100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 310 InterviewSolution

15451.	If `f(x)=(a^(x))/(a^(x)+sqrt(a))(agt0),g(n)=sum_(r=1)^(2n-1)2f((r)/(2n))`. Find te value `g(4)`
Answer» Correct Answer - 7 `f(x)=(a^(x))/(a^(x)+sqrt(a))(agt0)`,,,,. `f(x)+f(1-x)=(a^(x))/(a^(x)+sqrt(a))+(a^(1-x))/(a^(1-x)+sqrt(a))` `=(a^(x))/(a^(x)+sqrt(a))+((a)/(a^(x)))/((a)/(a^(x))+sqrt(a))` `=(a^(x))/(a^(x)+sqrt(a))+(a)/(a+sqrt(aa)^(x))` `=(a^(x))/(a^(x)+sqrt(a))+(sqrt(a))/(sqrt(a)+a^(x))=(a^(x)+sqrt(a))/(a^(x)+sqrt(a))=1` `impliessum_(r=1)^(2n-1)2f((r)/(2n))` `=2[(f((1)/(2n))+f((2)/(2n))+...+f((n-1)/(2n)+f((n)/(2n))+f((n+1)/(2n))+),(+f((2n-2)/(2n))+f((2n-1)/(2n))]` `=2[f((1)/(2n))+f((2n-1)/(2n))+f((2)/(2n))+f((2n-2)/(2n))+..+f((n-1)/(2n))+f((n+1)/(2n))+f((n)/(2n))]` `=2[1+1+1+.....(n-1)` times `+f((n)/(2n))]` `2[n-1+f((1)/(2))]` `f((1)/(2))=(a^((1)/(2)))/(a^((1)/(2))+sqrt(a))=(sqrt(a))/(2sqrt(a))=(1)/(2)` `sum_(r=1)^(n)2f((r)/(2n))=2(n-1+(1)/(2))` `=2(n-(1)/(2))=2n-1`

Discussion

15452.	What is the inverse of the function f(x) = ln (8x – 4) for {x: x ∈ R \| x > 0.5}?1. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x}}}{8}\)2. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} - 4}}{8}\)3. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 8}}{4}\)4. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 4}}{8}\)
Answer» Correct Answer - Option 4 : \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 4}}{8}\) CONCEPT: A function f: X → Y is defined to be invertible, if there exists a function g: Y → X such that g o f = I_x and f o g = I_Y. The function g is called the inverse of f and is denoted by f ^–1. For calculation of inverse of any function, it should be arranged in the terms of x = f(y) and then every y should be replaced with x. CALCULATIONS: Given function is f(x) = ln (8x – 4) = y (say) ∴ y = ln (8x – 4) e ^y = 8x – 4 ⇒ \(x = \;\frac{{{e^y} + 4}}{8} = {f^{ - 1}}\left( {\text{y}} \right)\) ∴ \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 4}}{8}\)

15452.

What is the inverse of the function f(x) = ln (8x – 4) for {x: x ∈ R | x > 0.5}?1. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x}}}{8}\)2. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} - 4}}{8}\)3. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 8}}{4}\)4. \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 4}}{8}\)

Answer» Correct Answer - Option 4 : \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 4}}{8}\)

CONCEPT:

A function f: X → Y is defined to be invertible, if there exists a function g: Y → X such that g o f = I_x and

f o g = I_Y. The function g is called the inverse of f and is denoted by f ^–1.

For calculation of inverse of any function, it should be arranged in the terms of x = f(y) and then every y should be replaced with x.

CALCULATIONS:

Given function is f(x) = ln (8x – 4) = y (say)

∴ y = ln (8x – 4)

e ^y = 8x – 4 ⇒ \(x = \;\frac{{{e^y} + 4}}{8} = {f^{ - 1}}\left( {\text{y}} \right)\)

∴ \({f^{ - 1}}\left( {\text{x}} \right) = \frac{{{e^x} + 4}}{8}\)

Discussion

15453.	Who among the following has been named as the 2016 ITF world Champions by the International Tennis Federation?A. Rafael NadalB. Andy MurrayC. Roger FedererD. Noval Dokovic
Answer» Correct Answer - A

Discussion

15454.	Let A = {3, 7}, B = {5, 11} and relation R = {(a, b): a ∈ A, b ∈ B and a + b is even} then inverse relation R-1 is 1. Identity relation2. Universal relation3. Empty relation4. Both 1 and 2
Answer» Correct Answer - Option 2 : Universal relation Concept: If R is a relation from set A to set B, then inverse relation of R to be denoted by R-1, is a relation from set B to set A. Symbolically R-1 = {(b, a): (a, b) ∈ R for all a ∈ A and b ∈ B}. A relation R on a set A is said to be identity relation on A if R = {(a, b): a ∈ A, b ∈ A and a = b}. A relation R from A to B is said to be the universal relation, if R = A ×. B A relation R from A to B is called an empty relation or a void relation from A to B if R = ø. Calculation: Given A = {3, 7}, B = {5, 11} Cartesian product \(A× B\) = {(3, 5), (3, 11), (7, 5), (7, 11)} Relation R = {(a, b): a ∈ A, b ∈ B and a + b is even} Roster form R = {(3, 5), (3, 11), (7, 5), (7, 11)} R^-1 = {(5, 3), (11, 3), (5, 7), (11, 7)} Inverse relation R^-1 = {(3, 5), (3, 11), (7, 5), (7, 11)} = A × B Hence relation R^-1 is universal relation

15454.

Let A = {3, 7}, B = {5, 11} and relation R = {(a, b): a ∈ A, b ∈ B and a + b is even} then inverse relation R-1 is 1. Identity relation2. Universal relation3. Empty relation4. Both 1 and 2

Answer» Correct Answer - Option 2 : Universal relation

Concept:

If R is a relation from set A to set B, then inverse relation of R to be denoted by R-1, is a relation from set B to set A.

Symbolically R-1 = {(b, a): (a, b) ∈ R for all a ∈ A and b ∈ B}.

A relation R on a set A is said to be identity relation on A if R = {(a, b): a ∈ A, b ∈ A and a = b}.

A relation R from A to B is said to be the universal relation, if R = A ×. B

A relation R from A to B is called an empty relation or a void relation from A to B if R = ø.

