Search Previous Year Questions

Concept:

 

The solution to the quadratic equation ax² + bx + c = 0 is given by: \(\rm x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\).

 

Calculation:

It is given that cos x = tan y, cot y = tan z and cot z = tan x.

∴ \(\rm \cos x=\tan y=\dfrac{1}{\cot y}=\dfrac{1}{\tan z}=\cot z=\tan x=\dfrac{\sin x}{\cos x}\)

⇒ cos² x = sin x

⇒ 1 - sin² x = sin x

⇒ sin² x + sin x - 1 = 0

Using the formula for the roots of a quadratic equation:

⇒ \(\rm \sin x = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)}\)

⇒ \(\rm \sin x = \dfrac{-1 +\sqrt{5}}{2}\) OR \(\rm \sin x = \dfrac{-1 -\sqrt{5}}{2}\)

From the given answer options, \(\rm \sin x = \dfrac{\sqrt{5}-1}{2}\).

Concept:

The identities of trigonometry are:

• \(\rm \tan a={\sin a\over \cos a}\)
• \(\rm \sin 2a = 2\sin a\cos a\)
• \(\rm \tan (a+b)={\tan a +\tan b\over{1-\tan a \tan b}}\)
• \(\rm \cos 2a=(\cos^2 a- \sin^2 a)= (2\cos^2 a-1)=(1-2\sin^2 a)\)

 

Calculation:

Let \(\rm \tan \left(45 + \frac{\theta}{2} \right ) =x\)

\(\rm ⇒ x ={\tan 45^o +\tan \frac{\theta}{2}\over{1-\tan 45^o\tan \frac{\theta}{2}}}\)

\(\rm ⇒ x={1 +\tan \frac{\theta}{2}\over{1-\tan \frac{\theta}{2}}}\)                     (∵ tan 45o = 1)

\(\rm⇒ x = {\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\over \cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\)

\(\boldsymbol {\rm⇒ x = {\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\over \cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\times{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\over \cos \frac{\theta}{2}+\sin \frac{\theta}{2}}}\)

\(\rm⇒ x = {(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})^2\over \cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}\)

\(\rm⇒ x = {\cos^2 \frac{\theta}{2}+\sin^2 \frac{\theta}{2}+2\cos\frac{\theta}{2}\sin\frac{\theta}{2}\over \cos\theta}\)

\(\rm⇒ x = {1+\sin \theta\over \cos \theta}\)                       (∵ sin2 a + cos2 a = 1)

\(\boldsymbol {\rm ⇒ x =\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta} =\sec\theta+\tan\theta}\)

Concept:

**Trigonometric Identities:**​

• sin 2θ = 2 sin θ cos θ

 

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (nπ + θ) = (-1)ⁿ sin θ.
• cos (nπ + θ) = (-1)ⁿ cos θ.
• \(\rm \sin \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]\) = (-1)ⁿ cos θ.
• \(\rm \cos \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]\) = (-1)ⁿ (-sin θ).

 

Calculation:

sin 10° sin 50° sin 70°

= sin 10° sin (90∘ - 40°) sin (90∘ - 20°)

= sin 10° cos 40° cos 20°

= \(\rm \left (\dfrac{2\sin 10^{\circ}\cos 10^{\circ}}{2\cos 10^{\circ}} \right ) \cos 40^{\circ}\cos 20^{\circ}\)

= \(\rm \left (\dfrac{\sin 20^{\circ}}{2\cos 10^{\circ}} \right ) \cos 40^{\circ}\cos 20^{\circ}\)

= \(\rm \left (\dfrac{2\sin 20^{\circ}\cos 20^{\circ}}{4\cos 10^{\circ}} \right ) \cos 40^{\circ}\)

= \(\rm \dfrac{2\sin 40^{\circ}\cos 40^{\circ}}{8\cos 10^{\circ}}\)

= \(\rm \dfrac{\sin 80^{\circ}}{8\cos 10^{\circ}}\)

= \(\rm \dfrac{\sin (90^{\circ}-10^{\circ})}{8\cos 10^{\circ}}\)

= \(\rm \dfrac{\cos 10^{\circ}}{8\cos 10^{\circ}}\)

= \(\rm \dfrac{1}{8}\).

