100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	`20 ml` of an `H_2 O_2` solution on reaction with excess of acidified `KMnO_4` released `224` cc of `O_2`. What is the volume strength of that `H_2 O_2` ?A. 5.6 volB. 11.2 volC. 22.4 volD. 2.8 vol
Answer» Correct Answer - A `2MnO_4^(-) + 5H_2 O_2 + 6H^+ rarr 2 Mn^(2+) + 5O_2 + 8 H_2O` `22,400 "ml of" O_2 -= 34 g of H_2 O_2` `224` ml of `O_2 -= 0.34 g of H_2 O_2` Normality `= (0.34)/(17) xx (1000)/(20) = 1` Volume strength `= (5.6) (N) = 5.6 Vol`.

Discussion

2.	Choose the word from the options given to form a compound word with “drop”. (a) heave (b) mock (c) eaves (d) out
Answer» Correct answer is (c) leaves

Discussion

3.	Choose the word from the options given to form a compound word with “out” , (a) in (b) wipe(c) weather (d) hind
Answer» Correct answer is (b) wipe

Discussion

4.	There is no bread left because we ______ it all. A) have eaten B) has been eaten C) had eaten D) have been eaten
Answer» Correct option is A) have eaten

Discussion

5.	[" of "0],[CH_(3)-c-NH_(2)+H_(2)O]
Answer» Reduce Zero Numerator `0 + 1 / 20` Remove Adding Zero `1 / 20` 0.05

Discussion

6.	What are the optimum temperature and pH at which enzymes are highly active ?
Answer» (Hint : Temperature 298-310 K and pH 5 to 7)

Discussion

7.	The men were to ______ but the manager decided to give them a second change. A) have been dismissed B) dismissed C) being dismissed D) be dismissing
Answer» Correct option is A) have been dismissed

Discussion

8.	Name the enzyme which converts milk into curd.
Answer» (Hint : Lactobacilli.)

Discussion

9.	(i) \( \int_{0}^{\pi} \theta \sin ^{2} \theta \cos \theta d \theta \)
Answer» \(\int_0^\pi \theta\,sin^2\theta\,cos\theta\,d\theta\) = \(\int_0^\pi \theta\,(1-cos^2\theta)\,cos\theta\,d\theta\) = \(\int_0^\pi \theta\,cos\theta\,d\theta\) - \(\int_0^\pi \theta\,cos^3\theta\,d\theta\) = \([\theta\int cos\theta\,d\theta]_0^{\pi}\) - \(\int_0^{\pi}(1.\int cos^3\theta\,d\theta)\) = \([\theta\,sin\theta]_0^{\pi}\) - \(\int_0^\pi \theta\,sin\theta\,d\theta\) - \([\theta(sin\theta - \frac{sin^3\theta}{3})]_0^{\pi}\) + \(\int_0^\pi(sin\theta - \frac{sin^3\theta}{3})d\theta\) (∵\(\int cos^3\theta\,d\theta\) = \(\int (1-sin^2\theta)cos\theta\,d\theta\) = \(\int cos\theta\,d\theta\) - \(\int sin^2\theta\,cos\theta\,d\theta\) = \(sin\theta-\frac{sin^3\theta}{3}\)) = (π sinπ - 0) + \([cos\theta]_0^{\pi}\) - (π(sinπ - \(\frac{sin^3\pi}{3}\)) - 0) - \([cos\theta]_0^{\pi}\) - \(\frac{2}{3}\)\(\int_0^{\pi/2} sin^3\theta\,d\theta\) = 0 - 0 - \(\frac{2}{3}\)\(\frac{\Gamma(\frac{3+1}{2})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{2})}\) (\(\int_0^{\pi/2} sin^m\theta\,cos^n \theta d\theta\) = \(\frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+1}{2})}\)) = \(\frac{2}{3}\)\(\frac{\Gamma(2)\Gamma(\frac{1}{2})}{2.\frac{3}{2}.\frac{1}{2}\Gamma(\frac{1}{2})}\) (∵ \(\Gamma(n+1)=n\Gamma(n)\)) = \(-\frac{2}{3}\times \frac{2}{3}\times 1!\) (\(\Gamma(2) = 1!=1\)) = \(\frac{-4}{9}\).

9.

(i) \( \int_{0}^{\pi} \theta \sin ^{2} \theta \cos \theta d \theta \)

Answer»

\(\int_0^\pi \theta\,sin^2\theta\,cos\theta\,d\theta\)

= \(\int_0^\pi \theta\,(1-cos^2\theta)\,cos\theta\,d\theta\)

= \(\int_0^\pi \theta\,cos\theta\,d\theta\) - \(\int_0^\pi \theta\,cos^3\theta\,d\theta\)

= \([\theta\int cos\theta\,d\theta]_0^{\pi}\) - \(\int_0^{\pi}(1.\int cos^3\theta\,d\theta)\)

= \([\theta\,sin\theta]_0^{\pi}\) - \(\int_0^\pi \theta\,sin\theta\,d\theta\) - \([\theta(sin\theta - \frac{sin^3\theta}{3})]_0^{\pi}\) + \(\int_0^\pi(sin\theta - \frac{sin^3\theta}{3})d\theta\)

(∵\(\int cos^3\theta\,d\theta\) = \(\int (1-sin^2\theta)cos\theta\,d\theta\)

= \(\int cos\theta\,d\theta\) - \(\int sin^2\theta\,cos\theta\,d\theta\)

= \(sin\theta-\frac{sin^3\theta}{3}\))

= (π sinπ - 0) + \([cos\theta]_0^{\pi}\) - (π(sinπ - \(\frac{sin^3\pi}{3}\)) - 0) - \([cos\theta]_0^{\pi}\) - \(\frac{2}{3}\)\(\int_0^{\pi/2} sin^3\theta\,d\theta\)

= 0 - 0 - \(\frac{2}{3}\)\(\frac{\Gamma(\frac{3+1}{2})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{2})}\)

(\(\int_0^{\pi/2} sin^m\theta\,cos^n \theta d\theta\) = \(\frac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+1}{2})}\))

= \(\frac{2}{3}\)\(\frac{\Gamma(2)\Gamma(\frac{1}{2})}{2.\frac{3}{2}.\frac{1}{2}\Gamma(\frac{1}{2})}\)

(∵ \(\Gamma(n+1)=n\Gamma(n)\))

= \(-\frac{2}{3}\times \frac{2}{3}\times 1!\)

(\(\Gamma(2) = 1!=1\))

= \(\frac{-4}{9}\).