Calculation:

Given A = {3, 7}, B = {5, 11}

Cartesian product \(A× B\) = {(3, 5), (3, 11), (7, 5), (7, 11)}

Relation R = {(a, b): a ∈ A, b ∈ B and a + b is even}

Roster form R = {(3, 5), (3, 11), (7, 5), (7, 11)}

R^-1 = {(5, 3), (11, 3), (5, 7), (11, 7)}

Inverse relation R^-1 = {(3, 5), (3, 11), (7, 5), (7, 11)} = A × B

Hence relation R^-1 is universal relation

Discussion

15455.	A relation defined as R = {(1, 2), (3, 4), (2, 4)} find R-1.1. {Ø, (2, 1), (4, 3), (4, 2)}2. {{Ø}, (2, 1), (4, 3), (4, 2)}3. {(2, 1), (4, 3), (4, 2)}4. None of these
Answer» Correct Answer - Option 3 : {(2, 1), (4, 3), (4, 2)} Concept: If R is a relation from set A to set B, then inverse relation of R to be denoted by R^-1, is a relation from set B to set A. Symbolically R^-1 = {(b, a): (a, b) ∈ R for all a ∈ A and b ∈ B}. Calculation: R = {(1, 2), (3, 4), (2, 4)} to find R-1 reverse the sequence (a, b) to (b, a). R^-1 = {(2, 1), (4, 3), (4, 2)} Hence, option 3 is the correct answer.

15455.

A relation defined as R = {(1, 2), (3, 4), (2, 4)} find R-1.1. {Ø, (2, 1), (4, 3), (4, 2)}2. {{Ø}, (2, 1), (4, 3), (4, 2)}3. {(2, 1), (4, 3), (4, 2)}4. None of these

Answer» Correct Answer - Option 3 : {(2, 1), (4, 3), (4, 2)}

Concept:

If R is a relation from set A to set B, then inverse relation of R to be denoted by R^-1, is a relation from set B to set A.

Symbolically R^-1 = {(b, a): (a, b) ∈ R for all a ∈ A and b ∈ B}.

Calculation:

R = {(1, 2), (3, 4), (2, 4)} to find R-1 reverse the sequence (a, b) to (b, a).

R^-1 = {(2, 1), (4, 3), (4, 2)}

Hence, option 3 is the correct answer.

Discussion

15456.	Out of total candidate 60% were female, 40% were male. 60% of total passed. Female passed was twice of males passed. A candidate who passes was chosen. Find probability that it was a female.
Answer» Correct answer is 2/3 Given, Total = 100 Female = 60 Male = 40 Passed = 60 Female passed = 40, Male passes = 20 Probability 40/60 = 2/3

15456.

Out of total candidate 60% were female, 40% were male. 60% of total passed. Female passed was twice of males passed. A candidate who passes was chosen. Find probability that it was a female.

Answer»

Correct answer is 2/3

Given,

Total = 100

Female = 60

Male = 40

Passed = 60

Female passed = 40, Male passes = 20

Probability 40/60 = 2/3

Discussion

15457.	Let A = {3, 4, 5} and relation R = {(a, b): a, b ∈ A, a divides b and b divides a} then relation R-1 is1. Identity relation2. Universal relation3. both a and b4. empty relation
Answer» Correct Answer - Option 1 : Identity relation Concept: If R is a relation from set A to set B, then inverse relation of R to be denoted by R-1, is a relation from set B to set A. Symbolically R-1 = {(b, a): (a, b) ∈ R for all a ∈ A and b ∈ B}. A relation R on a set A is said to be identity relation on A if R = {(a, b): a ∈ A, b ∈ A and a = b}. A relation R from A to B is said to be the universal relation, if R = A × B. A relation R from A to B is called an empty relation or a void relation from A to B if R = ø. Calculation: A = {3, 4, 5} Cartesian product \(A× A\) = {(3, 3), (3, 4), (3, 5), (4, 3), (4, 4),(4, 5), (5, 3), (5, 4), (5, 5)} Given, R = {(a, b): a, b ∈ A, a divides b and b divides a} R^-1 = {(b, a): a, b ∈ A, b divides a and a divides b} If b divides a and a divides b. it is only possible if a = b. Hence, relation R^-1 = {(3, 3), (4, 4), (5, 5)}. therefore R^-1 is an identity relation.

15457.

Let A = {3, 4, 5} and relation R = {(a, b): a, b ∈ A, a divides b and b divides a} then relation R-1 is1. Identity relation2. Universal relation3. both a and b4. empty relation

Answer» Correct Answer - Option 1 : Identity relation

Concept:

If R is a relation from set A to set B, then inverse relation of R to be denoted by R-1, is a relation from set B to set A.

Symbolically R-1 = {(b, a): (a, b) ∈ R for all a ∈ A and b ∈ B}.

A relation R on a set A is said to be identity relation on A if R = {(a, b): a ∈ A, b ∈ A and a = b}.

A relation R from A to B is said to be the universal relation, if R = A × B.

A relation R from A to B is called an empty relation or a void relation from A to B if R = ø.

Calculation:

A = {3, 4, 5}

Cartesian product \(A× A\) = {(3, 3), (3, 4), (3, 5), (4, 3), (4, 4),(4, 5), (5, 3), (5, 4), (5, 5)}

Given, R = {(a, b): a, b ∈ A, a divides b and b divides a}

R^-1 = {(b, a): a, b ∈ A, b divides a and a divides b}

If b divides a and a divides b. it is only possible if a = b.

Hence, relation R^-1 = {(3, 3), (4, 4), (5, 5)}. therefore R^-1 is an identity relation.

Discussion

15458.	If log x (y) = 10 and log2 x = 5, then the value of y is:1. 2102. 253. 2504. 215
Answer» Correct Answer - Option 3 : 250 CONCEPT : On removing the log from the equation the base of the log becomes base of the new exponent term formed i.e. log _a (x) = b, on removing log from the equation, we will get x = a^b. CALCULATION: Given that log x (y) = 10 and log2 x = 5 ∵ log2 x = 5 ⇒ x = 2⁵ Again log x (y) = 10 ⇒ y = x¹⁰ = y = (2⁵)¹⁰= 2⁵⁰ Therefore, option (3) is the correct answer.

15458.

If log x (y) = 10 and log2 x = 5, then the value of y is:1. 2102. 253. 2504. 215

Answer» Correct Answer - Option 3 : 250

CONCEPT :

On removing the log from the equation the base of the log becomes base of the new exponent term formed i.e. log _a (x) = b, on removing log from the equation, we will get x = a^b.

CALCULATION:

Given that log x (y) = 10 and log2 x = 5

∵ log2 x = 5

⇒ x = 2⁵

Again log x (y) = 10

⇒ y = x¹⁰

= y = (2⁵)¹⁰= 2⁵⁰

Therefore, option (3) is the correct answer.

Discussion

15459.	Find the value of x, if log5 (x - 2) + log5 (x + 3) = log5 14.1. 22. 33. 44. 5
Answer» Correct Answer - Option 3 : 4 CONCEPT: Logarithmic product rule is represented as log_p (x) + log _p(y) = log_p(x.y) Log is not defined for negative values. CALCULATION: Given that log₅ (x - 2) + log₅ (x + 3) = log₅ 14 Dropping log from the both sides we get- ⇒ (x - 2) × (x + 3) = 14 ⇒x² + x - 6 = 14 ⇒ x² + x - 20 = 0 On solving the quadratic equation x² + x - 20 = 0 ⇒ (x - 4) × (x + 5) = 0 ⇒ x = 4 and x = - 5 Therefore x = 4. (negative value of x will be excluded) Therefore, option (3) is the correct answer.