Concept:

• | csc θ = \(\rm \dfrac{1}{\sin \theta}\) | sec θ = \(\rm \dfrac{1}{\cos \theta}\) | | --- | --- | | tan θ = \(\rm \dfrac{\sin \theta}{\cos \theta}\) | cot θ = \(\rm \dfrac{\cos \theta}{\sin \theta}\) | | tan θ = \(\rm \dfrac{1}{\cot \theta}\) | |

• sin2 θ + cos2 θ = 1.

• a³ - b³ = (a - b)(a² + ab + b²).

 

Calculation:

Let tan A = m ⇒ cot A = \(\rm \dfrac{1}{m}\).

∴ \(\rm \dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A}\)

= \(\rm \dfrac{m}{1-\tfrac{1}{m}} + \dfrac{\tfrac{1}{m}}{1-m}\)

= \(\rm \dfrac{m^2}{m-1} - \dfrac{1}{m(m-1)}\)

= \(\rm \dfrac{m^3-1}{m(m-1)}\)

= \(\rm \dfrac{(m-1)(m^2+m+1)}{m(m-1)}\)

= \(\rm \dfrac{m^2+m+1}{m}\)

= \(\rm m+1+\dfrac{1}{m}\)

= \(\rm \left (\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right )+1\)

= \(\rm \left (\dfrac{\sin^2 A+\cos^2 A}{\sin A\cos A}\right )+1\)

= \(\rm \left (\dfrac{1}{\sin A\cos A}\right )+1\)

= sec A csc A + 1.

Concept:

If the angle of elevation from a point at distance D from tower of length L is θ 

tan θ = \(\rm L\over D\)

Calculation:

Let the length of the tower OP be x

From the figure,

OA = \(\rm {x\over\tan30}=\sqrt{3}x\)

OB = \(\rm {x\over\tan45}=x\)

OC = \(\rm {x\over\tan60}={x\over√{3}}\)

∴ AB = OA - OB = \(\rm \sqrt 3 x-x\)

BC = OB - OC = \(\rm x-\frac{x}{\sqrt 3}\)

⇒ \(\rm {AB\over BC}={\sqrt 3 x-x\over\rm x-\frac{x}{\sqrt 3}}\)

⇒ \(\rm {AB\over BC}={\sqrt 3(\sqrt 3 -1)\over \sqrt 3 -1} \)

⇒ \(\boldsymbol{\rm AB:BC=\sqrt{3} : 1}\)

Concept:

• Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.
• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \displaystyle \left|\int_a^b f(x)\ dx\right|\), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
• Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \displaystyle \int_a^b f(x)\ dx = [ g(x)]_a^b\) = g(b) - g(a).
• \(\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C\).

 

Calculations:

Let's first find the points of intersection/touching of the two curves y2 = x and y = |x|.

Using the condition for intersecting, we must have:

y2 - x = y - |x| = 0

CASE 1: For, x ≥ 0, |x| = x.

⇒ y2 - x = y - x = 0

⇒ y2 - y = 0

⇒ y(y - 1) = 0

⇒ y = 0 OR y - 1 = 0

⇒ y = 0 OR y = 1.

And, x = 0² = 0 OR x = 1² = 1.

The points of intersection are (0, 0) and (1, 1).

CASE 2: For x < 0, |x| = -x.

⇒ y2 - x = y - (-x) = 0

⇒ y2 - x = - y - x = 0

⇒ y2 + y = 0

⇒ y(y + 1) = 0

⇒ y = 0 OR y + 1 = 0

⇒ y = 0 OR y = -1.

∵ y = |x|, y cannot be negative.

∴ The only solution in this case y = 0 and x = 0² = 0.

 

Finally, the solutions (points of intersection/touching) of y2 = x and y = |x| are (0, 0) and (1, 1).

 

The graph of the curves is shown below:

The required (shaded) area is:

(Area under y² = x from x = 0 to x = 1.) - (Area under y = |x| from x = 0 to x = 1.)