Discussion

10.	(ii) \( \int_{0}^{\pi} x \sin ^{6} x \cos ^{4} x d x \)
Answer» Let I = \(\int\limits_0^\pi x\,sin^6x\,cos^4x\,dx\) ....(1) ∴ I = \(\int\limits_0^\pi (\pi -x)\,sin^6(\pi-x)\,cos^4(\pi-x)\,dx\) ....(2) ∴ 2I = \(\int\limits_0^\pi \pi\,sin^6x\,cos^4x\,dx\) (By adding (1) & (2)) ⇒ I = \(\frac{\pi}{2}\int\limits_0^\pi\,sin^6x\,cos^4x\,dx\) = \(\frac{\pi}{2}\) x 2\(\int\limits_0^\pi\,sin^6x\,cos^4x\,dx\) = π \(\frac{\Gamma(\frac{6+1}{2}).\Gamma(\frac{4+1}{2})}{2\Gamma(\frac{6+4+2}{2})}\)(By gamma function) = \(\frac{\pi}{2}\) \(\frac{\Gamma(\frac{7}{2}).\Gamma(\frac{5}{2})}{\Gamma(6)}\) = \(\frac{\pi}{2}\)\(\frac{\frac{5}{2}.\frac{3}{2}.\Gamma(\frac{1}{2})\times \frac{3}{2}.\frac{1}{2}.\Gamma(\frac{1}{2})}{5!}\) (\(\Gamma(n) = (n-1)!\) n ∈ N & \(\Gamma(n+1)=n\Gamma(n)\)) = \(\frac{\pi}{2}\) x \(\frac{45}{32}\) x \(\frac{1}{120}\) √π x √π (∵ \(\Gamma(\frac{1}{2})= \sqrt {\pi}\)) = \(\frac{3\pi^2}{512}\)

10.

(ii) \( \int_{0}^{\pi} x \sin ^{6} x \cos ^{4} x d x \)

Answer»

Let I = \(\int\limits_0^\pi x\,sin^6x\,cos^4x\,dx\) ....(1)

∴ I = \(\int\limits_0^\pi (\pi -x)\,sin^6(\pi-x)\,cos^4(\pi-x)\,dx\) ....(2)

∴ 2I = \(\int\limits_0^\pi \pi\,sin^6x\,cos^4x\,dx\) (By adding (1) & (2))

⇒ I = \(\frac{\pi}{2}\int\limits_0^\pi\,sin^6x\,cos^4x\,dx\)

= \(\frac{\pi}{2}\) x 2\(\int\limits_0^\pi\,sin^6x\,cos^4x\,dx\)

= π \(\frac{\Gamma(\frac{6+1}{2}).\Gamma(\frac{4+1}{2})}{2\Gamma(\frac{6+4+2}{2})}\)(By gamma function)

= \(\frac{\pi}{2}\) \(\frac{\Gamma(\frac{7}{2}).\Gamma(\frac{5}{2})}{\Gamma(6)}\)

= \(\frac{\pi}{2}\)\(\frac{\frac{5}{2}.\frac{3}{2}.\Gamma(\frac{1}{2})\times \frac{3}{2}.\frac{1}{2}.\Gamma(\frac{1}{2})}{5!}\)

(\(\Gamma(n) = (n-1)!\) n ∈ N & \(\Gamma(n+1)=n\Gamma(n)\))

= \(\frac{\pi}{2}\) x \(\frac{45}{32}\) x \(\frac{1}{120}\) √π x √π (∵ \(\Gamma(\frac{1}{2})= \sqrt {\pi}\))

= \(\frac{3\pi^2}{512}\)

Discussion

11.	Find the sum of `h+2e^(2)h+2e^(4)h+2e^(6)h +.....`
Answer» `h[1+2e^(2)(1+e^(2)+e^(4)+.....)]` `=h[1+2e^(2)((1)/(1-e^(2)))]` `=h((1+e^(2))/(1-e^(2)))`

Discussion

12.	Solve `x^(2)+x-2=0`
Answer» For quadratic equation ` ax^(2)+bx +c=0` `x=(-b+-sqrt(b^(2)-4ac))/(2a)` So, for given equation ` x^(2)+x-2=0` `x=(-1+-sqrt(1+8))/(2)=(-1+-3)/(2)=1,-2`

Discussion

13.	Find the directional derivative of q = 4x3 - 3x2 * y2 (2.-1.1) along the line which makes equal angles with Co-ordinate axes.
Answer» q = 4x³ - 3x²y² \(\vec ∇q\) = (\(\hat i \frac{\delta}{\delta x}\) + \(\hat j \frac{\delta}{\delta y}\) + \(\hat k \frac{\delta}{\delta z}\))q = \(\hat i \frac{\delta}{\delta x}\)(4x³ - 3x²y²) + \(\frac{j\delta}{\delta y}\)(4x³ - 3x²y²) + \(0\hat k\) = \(\hat i\) (12x² - 6xy²) + \(\hat j\)(-6x²y) Direction cosines of line which makes equal angles with coordinate axis are (±\(\frac{1}{\sqrt 3}\), ±\(\frac{1}{\sqrt 3}\),±\(\frac{1}{\sqrt 3}\)). ∴ Direction derivative of q in direction of line is = (\(\vec ∇q\)).(\(\frac{1}{\sqrt3}\hat i\) + \(\frac{1}{\sqrt3}\hat j\) + \(\frac{1}{\sqrt3}\hat k\)) = \(\hat i\) (\(12x^2-6xy^2\)) + \(\hat j\) (-6x²y)(\(\frac{1}{\sqrt3}\hat i\) + \(\frac{1}{\sqrt3}\hat j\) + \(\frac{1}{\sqrt3}\hat k\)) = \(\frac{1}{\sqrt 3}\)\((12x^2-6xy^2)\) - \(\frac{6x^2y}{\sqrt 3}\) Direction derivative at (2,-1,1) = \(\frac{1}{\sqrt 3}\)(12 x 4 - 6 x 2 x 1) - \(\frac{6\times 2^2\times -1}{\sqrt 3}\) = \(\frac{1}{\sqrt 3}\)(48 - 12) + \(\frac{24}{\sqrt 3}\) = \(\frac{60}{\sqrt 3}\) = \(20\sqrt 3\).

13.