15459.

Find the value of x, if log5 (x - 2) + log5 (x + 3) = log5 14.1. 22. 33. 44. 5

Answer» Correct Answer - Option 3 : 4

CONCEPT:

Logarithmic product rule is represented as log_p (x) + log _p(y) = log_p(x.y)
Log is not defined for negative values.

CALCULATION:

Given that log₅ (x - 2) + log₅ (x + 3) = log₅ 14

Dropping log from the both sides we get-

⇒ (x - 2) × (x + 3) = 14

⇒x² + x - 6 = 14

⇒ x² + x - 20 = 0

On solving the quadratic equation

x² + x - 20 = 0

⇒ (x - 4) × (x + 5) = 0

⇒ x = 4 and x = - 5

Therefore x = 4. (negative value of x will be excluded)

Therefore, option (3) is the correct answer.

Discussion

15460.	Find the value of p, if log cos 60° sin30° = p.1. 42. 23. 34. 1
Answer» Correct Answer - Option 4 : 1 CONCEPT: As we know that \(\operatorname{Sin} {30^ \circ } = \frac{1}{2}\) and \(\operatorname{cos} {60^ \circ } = \frac{1}{2}\). log _x (x) = 1 CALCULATION: Given log cos 60° sin30° = p On putting \(\operatorname{Sin} {30^ \circ } = \frac{1}{2}\) and \(\operatorname{cos} {60^ \circ } = \frac{1}{2}\) we get, \( ⇒ {\log _{\frac{1}{2}}}\frac{1}{2} = p\) ⇒ p = 1 Therefore, option (d) is the correct answer.

15460.

Find the value of p, if log cos 60° sin30° = p.1. 42. 23. 34. 1

Answer» Correct Answer - Option 4 : 1

CONCEPT:

As we know that \(\operatorname{Sin} {30^ \circ } = \frac{1}{2}\) and \(\operatorname{cos} {60^ \circ } = \frac{1}{2}\).
log _x (x) = 1

CALCULATION:

Given log cos 60° sin30° = p

On putting \(\operatorname{Sin} {30^ \circ } = \frac{1}{2}\) and \(\operatorname{cos} {60^ \circ } = \frac{1}{2}\) we get,
\( ⇒ {\log _{\frac{1}{2}}}\frac{1}{2} = p\)

⇒ p = 1

Therefore, option (d) is the correct answer.

Discussion

15461.	Find the smallest number which when divided by 5, 6, 8 and 9 gives a remainder 3 in each case.1. 3602. 3633. 3704. 365
Answer» Correct Answer - Option 2 : 363 Formula used - number = LCM + remainder in each case Solution - LCM of (5, 6, 8, 9) = 360 ⇒ Number = 360 + 3 = 363 ∴ number = 363.

15461.

Find the smallest number which when divided by 5, 6, 8 and 9 gives a remainder 3 in each case.1. 3602. 3633. 3704. 365

Answer» Correct Answer - Option 2 : 363

Formula used -

number = LCM + remainder in each case

Solution -

LCM of (5, 6, 8, 9) = 360

⇒ Number = 360 + 3 = 363

∴ number = 363.

Discussion

15462.	Find the value of x, if log2 (9x - 2) = 41. 12. 23. 34. 4
Answer» Correct Answer - Option 2 : 2 CALCULATION: Given that log₂ (9x - 2) = 4 On raising both the sides to the power of 2 we get ⇒ 9x - 2 = 2⁴ ⇒ 9x = 18 \( ⇒ x = \frac{{18}}{9} = 2\) Therefore, option (2) is the correct answer.

15462.

Find the value of x, if log2 (9x - 2) = 41. 12. 23. 34. 4

Answer» Correct Answer - Option 2 : 2

CALCULATION:

Given that log₂ (9x - 2) = 4

On raising both the sides to the power of 2 we get

⇒ 9x - 2 = 2⁴

⇒ 9x = 18

\( ⇒ x = \frac{{18}}{9} = 2\)

Therefore, option (2) is the correct answer.

Discussion

15463.

If a = log12 18, b = log24 54, then ab + 5(a - b) is1. 12. 03. 24. \(\dfrac{3}{2}\)

Answer» Correct Answer - Option 1 : 1

Concept:

Logarithm properties:

Product Rule	\(\rm \log (mn) =log(m)+log(n)\)
Quotient Rule	\(\rm \log (\frac{m}{n}) =log(m)-log(n)\)
Power Rule	\(\rm \log (m^n) =n\text{ }log(m)\)
Change of Base	\(\boldsymbol {\rm log_m(n)=\frac{1}{log_n(m)}=\frac {log(n)}{log(m)}}\)

Calculation:

a = \(\rm \frac {\log18}{\log12} = {\log (3^2×2)\over \log (2^2×3)}\)

⇒ a = \(\rm {2\log 3+\log2\over 2\log 2+\log3}\)

b = \(\rm \frac {\log54}{\log24} = {\log (3^3×2)\over \log (2^3×3)}\)

⇒ b = \(\rm {3\log 3+\log2\over 3\log 2+\log3}\)

Let log 2 = x and log 3 = y

S = ab + 5(a - b)

⇒ S = \(\rm {2\log 3+\log2\over 2\log 2+\log3}\times\rm {3\log 3+\log2\over 3\log 2+\log3}\) + 5\(\rm \left({2\log 3+\log2\over 2\log 2+\log3} - \rm {3\log 3+\log2\over 3\log 2+\log3}\right)\)

⇒ S = \(\rm {2y+x\over 2x+y}\times\rm {3y+x\over 3x+y}\) + 5 \(\rm\left( {2y+x\over 2x+y}-\rm {3y+x\over 3x+y}\right)\)

⇒ S = \(\rm {6y^2 +5xy+x^2\over 6x^2+5xy+y^2}\) + 5 \(\rm\left( {3x^2+7xy+2y^2-(2x^2 +7xy+3y^2)\over6x^2+5xy+y^2}\right)\)

⇒ S = \(\rm {6y^2 +5xy+x^2\over 6x^2+5xy+y^2}\) + 5 \(\rm\left( {x^2-y^2\over6x^2+5xy+y^2}\right)\)

⇒ S = \(\rm {6y^2 +5xy+x^2\over 6x^2+5xy+y^2}\) + \(\rm {5x^2-5y^2\over6x^2+5xy+y^2}\)

⇒ S = \(\rm {6y^2 +5xy+x^2+5x^2-5y^2\over 6x^2+5xy+y^2}\)

