\)mv² - \(\cfrac{1}2\)mu² = F.s \(\because\) W = F.s then \(\cfrac{1}2\)mv² - \(\cfrac{1}2\)mu² = W K_f- k_i = W work is done by a force on the body over a certain displacement. The change in kinetic energy of a particle is equal to the work done on it by the net force.

49.

State and prove work energy theorem in case of rectilinear motion under constant acceleration.

Answer»

work energy theorem state that change in kinetic energy of an object is equal to the net work done on it by the net force

W = K_f- k_i

where k_f = final kinetic energy

k_i= initial kinetic energy

for rectilinear motion and constant acceleration.

v² - u² = 2as......(1)

where v = final speed, u = initial speed.

multiplying equation (1) both side by \(\cfrac{m}2\)

\(\cfrac{1}2\)mv² - \(\cfrac{1}2\)mu² = \(\cfrac{m}2\) .2 as

\(\because\) F = ma

\(\cfrac{1}2\)mv² - \(\cfrac{1}2\)mu² = F.s

\(\because\) W = F.s then

\(\cfrac{1}2\)mv² - \(\cfrac{1}2\)mu² = W

K_f- k_i = W

work is done by a force on the body over a certain displacement. The change in kinetic energy of a particle is equal to the work done on it by the net force.

Discussion

50.	ANSI means .........
Answer» American National Standards Institute.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

A random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is ______. (a) 6 (b) 4 (c) 3 (d) 2

The operation * defined by a*b = ab/7 is not a binary operation on ______.(a) Q+ (b) Z (c) R (c) C

The random variable X has the probability density function f(x) = \(\begin{cases} ax\,+\,b, &amp; 0&lt;x&lt;1\\ 0,&amp;otherwise \end{cases}\)and E(X) = 7/12 , then a and b are respectively _______.(a) 1 and 1/2(b) 1/2 and 1(c) 2 and 1(d) 1 and 2

A binary operation on a set S is a function from ________. (a) S → S (b)(S x S) → S (c) S → (S x S) (d) (S x S) → (S x S)

Is cos-1 (-x) = π – cos-1 x true? Justify your answer.

If `b^(2) -ac lt 0 and a gt 0` then the value of the determinant isA. positiveB. negativeC. zeroD. `b^(2)+ae`

If f(x) is continuous and increasing functin such that the domain of `g(x) = sqrt(f(x)-x)` be R and `h(x) =(1)/(1-x)` then the domain of `phi(x)=sqrt(f(f(f(x)))-h(h(h(x))))`is -A. RB. {0,1}C. R-{0,1}D. `R^(+)-{1}`

Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.

If `f(x)={alpha+sin[x]/x , x > 0 and 2 ,x=0 and beta+[(sin x-x)/x^3] ,x < 0` (whlenotes the greatest integer function) if `f(x)` is continuous at `x = 0`. then `beta` is equal toA. `alpha-1`B. `alpha+1`C. `alpha+2`D. `alpha-2`

The polynomial `f(x)=x^4+a x^3+b x^3+c x+d`has real coefficients and `f(2i)=f(2+i)=0.`Find the value of `(a+b+c+d)dot`A. 1B. 4C. 9D. 10

Find the greatest value of the term independent of `x`in the expansion of `(xsinalpha+(cosalpha)/x)^(10)`, where `alpha in Rdot`A. `2^(5)`B. `(10!)/(5!)^(2)`C. `(1)/(2^(5))(10!)/(5!)^(2)`D. None of these

In ∆ABC and ∆DEF, ∠F = ∠C, ∠B = ∠E and AB = \(\frac12\)DE. Then the two triangle are(a) Congruent, but not similar.(b) Similar, but not congruent(c) Neither congruent nor similar(d) Congruent as well as similar

Show that the relation R on the set N x N defined by(a, b) R (c, d) if a2 + d2 = b2 + c2 ∀ a, b, c, d ∈ N, is an equivalence relation.

Write the value of (1–sin2θ)sec2θ

Which term of the AP 5, 9, 13, 17, …….. is 81?