Find the directional derivative of q = 4x3 - 3x2 * y2 (2.-1.1) along the line which makes equal angles with Co-ordinate axes.

Answer»

q = 4x³ - 3x²y²

\(\vec ∇q\) = (\(\hat i \frac{\delta}{\delta x}\) + \(\hat j \frac{\delta}{\delta y}\) + \(\hat k \frac{\delta}{\delta z}\))q

= \(\hat i \frac{\delta}{\delta x}\)(4x³ - 3x²y²) + \(\frac{j\delta}{\delta y}\)(4x³ - 3x²y²) + \(0\hat k\)

= \(\hat i\) (12x² - 6xy²) + \(\hat j\)(-6x²y)

Direction cosines of line which makes equal angles with coordinate axis are (±\(\frac{1}{\sqrt 3}\), ±\(\frac{1}{\sqrt 3}\),±\(\frac{1}{\sqrt 3}\)).

∴ Direction derivative of q in direction of line is

= (\(\vec ∇q\)).(\(\frac{1}{\sqrt3}\hat i\) + \(\frac{1}{\sqrt3}\hat j\) + \(\frac{1}{\sqrt3}\hat k\))

= \(\hat i\) (\(12x^2-6xy^2\)) + \(\hat j\) (-6x²y)(\(\frac{1}{\sqrt3}\hat i\) + \(\frac{1}{\sqrt3}\hat j\) + \(\frac{1}{\sqrt3}\hat k\))

= \(\frac{1}{\sqrt 3}\)\((12x^2-6xy^2)\) - \(\frac{6x^2y}{\sqrt 3}\)

Direction derivative at (2,-1,1) = \(\frac{1}{\sqrt 3}\)(12 x 4 - 6 x 2 x 1) - \(\frac{6\times 2^2\times -1}{\sqrt 3}\)

= \(\frac{1}{\sqrt 3}\)(48 - 12) + \(\frac{24}{\sqrt 3}\)

= \(\frac{60}{\sqrt 3}\)

= \(20\sqrt 3\).

Discussion

14.	If \( y=x^{\tan x}+(\sin x)^{\cos x} \), Find \( \frac{d y}{d x} \)
Answer» y = x^tanx+ (sinx)^cosx Let y₁ = x^tanx& y₂ = (sinx)^cosx ∴ \(\frac{dy}{dx}\)= \(\frac{dy_1}{dx}\) + \(\frac{dy_2}{dx}\) .........(1) Now, logy₁ = tanx loga & logy₂ = cosx log sinx (∵ log a^b = blog a) ∴ \(\frac{1}{y_1}\frac{dy_1}{dx}\) = \(\frac{tanx}{x}\) + sec²x logx ⇒ \(\frac{dy_1}{dx}\) = (\(\frac{tanx}{x}\) + sec²x logx) x^tanx & \(\frac{1}{y_2}\frac{dy_2}{dx}\) = \(\frac{cosx}{sinx}\times cosx\) - sinx log sinx ⇒ \(\frac{dy_2}{dx}\) = (cosx cotx - sinx log sinx)(sinx)^cosx From (1), \(\frac{dy}{dx}\) = (\(\frac{tanx}{x}\) + sec²x logx) x^tanx+ (cosx cotx - sinx log sinx)(sinx)^cosx

14.

If \( y=x^{\tan x}+(\sin x)^{\cos x} \), Find \( \frac{d y}{d x} \)

Answer»

y = x^tanx+ (sinx)^cosx

Let y₁ = x^tanx& y₂ = (sinx)^cosx

∴ \(\frac{dy}{dx}\)= \(\frac{dy_1}{dx}\) + \(\frac{dy_2}{dx}\) .........(1)

Now,

logy₁ = tanx loga & logy₂ = cosx log sinx

(∵ log a^b = blog a)

∴ \(\frac{1}{y_1}\frac{dy_1}{dx}\) = \(\frac{tanx}{x}\) + sec²x logx

⇒ \(\frac{dy_1}{dx}\) = (\(\frac{tanx}{x}\) + sec²x logx) x^tanx

& \(\frac{1}{y_2}\frac{dy_2}{dx}\) = \(\frac{cosx}{sinx}\times cosx\) - sinx log sinx

⇒ \(\frac{dy_2}{dx}\) = (cosx cotx - sinx log sinx)(sinx)^cosx

From (1),

\(\frac{dy}{dx}\) = (\(\frac{tanx}{x}\) + sec²x logx) x^tanx+ (cosx cotx - sinx log sinx)(sinx)^cosx

Discussion

15.	Simplify: \(\frac1{3-\sqrt8}-\frac1{\sqrt8-\sqrt7}+\frac1{\sqrt7-\sqrt6}\)\(-\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}\)
Answer» \(\frac1{3-\sqrt8}-\frac1{\sqrt8-\sqrt7}+\frac1{\sqrt7-\sqrt6}\)\(-\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}\) \(= \frac{3+\sqrt8}{9-8}-\frac{\sqrt8+\sqrt7}{8-7}+\frac{\sqrt7+\sqrt6}{7-6}\)\(-\frac{\sqrt6+\sqrt5}{6-5}+\frac{\sqrt5+2}{5-4}\) = (3 + √8) - (√8 + √7) + (√7 + √6) - (√6 - √5) + (√5 + 2) = 3 + 2 = 5

15.

Simplify: \(\frac1{3-\sqrt8}-\frac1{\sqrt8-\sqrt7}+\frac1{\sqrt7-\sqrt6}\)\(-\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}\)

Answer»

\(\frac1{3-\sqrt8}-\frac1{\sqrt8-\sqrt7}+\frac1{\sqrt7-\sqrt6}\)\(-\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}\)

\(= \frac{3+\sqrt8}{9-8}-\frac{\sqrt8+\sqrt7}{8-7}+\frac{\sqrt7+\sqrt6}{7-6}\)\(-\frac{\sqrt6+\sqrt5}{6-5}+\frac{\sqrt5+2}{5-4}\)

= (3 + √8) - (√8 + √7) + (√7 + √6) - (√6 - √5) + (√5 + 2)

= 3 + 2 = 5

Discussion

16.	Write four numbers which are divisible by 6 but not by 12.
Answer» Multiples of 6 are 6,12,18,24,30,36,42,48,54,60,66……………………… Multiples of 12 are 12,24,36,48,60,72,84,96,108,120,132……………….. Here 6,18,30,42,54,66………… are not multiples of 12. So if a number is divisible by 6,it is not always divisible by 12.