⇒ S = \(\rm {6x^2 +5xy+y^2\over 6x^2+5xy+y^2}\)

⇒ S = 1

Discussion

15464.	If function sgn(x2 – 9x + 20) = 1, then find the value of x.1. (4, 5) 2. (5, ∞)3. (-∞, 4) 4. None of these
Answer» Correct Answer - Option 3 : (-∞, 4) CONCEPT:The real function f: R → R defined by\(f\left( x \right) = \left\{ {\begin{array}{{20}{c}} {\frac{{\left\| x \right\|}}{x},\;\;\;x \ne 0} \\ {0,\;\;\;\;x = 0} \end{array}} \right. = \left\{ {\begin{array}{{20}{c}} {1,\;\;\;\;\;\;\;if\;\;x > 0} \\ {0,\;\;\;\;\;\;\;if\;\;x = 0\;} \\ { - 1,\;\;\;\;\;\;\;\;\;if\;x < 0} \end{array}} \right.\)is called the signum function. Domain of f = R, Range of f = {1, 0, – 1}CALCULATIONS:Given function is f (x) = sgn(x² – 9x + 20) = 1As we know that signum function gives 1, only when x > 0.∴ x² – 9x + 20 > 0⇒ (x – 4) (x – 5) > 0So, x ∈ (-∞, 4) ⋃ (5, ∞)

Discussion

15465.	If \|x2 – 12x + 32\| + \|x2 – 9x + 20\| = 0. Then find the value of x.1. x = 52. x = 43. x = 84. x = 3
Answer» Correct Answer - Option 2 : x = 4 CONCEPT: The modulus of a given function gives the magnitude of that function. Modulus Function is defined as the real valued function. The real function f: R → R defined by f (x) = \|x\|= x, x > 0, and f (x) = \|x\|= -x, if x < 0. ∀ x ∈ R is called the modulus function. CALCULATIONS: Given \|x² – 12x + 32\| + \|x² – 9x + 20\| = 0. Every modulus function is a non-negative function and if two non-negative functions add up to get zero then individual function itself equal to zero simultaneously. x² – 12x + 32 for x = 4 or 8 x² – 9x + 20 for x = 4 or 5 Both the equations are zero at x = 4 So, x = 4 is the only solution for this equation.

15465.

If |x2 – 12x + 32| + |x2 – 9x + 20| = 0. Then find the value of x.1. x = 52. x = 43. x = 84. x = 3

Answer» Correct Answer - Option 2 : x = 4

CONCEPT:

The modulus of a given function gives the magnitude of that function. Modulus Function is defined as the real valued function.

The real function f: R → R defined by f (x) = |x|= x, x > 0, and f (x) = |x|= -x, if x < 0. ∀ x ∈ R is called the modulus function.

CALCULATIONS:

Given |x² – 12x + 32| + |x² – 9x + 20| = 0.

Every modulus function is a non-negative function and if two non-negative functions add up to get zero then individual function itself equal to zero simultaneously.

x² – 12x + 32 for x = 4 or 8

x² – 9x + 20 for x = 4 or 5

Both the equations are zero at x = 4

So, x = 4 is the only solution for this equation.

Discussion

15466.	Let f, g: R → R be defined respectively, and f(x) = x2 - 12x + 32, g(x) = x -3. Find f o g (1).1. 302. 403. 504. 60
Answer» Correct Answer - Option 4 : 60 CONCEPT: Let f: A → B and g: B → C be two functions. Then, the composition of f and g, denoted by g o f, is defined as the function g o f: A → C given by g o f (x) = g (f (x)), ∀ x ∈ A. CALCULATIONS: Given functions are f(x) = x ²- 12x + 32, g(x) = x -3. f o g means g(x) function is in f(x) function. This means put x = x -3 in function f(x). f[g(x)] = (x - 3) ² - 12(x -3) + 32 f[g(x)] = x² – 18x + 77 now we have to find f o g (1). ∴ f [g (1)] = 1² – 18+ 77 = 60

15466.

Let f, g: R → R be defined respectively, and f(x) = x2 - 12x + 32, g(x) = x -3. Find f o g (1).1. 302. 403. 504. 60

Answer» Correct Answer - Option 4 : 60

CONCEPT:

Let f: A → B and g: B → C be two functions. Then, the composition of f and g, denoted by g o f, is defined as the function g o f: A → C given by g o f (x) = g (f (x)), ∀ x ∈ A.

CALCULATIONS:

Given functions are f(x) = x ²- 12x + 32, g(x) = x -3.

f o g means g(x) function is in f(x) function.

This means put x = x -3 in function f(x).

f[g(x)] = (x - 3) ² - 12(x -3) + 32

f[g(x)] = x² – 18x + 77

now we have to find f o g (1).

∴ f [g (1)] = 1² – 18+ 77 = 60

Discussion

15467.	If (x + k) is the HCF of (x2 + 12x + 35) and (x2 + 2x – 35), then what is the value of k?1. 52. 23. 34. 7
Answer» Correct Answer - Option 4 : 7 Given: Our polynomials are (x2 + 12x + 35) and (x2 + 2x – 35) Concept used: The HCF defines the greatest factor present in between two or more numbers. Calculation: Taking one polynomial (x2 + 12x + 35) ⇒ (x2 + 7x + 5x + 35) ⇒ (x + 7)(x + 5) Taking another polynomial (x2 + 2x – 35) ⇒ (x2 + 2x – 35) ⇒ (x2 + 7x – 5x – 35) ⇒ (x + 7)(x – 5) In the two polynomial the highest common factor is (x + 7) So the HCF is (x + 7) By comparing both the HCF, the value of k is 7 ∴ The value of k is 7

15467.

If (x + k) is the HCF of (x2 + 12x + 35) and (x2 + 2x – 35), then what is the value of k?1. 52. 23. 34. 7

Answer» Correct Answer - Option 4 : 7

Given:

Our polynomials are (x2 + 12x + 35) and (x2 + 2x – 35)

Concept used:

The HCF defines the greatest factor present in between two or more numbers.

Calculation:

Taking one polynomial (x2 + 12x + 35)

⇒ (x2 + 7x + 5x + 35)

⇒ (x + 7)(x + 5)

Taking another polynomial (x2 + 2x – 35)

⇒ (x2 + 2x – 35)

⇒ (x2 + 7x – 5x – 35)

⇒ (x + 7)(x – 5)

In the two polynomial the highest common factor is (x + 7)

So the HCF is (x + 7)

By comparing both the HCF, the value of k is 7

∴ The value of k is 7

Discussion

15468.	Find the smallest number which should be added to the given expression to make it a perfect square?789211 × 7892151. 42. 23. 164. 6
Answer» Correct Answer - Option 1 : 4 Given: 789211 × 789215 Concept: a² - b² = (a - b)(a + b) Calculation: Let 789213 be x, then 789211 = (a - 2) and 789215 = (a + 2) now, 789211 × 789215 = (a - 2)(a + 2) ⇒ a² - 4 ∴ To make a perfect square we added 4 in this expression.