If tan θ = 2/3, then the value of sec θ is(a) \(\frac{\sqrt{13}}{3}\)(b) \(\frac{\sqrt{5}}{3}\)(c) \(\sqrt{\frac{13}{3}}\)(d) \(\frac{3}{\sqrt{13}}\)

The base BC of an equilateral ΔABC lies on the y-axis. The co-ordinates of C are (0, -3). If the origin is the mid-point of the base BC, what are the co-ordinates of A and B ?(a) A(√3, 0), B(0, 3)(b) A\((\pm 3\sqrt{3}, 0)\), B(3, 0)(c) A\((\pm 3\sqrt{3}, 0)\), B(0, 3)(d) A(-√3, 0), B(3, 0)

If sec θ + tan θ = p, then tan θ is(a) \(\frac{p^2 + 1}{2p}\)(b) \(\frac{p^2 - 1}{2p}\)(c) \(\frac{p^2 - 1}{p^2 + 1}\)(d) \(\frac{p^2 + 1}{p^2 - 1}\)

In a ΔABC, ∠A = x°, ∠B = (3x - 2)°, ∠C = y°. Also ∠C - ∠B = 9°. The sum of the greatest and the smallest angles of this triangle is(a) 107°(b) 135°(c) 155°(d) 145°

(1 + sin θ/ 1 - sin θ) + (1 - sin θ/1 + sin θ)\(\frac{1+sin\theta}{1-sin\theta}+\frac{1-sin\theta}{1+sin\theta}\)

Find the derivative of y = sec4x - tan4x.

Find domain &amp; range of y = sec-1 x + cosec-1 x + tan-1 x.

Prove that:Sin-1 (1/√17) + cos-1 (9/√85) = tan-1 (1/2)

A particle is performing U.C.M. along the cirqumference of a circle of diameter 50 cm with frequency 2 Hz. The acceleration of the particle in `m//s^(2)` isA. `2 pi^(2)`B. ` 8 pi^(2)`C. `pi^(2)`D. `4pi^(2)`

Find the wrong statement from the following about the equation of stationary wave given by `Y = 0.04 cos(pix) sin(50 pit)` m where tis in second. Then for the stationary wave.A. Time period= 0.02 sB. Wavelength= 2 mC. Velocity = 50 m/sD. Amplitude= 0.02 m

An ammeter has resistance `R_(0)` and range l what resistance should be connected in parallel with it to increase its range by nl ?A. a series resistance of `(G)/(n +1) Omega`B. a shut of `(G)/(n +1) Omega`C. a shut of `(G)/(n +1) Omega`D. a series resistance of `(G)/(n -1) Omega`

The first and the last terms of an AP are 8 and 350 respectively. If its commondifference is 9, how many terms are there and what is their sum?

Prove that (3 − √5) is irrational.

150 is a term of the AP : \( 11,8,5,2 \ldots \)8. An AP consists of 50 terms of which 3 rd term is 12 and the last tem is 736 . tem. zero?

A computerized system is not complete after the execution and testing phase? What is the next phase to complete the system?

The roots of the equation a2q2x2-p2 = 0 will be(a) a2p2/q2(b) a2/ap(c) ± p/aq(d) aq/p

Which of the following is a prime number?(a) 6 (b) 9 (c) 15 (d) 11

Total surface area of a sphere is(a) 2πr2 Sq unit(b) 3πr2 Sq unit(c) 4πr2 Sq unit(d) πr2 Sq unit

The administrator in a organization consider a microcomputer asA. a simple electronic machineB. as important component of information systemC. a powerfull tool of productivityD. a calculating machine

The disk-caching feature improves theA. performance of hard diskB. speed of processorC. performance of monitorD. performance of CD drive

The symbol used for copy right is a ....... (a) @ (b) Copy (c) &amp; (d) ©

The symbol used for copy right is a ........ (a) &amp; (b) Copy (c) # (d) ©

निम्न में से कौन-सा लूप स्टेटमेण्ट नहीं है? (a) if(b) do-while(c) while(d) for

&lt;TABLE&gt; टैग के निम्न एट्रिब्यूट्स को परिभाषित कीजिए।(i) cellpadding(ii) bordercolor

Some of the components in the phases of programming are given below. Write them in order of their occurrence. a) Translation b) Documentation c) Problem identification d) Coding of a program

Program written in HLL is known as .......

Mr. George wants to check a number is greater than zero and to perform an operation while drawing a flow chart which symbol is used for this?

ANSI means ......

Following are the advantages of flowcharts one among them is wrong. Find it. (a) Better communication (b) Effective analysis (c) proper program documentation (d) Modification easy

State and prove work energy theorem in case of rectilinear motion under constant acceleration.

ANSI means .........

The random variable X has the probability density function f(x) = \(\begin{cases} ax\,+\,b, & 0<x<1\\ 0,&otherwise \end{cases}\)and E(X) = 7/12 , then a and b are respectively _______.(a) 1 and 1/2(b) 1/2 and 1(c) 2 and 1(d) 1 and 2

Find domain & range of y = sec-1 x + cosec-1 x + tan-1 x.

The symbol used for copy right is a ....... (a) @ (b) Copy (c) & (d) ©

The symbol used for copy right is a ........ (a) & (b) Copy (c) # (d) ©

<TABLE> टैग के निम्न एट्रिब्यूट्स को परिभाषित कीजिए।(i) cellpadding(ii) bordercolor