16.

Write four numbers which are divisible by 6 but not by 12.

Answer»

Multiples of 6 are 6,12,18,24,30,36,42,48,54,60,66………………………

Multiples of 12 are 12,24,36,48,60,72,84,96,108,120,132………………..

Here 6,18,30,42,54,66………… are not multiples of 12.

So if a number is divisible by 6,it is not always divisible by 12.

Discussion

17.	Subtract:4x by x2 - 1 - x + 1 by x - 1.
Answer» \(\frac{4x}{x^2-1}-\frac{x+1}{x-1}\) \(=\frac{4x-(x+1)^2}{x^2-1}\) \(=\frac{4x-(x^2+2x+1)}{x^2-1}\) \(=\frac{-(x^2+2x+1-4x)}{x^2-1}\) \(=-\frac{x^2-2x+1}{x^2-1}\) \(=-\frac{(x-1)^2}{x^2-1}\) \(=\frac{(x-1)^2}{1-x^2}\)

17.

Subtract:4x by x2 - 1 - x + 1 by x - 1.

Answer»

\(\frac{4x}{x^2-1}-\frac{x+1}{x-1}\)

\(=\frac{4x-(x+1)^2}{x^2-1}\)

\(=\frac{4x-(x^2+2x+1)}{x^2-1}\)

\(=\frac{-(x^2+2x+1-4x)}{x^2-1}\)

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

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

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

Discussion

18.	Smallest prime number
Answer» 2 is the smallest prime number.

Discussion

19.	Given, `y=sin 2x` . Then find `(dy)/(dx)` .
Answer» `(dy)/(dx)=(d)/(dx) (sin 2x)` `=(d(sin u))/(du)xx(du)/(dx)` Here , u=2x `:. " " (du)/(dx)=(d(2x))/(dx)=2 "and " (d(sin u))/(du)=cos u` `:. " " (dy)/(dx) =cos u xx2 ` `=2 cos 2x ` Or `(dy)/(dx)=(d)/(dx)(sin 2x)` `=(d)/(d(2x)) sin (2x) xx (d(2x))/(dx)` ` cos 2x xx 2 = cos 2x`

Discussion

20.	Solve: x4 - 13x2 + 42 = 0
Answer» x⁴ - 13x² + 42 = 0 x⁴ - 7x² - 6x² + 42 = 0 ⇒ x²(x² - 7) - 6(x² - 7) = 0 ⇒ (x² - 7) (x² - 6) = 0 ⇒ (x² - 7) = 0 or x² - 6 = 0 ⇒ x = \(\pm\sqrt7\) or x = \(\pm\sqrt6\) Hence, roots or given equation are -√6, √6, -√7 & √7.

20.

Solve: x4 - 13x2 + 42 = 0

Answer»

x⁴ - 13x² + 42 = 0

x⁴ - 7x² - 6x² + 42 = 0

⇒ x²(x² - 7) - 6(x² - 7) = 0

⇒ (x² - 7) (x² - 6) = 0

⇒ (x² - 7) = 0 or x² - 6 = 0

⇒ x = \(\pm\sqrt7\) or x = \(\pm\sqrt6\)

Hence, roots or given equation are -√6, √6, -√7 & √7.

Discussion

21.	7. The length of a field is 10 meter more than the breadth of the field. If perimeter of the field is 100 \( m \), find the dimensions of the field.
Answer» please answer here

Discussion

22.	Given, `y=(ax+b)^(2)` , evaluate ` (dy)/(dx)`.
Answer» Method I : Substituting (ax+b)=u Then `" "(du)/(dx)=(d(ax+b))/(dx)=a` and `" "(dy)/(du)=(d(ax+b)^(2))/(du)=(d(u)^(2))/(du)=2u` `:. " " (dy)/(dx)=(dy)/(du) xx (du)/(dx)=2u xx a =2 (ax+b)a` `" " =2a(ax+b)` Method II : `y=(ax+b)^(2)` `=a^(2)x^(2)+b^(2)+2abx` `"Then" " " (dy)/(dx)=(d)/(dx)[a^(2)x^(2)+b^(2) + 2abx]` `=(d)/(dx) (a^(2)x^(2))+(d)/(dx)(b^(2)) +(d)/(dx) (2abx)` `=a^(2)=(d(x)^(2))/(dx)+ 0+ 2ab(dx)/(dx)` `=a^(2)xx2x+2ab xx1` `=2a(ax+b)`

Discussion

23.	)42) If \( \alpha \) and \( \beta \) are the roots of the equation \( x^{2}+x-4=0 \) form the quadratic equation whose roots are \( \alpha^{2} \) and \( \beta^{2} \).
Answer» given-: a+b=-1............ eqn ¹ ab=-4. ............. eqn ² on squaring eqn ¹ a²+b²+2ab=1 a²+b²=9 on squaring eqn ² a²b²=16 hence the eqn is x²+9x+16=0 ♥

Discussion

24.	PBA and PDC are two secants. AD is the diameter of the circle with center at O. ∠A = 40°, ∠P = 20°. Find the measure of ∠DBC.(a) 30° (b) 45° (c) 50° (d) 40°
Answer» Answer : (a) 30º In ∆ ADP, ext. ∠ADC = Int. ∠s (∠A + ∠P) = 40° + 20° = 60°. ∠ABC = ∠ADC = 60° (Angles in the same segment) \(\because\) AD is the diameter, ∠ABD = 90° ∴ ∠DBC = ∠ABD – ∠ABC = 90° – 60° = 30°

24.

PBA and PDC are two secants. AD is the diameter of the circle with center at O. ∠A = 40°, ∠P = 20°. Find the measure of ∠DBC.(a) 30° (b) 45° (c) 50° (d) 40°

Answer»

Answer : (a) 30º

In ∆ ADP, ext. ∠ADC = Int. ∠s (∠A + ∠P)

= 40° + 20° = 60°.

∠ABC = ∠ADC = 60° (Angles in the same segment)

\(\because\) AD is the diameter, ∠ABD = 90°

∴ ∠DBC = ∠ABD – ∠ABC = 90° – 60°

= 30°

Discussion

25.	If tanA = \(\frac{3}{4}\), find the value of \(\frac{1}{sinA}+\frac{1}{cosA}\)
Answer» tanA = \(\frac{3}{4}\) = \(\frac{3k}{4k}\) sinA = \(\frac{3k}{5k}\) = \(\frac{3}{5}\), cosA = \(\frac{4k}{5k}\) = \(\frac{4}{5}\) \(\frac{1}{sinA}+\frac{1}{cosA}\) = \(\frac{5}{3}+\frac{5}{4}\) = \(\frac{(20+15)}{12}\) = \(\frac{35}{12}\)

25.