15468.

Find the smallest number which should be added to the given expression to make it a perfect square?789211 × 7892151. 42. 23. 164. 6

Answer» Correct Answer - Option 1 : 4

Given:

789211 × 789215

Concept:

a² - b² = (a - b)(a + b)

Calculation:

Let 789213 be x, then

789211 = (a - 2) and

789215 = (a + 2)

now,

789211 × 789215 = (a - 2)(a + 2)

⇒ a² - 4

∴ To make a perfect square we added 4 in this expression.

Discussion

15469.	Which one of the following is the largest divisor of 7x + 7x + 1 + 7x + 2, if x is any natural number? (from the given option)1. 72. 573. 3994. 2793
Answer» Correct Answer - Option 4 : 2793 Given: Our given expression is 7x + 7x + 1 + 7x + 2 Concept used: A natural number is the counting number starting from 1. Calculation: Our given expression is 7x + 7x + 1 + 7x + 2 ⇒ 7x(1 + 7 + 72) ⇒ 7x(57) Taking the value of x = 1, 2, 3, … and so on then, The number is 399, 2793, … and so on So, the greatest divisor when x = 1, will be 399. When x = 2, greatest divisor will be 2793 ...and so on. ∴ The correct answer from the option is 2793.

15469.

Which one of the following is the largest divisor of 7x + 7x + 1 + 7x + 2, if x is any natural number? (from the given option)1. 72. 573. 3994. 2793

Answer» Correct Answer - Option 4 : 2793

Given:

Our given expression is 7x + 7x + 1 + 7x + 2

Concept used:

A natural number is the counting number starting from 1.

Calculation:

Our given expression is 7x + 7x + 1 + 7x + 2

⇒ 7x(1 + 7 + 72)

⇒ 7x(57)

Taking the value of x = 1, 2, 3, … and so on then,

The number is 399, 2793, … and so on

So, the greatest divisor when x = 1, will be 399.

When x = 2, greatest divisor will be 2793 ...and so on.

∴ The correct answer from the option is 2793.

Discussion

15470.	Express - 8 by 11 as a rational number with denominator 22 and 55
Answer» \(\frac{-8}{11} = \frac{-8}{11} \times \frac 22 = \frac{-16}{22}\) \(\frac{-8}{11} = \frac{-8}{11} \times \frac 55 = \frac{-40}{55}\)

15470.

Express - 8 by 11 as a rational number with denominator 22 and 55

Answer»

\(\frac{-8}{11} = \frac{-8}{11} \times \frac 22 = \frac{-16}{22}\)

\(\frac{-8}{11} = \frac{-8}{11} \times \frac 55 = \frac{-40}{55}\)

Discussion

15471.	\( \left\{\frac{5}{6} \times \frac{-1}{4}\right\}-\left\{\frac{2}{3} \times \frac{5}{6}\right\}+\left\{\frac{5}{6} \times \frac{-5}{6}\right\} \)
Answer» {5/6 x −1/4} − {2/3 x 5/6} + { 5/6 x −5/6} = \(-\frac{5}{24}-\frac{10}{24}-\frac{25}{36}\) \(=\frac{-15-40-50}{72}\) \(=\frac{-105}{72}=\frac{-35}{24}\)

15471.

\( \left\{\frac{5}{6} \times \frac{-1}{4}\right\}-\left\{\frac{2}{3} \times \frac{5}{6}\right\}+\left\{\frac{5}{6} \times \frac{-5}{6}\right\} \)

Answer»

{5/6 x −1/4} − {2/3 x 5/6} + { 5/6 x −5/6}

= \(-\frac{5}{24}-\frac{10}{24}-\frac{25}{36}\)

\(=\frac{-15-40-50}{72}\)

\(=\frac{-105}{72}=\frac{-35}{24}\)

Discussion

15472.	Length of a rectangular tank is twice its breadth. If the depth of the tank is 3 m and area of its four walls is 108m2 , find the length of the tank.
Answer» Let breadth of the tank be x. ∴ Length of the tank = 2x. Area of the walls of the tank = 2(length + breadth) × depth. ∴ 108 = 2(2x + x)× 3 ∴ 108 = 18 x ∴ x = 6 ∴ 2x = 12 ∴ Length of the tank = 12m. tan R = PQ / QR = 3/4

15472.

Length of a rectangular tank is twice its breadth. If the depth of the tank is 3 m and area of its four walls is 108m2 , find the length of the tank.

Answer»

Let breadth of the tank be x.

∴ Length of the tank = 2x.

Area of the walls of the tank = 2(length + breadth) × depth.

∴ 108 = 2(2x + x)× 3

∴ 108 = 18 x ∴ x = 6 ∴ 2x = 12

∴ Length of the tank = 12m.

tan R = PQ / QR = 3/4

Discussion

15473.	In the interval `2/pi
Answer» `A = [[2sinx,3],[1,2sinx]]` Matrix `A` will be singular if `\|A\| = 0` Here, `\|A\| = 2sinx(2sinx) -3(1) = 4sin^2x -3` `:. 4sin^2x-3 = 0 =>sin^2x = 3/4` `=>sinx = +-sqrt3/2 ` As, `x in (pi/2,pi)`, `:. x = (pi-pi/3) = (2pi)/3`

Discussion

15474.	Prove that `2tan^(-1)(1/2)+tan^(-1)(1/7)=sin^(-1)((31)/(25sqrt(2)))`
Answer» `L.H.S. = 2tan^-1(1/2)+tan^-1(1/7)` `=tan^-1((2*1/2)/(1-(1/2)^2))+tan^-1(1/7)` `=tan^-1((1)/(3/4))+tan^-1(1/7)` `=tan^-1(4/3)+tan^-1(1/7)` `=tan^-1((4/3+1/7)/(1-(4/3)(1/7)))` `=tan^-1((31/21)/(17/21))` `=tan^-1(31/17)` Let `tan^-1(31/17) = theta` Then, `tantheta = 31/17` `cot theta = 17/31` `cosectheta = sqrt(1+(17/31)^2) = sqrt1250/31 = (25sqrt2)/31` `:. sintheta = 31/(25sqrt2)` `=>theta = sin^-1(31/(25sqrt2))` `:. tan^-1(31/17) = sin^-1(31/(25sqrt2)) = R.H.S.`

Discussion

15475.	If `x=a e^t(sint+cost)`and `y=a e^t(sint-cost),`prove that `(dy)/(dx)=(x+y)/(x-y)dot`
Answer» `x = ae^t(sint+cost), y = ae^t(sint-cost) ` `=> dx/dt = (sint+cost)ae^t +ae^t(cost-sint) = x-y` `=>dy/dt = (sint-cost)ae^t +ae^t(cost+sint) = x+y` `:. dy/dx = (dy/dt)/(dx/dt) = (x+y)/(x-y)` `=>dy/dx = (x+y)/(x-y)`

Discussion

15476.	Two schools A and B decided to award prizes to their students for threevalues, team spirit, truthfulness and tolerance at the rate of `R sdotx ,R sdoty`and `R sdotz`per student respectively. School `A`, decided to award a total of Rs. 1,100 for the three values to 3, 1 and 2 students respectively whileschool B decided to award Rs. 1,400 for the three values to 1, 2 and 3 students respectively. If oneprize for all the three values together amount to Rs. 600 then(i) Represent the above situation by a matrix equation after forminglinear equations.(ii) Is it possible to solve the system of equations so obtained usingmatrices ?(iii) Which value you prefer to be rewarded most and why ?
Answer» `3x+10y+2xz=1100` `x+2xy+3xz=1400` `x+y+z=600` `[[3,1,2],[1,2,3],[1,1,1]][[x],[y],[z]]=[[1100],[1400],[600]]` `\|A\|=\|[1,2],[2,3]\|-1\|[3,2],[1,3]\|+\|[3,1],[1,2]\|` `=(3-4)-(9-2)+(6-1)` `=-1-7+5=-3!=0`.

Discussion

15477.	Find the area of the region in the first quadrant enclosed by the y-axis,the line `y=x`and the circle `x^2+y^2=32 ,`using integration.
Answer» `r=sqrt32=4sqrt2` `y=r=4sqrt2` `2y^2=32` `y^2=16` `y=4` `area=int_0^4sqrt(32-x^2)-xdx` `=int_0^4sqrt((4R)^2-x^2)dx-int_0^4xdx` `=(4pi-8)-[x^2/2]_0^4` `=4pi-8-[(16-0)/2]` `=4pi-8-8` `=4pi-16`.

Discussion

15478.	If a line makes angles `alpha,beta,gamma`with the positive direction ofcoordinate axes, then write the value of `sin^2alpha+sin^2beta+sin^2gammadot`
Answer» Direction cosines of the given line is `cos alpha, cos beta,cos gamma`. Also, we know, sum of squares of direction cosines is `1`. `:. cos^2alpha+cos^2beta+cos^2gamma = 1` `=>1-sin^2alpha+1-sin^2beta+1-sin^2gamma = 1` `=>sin^2alpha+sin^2beta+sin^2gamma = 3-1` `=>sin^2alpha+sin^2beta+sin^2gamma = 2` So, required value is `2`.

Discussion

15479.	Let `L` denots value of `cos^(2)(alpha - beta)` if `sin2alpha + sin2beta = (1)/(2), cos2alpha + cos2beta = (sqrt(3))/(2)` and `M` denotes value of `(1)/(log_(xy)xyz) + (1)/(log_(yz)xyz) + (1)/(log_(zx)xyz)` then `16L^(2) + M^(2)` is
Answer» Correct Answer - 5 `1 + 1 + 2cos2(alpha - beta) = (1)/(4) + (3)/(4) = 1` `2 + 2(2cos^(2)(alpha - beta) - 1) = 1` `cos^(2)(alpha - beta) = (1)/(4) = L rArr L = (1)/(4)` `m = log_(xyz)xy + log_(xyz)yz + log_(xyz)(xyz)^(2) = 2` `:. 16L^(2) + m^(2) = 1 + 4 = 5`

Discussion

15480.	How much `AgBr` could dissolve in `1.0L` of `0.4M NH_(3)`? Assume that `Ag(NH_(3))_(2)^(o+)` is the only complex formed. Given: the dissociation constant for `Ag(NH_(3))_(2)^(o+) hArr Ag^(o+) xx 2NH_(3)`, `K_(d) = 6.0 xx 10^(-8)` and `K_(sp)(AgBr) =5.0 xx 10^(-13)`.A. `2.82xx10^(-3)M`B. `2.22xx10^(-4)`C. `2.22xx10^(-3)`D. `2.82xx10^(-4)`
Answer» Correct Answer - A `AgBr hArrAg_(("aq."))^(+)+Br_(("aq"))^(-)` `K_(sp)=[Ag^(+)][Br^(-)]` `Ag^(+) +2NH_(3) hArr[Ag(NH_(3))_(2)]^(+)` `K_(f)=([Ag(NH_(3))_(2)]^(+))/([Ag^(+)][NH_(3)]^(2))=1xx10^(8)` Let `s` be the solubility of `AgBr` in `NH_(3)` then `[Br^(-)]=[Ag^(+)]+[Ag(NH_(3))_(2)]^(+)=S` Since, almost all `Ag^(+)` in solution state passes to `[Ag(NH_(3))_(2)]^(+)` as `K_(f)` is very high. Thus, `[Ag^(+)ltltlt[Ag(NH_(3))_(2)]^(+)` `:.[Br^(-)]=[Ag(NH_(3))_(2)]^(+)=S` Now `K_(sp)=[Ag^(+)][Br^(-)]=([[Ag(NH_(3))_(2)]^(+)])/(K_(f)xx [NH_(3)]^(2))xx[Br^(-)]` Since `[NH_(3)]=0.4M` `5xx10^(-13)=(SxxS)/(1.0xx10^(8)xx(0.4)^(2))` `S^(2)=8xx10^(-6)` `S=2.82xx10^(-3)M`

Discussion

15481.	ABCD is a rhombus with ∠BAC = 56°. Determine ∠ ACD de
Answer» angle acd will be 56 because the opposite side of rhombus are parallel so by alternate angle these angle will be equal

Discussion

15482.	Find the inradius of an equilateral triangle of side 10 cm user r = A/s
Answer» Side of equilateral triangle is a = 10 cm. ∴ s = semi perimeter of triangle \(=\frac{a+b+c}{2}=\frac{10+10+10}{2}\) = 30/2 = 15 cm And K = Area of equilateral triangle \(=\frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\times10^2\) = 25√3 cm² Now, inradius of the equilateral triangle is r = K/s \(=\frac{25\sqrt3}{15}=\frac{5\sqrt3}{3}\,cm.\) or inradius = r \(=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\) \(=\sqrt{\frac{(15-10)(15-10)(15-10)}{15}}\) \(=\sqrt{\frac{5\times5\times5}{15}}=\sqrt{25/3}\) = 5/√3 = 5√3/3 cm

15482.

Find the inradius of an equilateral triangle of side 10 cm user r = A/s

Answer»

Side of equilateral triangle is a = 10 cm.