If tanA = \(\frac{3}{4}\), find the value of \(\frac{1}{sinA}+\frac{1}{cosA}\)

Answer»

tanA = \(\frac{3}{4}\) = \(\frac{3k}{4k}\)

sinA = \(\frac{3k}{5k}\) = \(\frac{3}{5}\), cosA = \(\frac{4k}{5k}\) = \(\frac{4}{5}\)

\(\frac{1}{sinA}+\frac{1}{cosA}\)

= \(\frac{5}{3}+\frac{5}{4}\)

= \(\frac{(20+15)}{12}\)

= \(\frac{35}{12}\)

Discussion

26.	40) Find the square root of: \( \left(6 x^{2}+x-1\right)\left(3 x^{2}+2 x-1\right)\left(2 x^{2}+3 x+1\right) \)
Answer» Let f(x) = (6x²+ x - 1) (3x² + 2x - 1) (2x² + 3x +1) = (3x-1)(2x+1)(3x-1)(x+1)(2x+1)(x+1) = (3x+1)²(2x+1)²(x+1)² ∴ square roots f(x) is \(\sqrt{f(x)}\) = (3x-1)(2x+1)(x+1).

26.

40) Find the square root of: \( \left(6 x^{2}+x-1\right)\left(3 x^{2}+2 x-1\right)\left(2 x^{2}+3 x+1\right) \)

Answer»

Let f(x) = (6x²+ x - 1) (3x² + 2x - 1) (2x² + 3x +1)

= (3x-1)(2x+1)(3x-1)(x+1)(2x+1)(x+1)

= (3x+1)²(2x+1)²(x+1)²

∴ square roots f(x) is \(\sqrt{f(x)}\) = (3x-1)(2x+1)(x+1).

Discussion

27.	39) Simplify: \( \frac{a^{2}-16}{a^{3}-8} \times \frac{2 a^{2}-3 a-2}{2 a^{2}+9 a+4} \div \frac{3 a^{2}-11 a-4}{a^{2}+2 a+4} \)
Answer» \(\frac{a^2+6}{a^3-8}\) x \(\frac{2a^2-3a-2}{2a^2+9a+4}\) ÷ \(\frac{3a^2-11a-4}{a^2+2a+4}\) = \(\frac{(a+4)(a-4)}{(a-2)(a^2+2a+4)}\) x \(\frac{(2a+1)(a-2)}{(2a+1)(a+4)}\) x \(\frac{(a+2)^2}{(3a+1)(a-4)}\) = \(\frac{(a+2)^2}{(a^2+2a+4)(3a+1)}\)

27.

39) Simplify: \( \frac{a^{2}-16}{a^{3}-8} \times \frac{2 a^{2}-3 a-2}{2 a^{2}+9 a+4} \div \frac{3 a^{2}-11 a-4}{a^{2}+2 a+4} \)

Answer»

\(\frac{a^2+6}{a^3-8}\) x \(\frac{2a^2-3a-2}{2a^2+9a+4}\) ÷ \(\frac{3a^2-11a-4}{a^2+2a+4}\)

= \(\frac{(a+4)(a-4)}{(a-2)(a^2+2a+4)}\) x \(\frac{(2a+1)(a-2)}{(2a+1)(a+4)}\) x \(\frac{(a+2)^2}{(3a+1)(a-4)}\)

= \(\frac{(a+2)^2}{(a^2+2a+4)(3a+1)}\)

Discussion

28.	Find the area of a square,whose perimeter is 64 m.
Answer» The perimeter of square is 4a=64 a=16 The area of square is a² =16×16 =256cm²

Discussion

29.	The age of sita and her sister are in the ratio of 3:8. after 12 years the ratio of their ages is 7:12. find there present ages.
Answer» Let the age of Sita and her sister be 3x and 8x respectively. After 12 years, Sita's age = 3x + 12 and her sister's age = 8x + 12 A/Q, 3x + 12 / 8x + 12 = 7 /12 36x + 144 = 56x + 84 60 = 20x x = 3, Sita's age = 3x = 9 years Sister's age = 8x = 24 years. Therefore, the ages of Sita and her sister are 9 and 24 years respectively.

29.

The age of sita and her sister are in the ratio of 3:8. after 12 years the ratio of their ages is 7:12. find there present ages.

Answer»

Let the age of Sita and her sister be 3x and 8x respectively.

After 12 years,

Sita's age = 3x + 12 and her sister's age = 8x + 12

A/Q,

3x + 12 / 8x + 12 = 7 /12

36x + 144 = 56x + 84

60 = 20x

x = 3,

Sita's age = 3x = 9 years

Sister's age = 8x = 24 years.

Therefore, the ages of Sita and her sister are 9 and 24 years respectively.

Discussion

30.	\( =\left(\frac{3}{5}\right)^{11} \div\left(\frac{3}{5}\right)^{6} \)
Answer» = a^m÷ aⁿ= am-n = (3/5) ^11-6 = (3/5)⁵

Discussion

31.	38 If \( A=\frac{x}{x+1} B=\frac{1}{x+1} \) prove that \( \frac{(A+B) 2+(A-B) 2}{A \div B}=\frac{2(x 2+1)}{x(x+1) 2} \)
Answer» We have, A = \(\frac{x}{x+1}\), B = \(\frac{1}{x+1}\) \(\frac{(A+B)^2+(A-B)^2}{A\,÷\,B}\) = \(\frac{B[(A^2+B^2+2AB)+(A^2+B^2-2AB)]}{A}\) = \(\frac{2B(A^2+B^2)}{A}\) = \(\frac{\frac{2}{x+1}((\frac{x}{x+1})^2+(\frac{1}{x+1})^2)}{\frac{x}{x+1}}\) = \(\frac{2}{x}\)\((\frac{x^2+1}{(x+1)^2})\) = \(\frac{2(x^2+1)}{x(x+1)^2}\) Hence proved.

31.