∴ s = semi perimeter of triangle \(=\frac{a+b+c}{2}=\frac{10+10+10}{2}\)

= 30/2 = 15 cm

And K = Area of equilateral triangle \(=\frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\times10^2\) = 25√3 cm²

Now, inradius of the equilateral triangle is r = K/s \(=\frac{25\sqrt3}{15}=\frac{5\sqrt3}{3}\,cm.\)

or

inradius = r \(=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\)

\(=\sqrt{\frac{(15-10)(15-10)(15-10)}{15}}\)

\(=\sqrt{\frac{5\times5\times5}{15}}=\sqrt{25/3}\) = 5/√3 = 5√3/3 cm

Discussion

15483.	Δ lf the perimeter of ΔABC is 10cm, calculate the perimeter of triangle PQR?
Answer» Perimeter of ΔPQR = 2 × 10 = 20 One side = 20/3 Area of ΔPQR = \(A=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}(\frac{20}{3})^2=\frac{100}{3\sqrt{3}}\)

15483.

Δ lf the perimeter of ΔABC is 10cm, calculate the perimeter of triangle PQR?

Answer»

Perimeter of ΔPQR = 2 × 10 = 20

One side = 20/3

Area of ΔPQR = \(A=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}(\frac{20}{3})^2=\frac{100}{3\sqrt{3}}\)

Discussion

15484.	Find the inradius of an equilateral triangle of side 10cm. User r = A/s
Answer» \(s=\frac{10+10+10}{2}=\frac{30}{2}=15\) \(A=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}10^2=25\sqrt{3}\) \(r=\frac{A}{s}\) \(=\frac{25\sqrt{3}}{15}=\frac{5}{\sqrt{3}}=2.9\) cm

15484.

Find the inradius of an equilateral triangle of side 10cm. User r = A/s

Answer»

\(s=\frac{10+10+10}{2}=\frac{30}{2}=15\)

\(A=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}10^2=25\sqrt{3}\)

\(r=\frac{A}{s}\)

\(=\frac{25\sqrt{3}}{15}=\frac{5}{\sqrt{3}}=2.9\) cm

Discussion

15485.	Area of a right triangle is 60 sq. centimetres and its inradius 3 cm. What is the length of its hypotenuse?
Answer» Area = 60 sq.cm, radius = 3 ∴ Perimeter = \(2\times\frac{60}{3}=40\) cm. Hypotenuse = \(\frac{40-6}{2}=17\) cm.

15485.

Area of a right triangle is 60 sq. centimetres and its inradius 3 cm. What is the length of its hypotenuse?

Answer»

Area = 60 sq.cm, radius = 3

∴ Perimeter = \(2\times\frac{60}{3}=40\) cm.

Hypotenuse = \(\frac{40-6}{2}=17\) cm.

Discussion

15486.	Let PQ be a tangent to a circle at A and AB be a chord. Let C be a point on the circle such that ∠BAC = 54° and ∠BAQ= 62°. Find ∠ABC.
Answer» ∠ABC = 180° – (∠BAC + ∠ACB) ∠ABC = 180° – (54° + 62°) = 64°

15486.

Let PQ be a tangent to a circle at A and AB be a chord. Let C be a point on the circle such that ∠BAC = 54° and ∠BAQ= 62°. Find ∠ABC.

Answer»

∠ABC = 180° – (∠BAC + ∠ACB)

∠ABC = 180° – (54° + 62°) = 64°

Discussion

15487.	Prove that the angles formed by the tangents from the endpoints of a chord are equal.
Answer» Each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord. [Angles in the alternate segments are equal]. ∴ ∠P = x ie ∠RBA = x. ie ∠QAB = ∠RBA

15487.

Prove that the angles formed by the tangents from the endpoints of a chord are equal.

Answer»

Each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord.

[Angles in the alternate segments are equal].

∴ ∠P = x ie ∠RBA = x.

ie ∠QAB = ∠RBA

Discussion

15488.	AB and AC are the tangents of the circle with center O.Show that the quadrilateral ABOC is a cyclic quadrilateral.
Answer» AB and AC are tangents ∠B = ∠C = 90° ∠B + ∠C = 180° The sum of the four angles of a quadrilateral is 360°. ie, ∠A + ∠O = 180° The opposite angles of quadrilateral ABOC are complementary. ∴ ABOC is a cyclic quadrilateral.

15488.

AB and AC are the tangents of the circle with center O.Show that the quadrilateral ABOC is a cyclic quadrilateral.

Answer»

AB and AC are tangents

∠B = ∠C = 90°

∠B + ∠C = 180°

The sum of the four angles of a quadrilateral is 360°.

ie, ∠A + ∠O = 180°

The opposite angles of quadrilateral ABOC are complementary.

∴ ABOC is a cyclic quadrilateral.

Discussion

15489.	In the following figure, PQ is the tangent and PB is the scent. Prove that PQ2 = PA × PB
Answer» Considering the triangles PAQ and PQB. ∠APQ = ∠BPQ (Common angle) ∠PQA = ∠QBP (Angle between tangent and chord = the angle made by the chord on its . complimentary arc) ΔPAQ ~ ΔPQB [A.A Similarly] ∴ \(\frac{PA}{PQ}=\frac{AQ}{QB}=\frac{PQ}{PB}\) [Ratio of the similar sides are equal] ∴ \(\frac{PA}{PQ}=\frac{PQ}{PB}\) PQ² = PA × PB

15489.

In the following figure, PQ is the tangent and PB is the scent. Prove that PQ2 = PA × PB

Answer»

Considering the triangles PAQ and PQB.

∠APQ = ∠BPQ (Common angle)

∠PQA = ∠QBP (Angle between tangent and chord = the angle made by the chord on its . complimentary arc)

ΔPAQ ~ ΔPQB

[A.A Similarly]

∴ \(\frac{PA}{PQ}=\frac{AQ}{QB}=\frac{PQ}{PB}\)

[Ratio of the similar sides are equal]

∴ \(\frac{PA}{PQ}=\frac{PQ}{PB}\)

PQ² = PA × PB

Discussion

15490.	In the figure, angles formed by the radius segment of the meeting points of the tangent to incircle are given. Find all angles of the triangle.
Answer» Angles 180 – 120 = 60° 180 -130 = 50° Third angle 180 – (60 + 50) = 70°

15490.

In the figure, angles formed by the radius segment of the meeting points of the tangent to incircle are given. Find all angles of the triangle.

Answer»

Angles 180 – 120 = 60°

180 -130 = 50°

Third angle 180 – (60 + 50) = 70°

Discussion

15491.	In the figure PA, PB and RS are the three tangents to the circle. Prove that the perimeter of ΔPRS is the sum of the lengths PA and PB.
Answer» PA = PB (Tangents from an outside point are equal) RA = RT ST = SB Perimeter of ΔPRS = PR + RS + PS = PR + (RT + ST) + PS = (PR + RA) + (SB + PS) (Since RT = RA, ST= SB) = PA + PB

15491.

In the figure PA, PB and RS are the three tangents to the circle. Prove that the perimeter of ΔPRS is the sum of the lengths PA and PB.