38 If \( A=\frac{x}{x+1} B=\frac{1}{x+1} \) prove that \( \frac{(A+B) 2+(A-B) 2}{A \div B}=\frac{2(x 2+1)}{x(x+1) 2} \)

Answer»

We have,

A = \(\frac{x}{x+1}\),

B = \(\frac{1}{x+1}\)

\(\frac{(A+B)^2+(A-B)^2}{A\,÷\,B}\) = \(\frac{B[(A^2+B^2+2AB)+(A^2+B^2-2AB)]}{A}\)

= \(\frac{2B(A^2+B^2)}{A}\)

= \(\frac{\frac{2}{x+1}((\frac{x}{x+1})^2+(\frac{1}{x+1})^2)}{\frac{x}{x+1}}\)

= \(\frac{2}{x}\)\((\frac{x^2+1}{(x+1)^2})\)

= \(\frac{2(x^2+1)}{x(x+1)^2}\)

Hence proved.

Discussion

32.	Find the area of each of the following triangles:(i)(ii)(iii)
Answer» "QUESTION INCOMPLETE "

Discussion

33.	A force F=(a+bx) acts on a particle in x direction where a and b are constants . Find the work done by this force during displacement from `x_(1)` to `x_(2)` .
Answer» Work done for displacement dx `dW=Fdx=(a+bx)dx` Total work done `W= int dW` `=int_(x_(1))^(x_(2))(a+bx)dx` `=[ax+(bx^(2))/(2)]_(x_(1))^(x_(2))` ` =a(x_(2)-x_(1))+(b)/(2)(x_(2)^(2) -x_(1)^(2))`

Discussion

34.	Find value of `(104)^(1//2)` using binomial approx.
Answer» `(104)^(1//2)=(100 +4)^(1//2)=10 [ 1 + 0.04]^(1//2)` ` =10 [ 1+0.02]=10.2`

Discussion

35.	Find a quadratic polynomial whose zeroes are 5 - 3√2 and 5 + 3√2.
Answer» Sum of zeroes = 5 - 3√2 + 5 + 3√2 = 10 Product of zeroes= (5 - 3√2)(5 + 3√2) = 7 P(x)= x² - 10x + 7

35.

Find a quadratic polynomial whose zeroes are 5 - 3√2 and 5 + 3√2.

Answer»

Sum of zeroes = 5 - 3√2 + 5 + 3√2 = 10

Product of zeroes= (5 - 3√2)(5 + 3√2) = 7

P(x)= x² - 10x + 7

Discussion

36.	Find the point on x-axis which is equidistant from the points (2,-2) and (-4,2)
Answer» Let P(x,0) be a point on X-axis PA = PB PA² = PB² (x - 2)²+ (0 + 2)² = (x + 4)² + (0 - 2)² x² + 4 - 4x + 4 = x² + 16 + 8x + 4 -4x + 4 = 8x + 16 x = -1 P(-1,0)

36.

Find the point on x-axis which is equidistant from the points (2,-2) and (-4,2)

Answer»

Let P(x,0) be a point on X-axis

PA = PB

PA² = PB²

(x - 2)²+ (0 + 2)² = (x + 4)² + (0 - 2)²

x² + 4 - 4x + 4

= x² + 16 + 8x + 4

-4x + 4 = 8x + 16

x = -1

P(-1,0)

Discussion

37.	A ball of mass `m` inside a smooth spherical shell of radius `R` with velocity `sqrt(2qR)` at `A` What is the direction of force acting on the ball, when it reaches `B` ? A. BPB. BQC. BRD. BS
Answer» Correct Answer - A

Discussion

38.	Find the square root of 21 – 2√54.1. 3√2 – √3 2. √2 – √3 3. 18 – 3√3 4. 9 – 3√3
Answer» Correct Answer - Option 1 : 3√2 – √3 Given: The number is 21 – 2√54 Formula Used: (a + b)² = a² + b² + 2 × a × b Calculation: The square root of 21 – 2√54 ⇒ √(21 – 2√54) ⇒ √(21 – 2 × √18 × √3) ⇒ √(18 + 3 – 2 × √18 × √3) ⇒ √[(√18)² + (√3)² – 2 × √18 × √3] ⇒ √(√18 – √3)² ⇒ √18 – √3 ⇒ 3√2 – √3 ∴ The square root of 21 – 2√54 is 3√2 – √3.

38.

Find the square root of 21 – 2√54.1. 3√2 – √3 2. √2 – √3 3. 18 – 3√3 4. 9 – 3√3

Answer» Correct Answer - Option 1 : 3√2 – √3

Given:

The number is 21 – 2√54

Formula Used:

(a + b)² = a² + b² + 2 × a × b

Calculation:

The square root of 21 – 2√54

⇒ √(21 – 2√54)

⇒ √(21 – 2 × √18 × √3)

⇒ √(18 + 3 – 2 × √18 × √3)

⇒ √[(√18)² + (√3)² – 2 × √18 × √3]

⇒ √(√18 – √3)²

⇒ √18 – √3

⇒ 3√2 – √3

∴ The square root of 21 – 2√54 is 3√2 – √3.

Discussion

39.	The product of the two integers x and y where x > 65 and y > 80, is 5780 and their HCF is 17 then what is the sum of the two integers?1. 682. 853. 1534. 136
Answer» Correct Answer - Option 3 : 153 Given: The product of the two number = 5780 HCF = 17 Concept used: The HCF defines the greatest factor present in between two or more numbers. Calculation: Let the two number be 17x and 17y The product of the two number = 5780 ⇒ 17x × 17y = 5780 ⇒ 289 xy = 5780 ⇒ xy = 20 The possible value of x and y is (1, 20), (20, 1), (2, 10), (10, 2), (4, 5), (5, 4) So the value which satisfied above condition is 4, 5 The first integer is 68 and the second integer is 85 Sum = 68 + 85 = 153 ∴ The sum of the two integers is 153

39.

The product of the two integers x and y where x > 65 and y > 80, is 5780 and their HCF is 17 then what is the sum of the two integers?1. 682. 853. 1534. 136

Answer» Correct Answer - Option 3 : 153

Given:

The product of the two number = 5780

HCF = 17

Concept used:

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

Calculation:

Let the two number be 17x and 17y

The product of the two number = 5780

⇒ 17x × 17y = 5780

⇒ 289 xy = 5780

⇒ xy = 20

The possible value of x and y is (1, 20), (20, 1), (2, 10), (10, 2), (4, 5), (5, 4)

So the value which satisfied above condition is 4, 5

The first integer is 68 and the second integer is 85

Sum = 68 + 85 = 153

∴ The sum of the two integers is 153

Discussion

40.	One is asked to say a two digit number. How many two digit numbers are there. What is the probability of one of the digit being 1. What is the probability of the product of the digits being a prime number.
Answer» prime number=1/36 = 1/6 Hence, the required probability is 16.

40.

One is asked to say a two digit number. How many two digit numbers are there. What is the probability of one of the digit being 1. What is the probability of the product of the digits being a prime number.

Answer»

prime number=1/36 = 1/6

Hence, the required probability is 16.

Discussion

41.	Debt equity ratio is a measure of
Answer» Answer: Long term solvency

Discussion

42.	write the expression an-ak for the AP a, a+d,a+2d....hence find the d of the AP for which 25 term is 10 morethan the 23rd term ?????
Answer» Actually this is not trigonometry question check question correctly

Discussion

43.	If x = cosθ, then find the value of (x6 + 1/x6)
Answer» ∵ x = cosθ + isinθ ⇒ x⁶ = (cos θ + isinθ)⁶ = cos6θ + isin6θ ⇒ 1/x⁶ = cos6θ – isin6θ ∴ x⁶ + 1/x⁶ = 2cos6θ

43.

If x = cosθ, then find the value of (x6 + 1/x6)

Answer»

∵ x = cosθ + isinθ

⇒ x⁶ = (cos θ + isinθ)⁶ = cos6θ + isin6θ

⇒ 1/x⁶ = cos6θ – isin6θ

∴ x⁶ + 1/x⁶ = 2cos6θ

Discussion

44.	The opposite angular points of a square are (5,4) and (-1,6). Find the coordinates of remaining two vertices.
Answer» We know, properties of square : (1) all sides are equal (2) angle between two sides = 90° use this concept here, Let one unknown point of square is C(P, Q) given, A( 5, 4) , D(-1, 6) AC = DC AC² = DC² (P-5)² + (Q -4)² = (P+1)² + (Q -6)² -10P +25 -8Q + 16 = 2P +1 -12Q + 36 -12P + 4Q + 4 = 0 -3P + Q + 1 = 0 ________(1) AC² + DC² = AD² (P+1)² + (Q -6)² + (P-5)² + (Q -4)² = (6)² + (2)² 2P² + 2Q² -8P -20Q + 78 = 40 P² + Q² -4P - 10Q + 19 = 0-----------(2) from (1) and (2) P² + (3P-1)² -4P -10(3P-1) + 19 = 0 10P² +1 - 6P -4P -30P +10 + 19 = 0 10P² - 40P + 30 = 0 P² -4P + 3 = 0 P = 1 and 3 Q = 3P -1 = 2 and 8 hence, two unknown points of square are ( 1 , 2) and (3, 8)

Discussion

45.	If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x2 + 1, then find fog(2).
Answer» Given f(x) = 3x –1 and g(x) = x² +1 (fog) (2) = f [g (2)] = f [2² + 1] = f (5) = 3 (5) – 1 = 14 ∴ (fog) (2) = 14

45.

If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x2 + 1, then find fog(2).

Answer»

Given f(x) = 3x –1 and g(x) = x² +1

(fog) (2) = f [g (2)]

= f [2² + 1] = f (5)

= 3 (5) – 1

= 14

∴ (fog) (2) = 14

Discussion

46.	For what values of m, the equation x2 – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?
Answer» The given equation will have equal roots if its discriminant is 0. Here ∆ =[(–2(1 + 3m)]² – 4 (1) [7 + (3 + 2m)] = 4(1 + 9m² + 6m) – 28(3 + 2m) = 9m² – 8m – 20 = (m – 2) (9m + 10) Hence ∆ =0 ⇔ m = 2, −10/9

46.

For what values of m, the equation x2 – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?

Answer»

The given equation will have equal roots if its discriminant is 0.

Here ∆ =[(–2(1 + 3m)]² – 4 (1) [7 + (3 + 2m)]

= 4(1 + 9m² + 6m) – 28(3 + 2m)

= 9m² – 8m – 20

= (m – 2) (9m + 10)

Hence ∆ =0 ⇔ m = 2, −10/9

Discussion

47.	If two coins are tossed simultaneously. Find the probability of getting 2 heads.
Answer» P ( Two Head) `= (1)/(4)`

Discussion

48.	Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.
Answer» In a given, word, [Hint : Vowels – A, E, I, O U] number of vowels is 3 number of consonants is 5 C C V V C C C V Since the relative positions of the vowels and consonants are not disturbed, the 3 vowels can be arranged in their relative positions in 3! ways and the 5 consonants can be arranged in their relative positions in 5! ways. ∴ The number of required arrangements = (3!) (5!) = (6) (120) = 720.

48.

Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.

Answer»

In a given, word, [Hint : Vowels – A, E, I, O U]

number of vowels is 3

number of consonants is 5

C C V V C C C V

Since the relative positions of the vowels and consonants are not disturbed, the 3 vowels can be arranged in their relative positions in 3! ways and the 5 consonants can be arranged in their relative positions in 5! ways.

∴ The number of required arrangements = (3!) (5!) = (6) (120) = 720.

Discussion

49.	Find the inverse of the function f : R → R defined by f(x) = ax + b, (a ≠ 0), a, b ∈ R.
Answer» Given f(x) = ax + b Let f(x) = y Then ax + b = y ⇒ ax = y – b Since a ≠ 0 ⇒ x = y - b/a ⇒ f^-1(y) = y - b/a Hence f^-1(x) = x - b/a

49.

Find the inverse of the function f : R → R defined by f(x) = ax + b, (a ≠ 0), a, b ∈ R.

Answer»

Given f(x) = ax + b

Let f(x) = y

Then ax + b = y

⇒ ax = y – b

Since a ≠ 0

⇒ x = y - b/a

⇒ f^-1(y) = y - b/a

Hence f^-1(x) = x - b/a

Discussion

50.	Find the sum of the squares and the sum of the cubes of the roots of the equation x3 – px2 + qx – r = 0 in terms of p, q, r.
Answer» Let α, β, γ be the roots of the given equation then α + β + γ = p, αβ + βγ + γα = q, αβγ = r Sum of the squares of the roots is α² + β² + γ² = (α + β + γ)² – 2 (αβ + βγ + γα) = p2 – 2q Sum of the cubes of the roots is α³ + β³ + γ³ = (α + β + γ) (α² + β² + γ² – αβ – βγ – γα) + 3αβγ = p(p² – 2q – q) + 3r = p(p² – 3q) + 3r

50.