Answer»

PA = PB (Tangents from an outside point are equal)

RA = RT

ST = SB

Perimeter of ΔPRS = PR + RS + PS

= PR + (RT + ST) + PS

= (PR + RA) + (SB + PS)

(Since RT = RA, ST= SB)

= PA + PB

Discussion

15492.	In the figure PA, PB are tangents through A and B of a circle with center O. If the radius of the circle is r, then prove that OP × OQ = r2.
Answer» Δ OQA, ΔOPA are triangle with equal angles. The ratio of sides opposite to the equal angles. \(\frac{OA}{OP}=\frac{OQ}{OA}\,\)\(\Rightarrow \frac{r}{OP}=\frac{OQ}{r}\) \(r^2=OP\times OQ\) \(OP\times OQ=r^2\)

15492.

In the figure PA, PB are tangents through A and B of a circle with center O. If the radius of the circle is r, then prove that OP × OQ = r2.

Answer»

Δ OQA, ΔOPA are triangle with equal angles. The ratio of sides opposite to the equal angles.

\(\frac{OA}{OP}=\frac{OQ}{OA}\,\)\(\Rightarrow \frac{r}{OP}=\frac{OQ}{r}\)

\(r^2=OP\times OQ\)

\(OP\times OQ=r^2\)

Discussion

15493.	AB, BC, CD and AD are the tangents of the circle. If AP = x, BP = y, CR = z, SD =w then show that the perimeter of the quadrilateral ABCD is 2(x + y + z + w).
Answer» AP = AS = x; BP = BQ = y CR = CQ = z; SD = DR = w ie, AB = x + y BC = y + z; CD = z + w AD = x + z ∴ Perimeter = 2x + 2y + 2z + 2w = 2(x + y + z + w)

15493.

AB, BC, CD and AD are the tangents of the circle. If AP = x, BP = y, CR = z, SD =w then show that the perimeter of the quadrilateral ABCD is 2(x + y + z + w).

Answer»

AP = AS = x;

BP = BQ = y

CR = CQ = z;

SD = DR = w

ie, AB = x + y

BC = y + z;

CD = z + w

AD = x + z

∴ Perimeter = 2x + 2y + 2z + 2w = 2(x + y + z + w)

Discussion

15494.	In the figure, AC and BC are tangents to the circle from C. Centre of the circle O. i. Find ∠A ii. If ∠C is 2 times ∠O, then what is ∠C ?
Answer» i. ∠A= 90° ii. ∠C + ∠O = 180° ∠C = 60°

15494.

In the figure, AC and BC are tangents to the circle from C. Centre of the circle O. i. Find ∠A ii. If ∠C is 2 times ∠O, then what is ∠C ?

Answer»

i. ∠A= 90°

ii. ∠C + ∠O = 180°

∠C = 60°

Discussion

15495.	The radius of a circle touching all sides of an equilateral triangle is 3centimetres. Draw this triangle.
Answer» Draw a circle of radius 3 cm. Mark 120° at the center of circle. Complete the equilateral triangle.

15495.

The radius of a circle touching all sides of an equilateral triangle is 3centimetres. Draw this triangle.

Answer»

Draw a circle of radius 3 cm.

Mark 120° at the center of circle.

Complete the equilateral triangle.

Discussion

15496.	The density of a pure liquid (molecular mass =80) is 1.5 gm/ml. if 4 ml of liquid contains 60 drops then the number of molecules per drops of liquid is give by : [Given `N_(A)=6xx10^(23)`]A. `7.5xx10^(20)`B. `1.33 xx10^(21)`C. `4xx10^(23)`D. `2xx10^(22)`
Answer» Correct Answer - A `d=M/V` M=4x 1.5 =6gm 6 gm have 60 drops 1 drop will have 0.1 gm `n=0.1/8xx6xx10^(23) =7.5xx10^(20)` molecules

Discussion

15497.	A sample of oxygen atoms contain only `._(8)O^(16)` and `._(8)O^(18)` isotopes. If the average atomic mass of the sample is 16.8, then identify the options which is/are correct ?A. mass % of `O^(16)` is less then 60 %B. mole % of `O^(18)` is 40%C. average number of neutron/atom is equal to 8.8D. average number of neutron/proton is equal to 1.1
Answer» Correct Answer - ABCD `{:(._(8)OO^(16),:,._(8)O^(18)),(x,:,100-x),(60,:,40):}` `M_(avg.) =16.8=(xxx15+18(100-x))/100` x=60 Average number of neutron `=(8xx6+16xx4)/10 =8.8` Average number of proton =8 `=n/p=8.8/8=1.1`

Discussion

15498.	Two different formulas are used in order to represent composition of any miolecule, empirical formula and molecular formula . While the fomer gives an idea of relative ratio of number of atoms, latter gives the exact number of atoms in the molecule. 4.6 gm of an organic compound on complete combustion gave 8.8 gm of `CO_(2)(g)` and 5.4 gm of `H_(2)O(g)` only and no other products . what will be the empirical formula of the hydrocarbon?A. `CH_(3)`B. `C_(2)H_(6)O`C. `CH_(2)O`D. `CH_(2)`
Answer» Correct Answer - A or B `{:(C_(2)H_(6)O,+3O_(2) to ,2CO_(2)(g),+3H_(2)O),("0.1 mole","0.2 mole","0.3 mole"):}`

Discussion

15499.	Two different formulas are used in order to represent composition of any miolecule, empirical formula and molecular formula . While the fomer gives an idea of relative ratio of number of atoms, latter gives the exact number of atoms in the molecule. A 62 gm sample of a substance consist of 2 gm hydrogen , 28 gm nitrogen and remaining oxygen . What will b its empirical formula?A. `HNO_(2)`B. HNOC. `HNO_(3)`D. `HNO_(4)`
Answer» Correct Answer - B `H:N:O` `2/1:26/14:32/16` 2:2:2 H:N:O

Discussion

15500.	Two different formulas are used in order to represent composition of any miolecule, empirical formula and molecular formula . While the fomer gives an idea of relative ratio of number of atoms, latter gives the exact number of atoms in the molecule. An organic compound contains C N and O . The number of oxygen atom is same as that of nitogen atom which is one third of number of carbon atoms and number of hydrogen atoms is approximately 2.33 times of carbon atoms. If vapour density of the compound is 73 then molecular formula of the compound will be :A. `C_(3)H_(7)NO`B. `C_(2)H_(5)NO`C. `C_(6)H_(14)N_(2)O_(2)`D. `C_(5)H_(12)N_(3)O_(2)`
Answer» Correct Answer - C `O=N=C/3` H=2.33 C `H=7/3 C` `H/C=7/3` EFM =73 MFM =2 x 73 Molecular formula `=C_(6)H_(14)N_(2)O_(2)`

Discussion

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