Find the sum of the squares and the sum of the cubes of the roots of the equation x3 – px2 + qx – r = 0 in terms of p, q, r.

Answer»

Let α, β, γ be the roots of the given equation then

α + β + γ = p, αβ + βγ + γα = q,

αβγ = r

Sum of the squares of the roots is α² + β² + γ²

= (α + β + γ)² – 2 (αβ + βγ + γα)

= p2 – 2q

Sum of the cubes of the roots is α³ + β³ + γ³

= (α + β + γ) (α² + β² + γ² – αβ – βγ – γα) + 3αβγ

= p(p² – 2q – q) + 3r

= p(p² – 3q) + 3r

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

`20 ml` of an `H_2 O_2` solution on reaction with excess of acidified `KMnO_4` released `224` cc of `O_2`. What is the volume strength of that `H_2 O_2` ?A. 5.6 volB. 11.2 volC. 22.4 volD. 2.8 vol

Choose the word from the options given to form a compound word with “drop”. (a) heave (b) mock (c) eaves (d) out

Choose the word from the options given to form a compound word with “out” , (a) in (b) wipe(c) weather (d) hind

There is no bread left because we ______ it all. A) have eaten B) has been eaten C) had eaten D) have been eaten

[" of "0],[CH_(3)-c-NH_(2)+H_(2)O]

What are the optimum temperature and pH at which enzymes are highly active ?

The men were to ______ but the manager decided to give them a second change. A) have been dismissed B) dismissed C) being dismissed D) be dismissing

Name the enzyme which converts milk into curd.

(i) \( \int_{0}^{\pi} \theta \sin ^{2} \theta \cos \theta d \theta \)

(ii) \( \int_{0}^{\pi} x \sin ^{6} x \cos ^{4} x d x \)

Find the sum of `h+2e^(2)h+2e^(4)h+2e^(6)h +.....`

Solve `x^(2)+x-2=0`

Find the directional derivative of q = 4x3 - 3x2 * y2 (2.-1.1) along the line which makes equal angles with Co-ordinate axes.

If \( y=x^{\tan x}+(\sin x)^{\cos x} \), Find \( \frac{d y}{d x} \)

Simplify: \(\frac1{3-\sqrt8}-\frac1{\sqrt8-\sqrt7}+\frac1{\sqrt7-\sqrt6}\)\(-\frac1{\sqrt6-\sqrt5}+\frac1{\sqrt5-2}\)

Write four numbers which are divisible by 6 but not by 12.

Subtract:4x by x2 - 1 - x + 1 by x - 1.

Smallest prime number

Given, `y=sin 2x` . Then find `(dy)/(dx)` .

Solve: x4 - 13x2 + 42 = 0

7. The length of a field is 10 meter more than the breadth of the field. If perimeter of the field is 100 \( m \), find the dimensions of the field.

Given, `y=(ax+b)^(2)` , evaluate ` (dy)/(dx)`.

)42) If \( \alpha \) and \( \beta \) are the roots of the equation \( x^{2}+x-4=0 \) form the quadratic equation whose roots are \( \alpha^{2} \) and \( \beta^{2} \).

PBA and PDC are two secants. AD is the diameter of the circle with center at O. ∠A = 40°, ∠P = 20°. Find the measure of ∠DBC.(a) 30° (b) 45° (c) 50° (d) 40°

If tanA = \(\frac{3}{4}\), find the value of \(\frac{1}{sinA}+\frac{1}{cosA}\)

40) Find the square root of: \( \left(6 x^{2}+x-1\right)\left(3 x^{2}+2 x-1\right)\left(2 x^{2}+3 x+1\right) \)

39) Simplify: \( \frac{a^{2}-16}{a^{3}-8} \times \frac{2 a^{2}-3 a-2}{2 a^{2}+9 a+4} \div \frac{3 a^{2}-11 a-4}{a^{2}+2 a+4} \)

Find the area of a square,whose perimeter is 64 m.

The age of sita and her sister are in the ratio of 3:8. after 12 years the ratio of their ages is 7:12. find there present ages.

\( =\left(\frac{3}{5}\right)^{11} \div\left(\frac{3}{5}\right)^{6} \)

38 If \( A=\frac{x}{x+1} B=\frac{1}{x+1} \) prove that \( \frac{(A+B) 2+(A-B) 2}{A \div B}=\frac{2(x 2+1)}{x(x+1) 2} \)

Find the area of each of the following triangles:(i)(ii)(iii)

A force F=(a+bx) acts on a particle in x direction where a and b are constants . Find the work done by this force during displacement from `x_(1)` to `x_(2)` .

Find value of `(104)^(1//2)` using binomial approx.

Find a quadratic polynomial whose zeroes are 5 - 3√2 and 5 + 3√2.

Find the point on x-axis which is equidistant from the points (2,-2) and (-4,2)

A ball of mass `m` inside a smooth spherical shell of radius `R` with velocity `sqrt(2qR)` at `A` What is the direction of force acting on the ball, when it reaches `B` ? A. BPB. BQC. BRD. BS

Find the square root of 21 – 2√54.1. 3√2 – √3 2. √2 – √3 3. 18 – 3√3 4. 9 – 3√3

The product of the two integers x and y where x &gt; 65 and y &gt; 80, is 5780 and their HCF is 17 then what is the sum of the two integers?1. 682. 853. 1534. 136

One is asked to say a two digit number. How many two digit numbers are there. What is the probability of one of the digit being 1. What is the probability of the product of the digits being a prime number.

Debt equity ratio is a measure of

write the expression an-ak for the AP a, a+d,a+2d....hence find the d of the AP for which 25 term is 10 morethan the 23rd term ?????

If x = cosθ, then find the value of (x6 + 1/x6)

The opposite angular points of a square are (5,4) and (-1,6). Find the coordinates of remaining two vertices.

If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x2 + 1, then find fog(2).

For what values of m, the equation x2 – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?

If two coins are tossed simultaneously. Find the probability of getting 2 heads.

Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.

Find the inverse of the function f : R → R defined by f(x) = ax + b, (a ≠ 0), a, b ∈ R.

Find the sum of the squares and the sum of the cubes of the roots of the equation x3 – px2 + qx – r = 0 in terms of p, q, r.

The product of the two integers x and y where x > 65 and y > 80, is 5780 and their HCF is 17 then what is the sum of the two integers?1. 682. 853. 1534. 136