= \(\rm \displaystyle \left|\int_0^1 \sqrt x\ dx\right| -\left|\int_0^1 x\ dx\right|\)

= \(\rm \left|\dfrac{2}{3}\left[x^{\tfrac{3}{2}} \right ]_0^1\right|-\left|\dfrac{1}{2}\left[x^{2} \right ]_0^1\right|\)

= \(\rm \left|\dfrac{2}{3}[1^{\tfrac{3}{2}}-0^{\tfrac{3}{2}}]\right|-\left|\dfrac{1}{2}[1^2-0^2]\right|\)

= \(\rm \dfrac 23 - \dfrac 12\)

= \(\dfrac16\) sq. units.

Concept:

The function f(x) is continuous at an if all the below condition follows:

1. f(a) is real and finite. 2. \(\rm \lim_{\alpha \to 0}f(a-\alpha)\text{ and }\lim_{\alpha \to 0}f(a+\alpha)\) are real and finite. 3. \(\rm \lim_{\alpha \to 0}f(a-\alpha)=\lim_{\alpha \to 0}f(a+\alpha)=f(a)\)

Calculation:

f(x) = \(\left{ {\begin{array}{*{20}{c}} {\frac{5}{2} - \rm x;;;;;;x < 2}\ {1;;;;;;;;;;;;;\rm x = 2}\ {\rm x - ;\frac{3}{2};;;;;x > 2} \end{array}} \right.\) 

\(\rm \lim_{\alpha \to 0}f(x-\alpha)={\frac{5}{2} - \rm x}\)

For x = 2,

⇒ \(\rm \lim_{\alpha \to 0}f(2-\alpha)=\lim_{\alpha \to 0}\left[\frac{5}{2} - (2-\alpha)\right]\)

⇒ \(\rm \lim_{\alpha \to 0}f(2-\alpha)=\rm \lim_{\alpha \to 0}\left(\frac{1}{2}+\alpha\right) =\boldsymbol{\rm \frac{1}{2}}\)

\(\rm \lim_{\alpha \to 0}f(x+\alpha)=x-\frac{3}{2}\)

For x = 2,

⇒ \(\rm \lim_{\alpha \to 0}f(2+\alpha)=\lim_{\alpha \to 0}\left[(2+\alpha)-\frac{3}{2}\right]\)

⇒ \(\rm \lim_{\alpha \to 0}f(2+\alpha)=\rm \lim_{\alpha \to 0}\left(\frac{1}{2}+\alpha\right) = \boldsymbol{\rm \frac{1}{2}}\)

f(x) = f(2) = 1

\(\rm \lim_{\alpha \to 0}f(2-\alpha)=\lim_{\alpha \to 0}f(2+\alpha)\neq f(2)\)

∴ The function is not continuous at x = 2

Concept:

Trigonometry Identities:

• \(\rm \tanθ ={1\over\cotθ}={\sinθ\over\cosθ}\)
• cosec θ \(\rm={1\over\sinθ} \)
• sec θ \(\rm={1\over\cosθ} \)
• cos θ = sin (90 - θ)
• cot θ = tan (90 - θ)
• cos-1 x = 90o - sin-1 x
• cot^-1 x = 90^o - tan^-1 x
• \(\rm \sin 2θ = 2\sin θ\cos θ\)
• \(\rm \cos 2θ=(\cos^2θ- \sin^2θ)= (2\cos^2θ-1)=(1-2\sin^2θ)\)

 

Calculation:

S = 2 tan-1[cosec(tan-1x) - tan(cot-1x)]

⇒ S = 2 tan-1[cosec(tan-1x) - tan(90 - tan-1x)]

Let tan-1x = θ

⇒ S = 2 tan-1[cosec θ - tan(90 - θ)]

⇒ S = 2 tan-1[cosec θ - cot θ]

⇒ \(\rm S=2\tan^{-1}\left[{1\over\sinθ}-{\cosθ\over\sinθ}\right]\)

⇒ \(\rm S=2\tan^{-1}\left[{1-\cosθ\over\sinθ}\right]\)

⇒ \(\rm S=2\tan^{-1}\left[{2\sin^2{θ\over2}\over2\sin{θ\over2}\cos{θ\over2}}\right]\)

⇒ \(\rm S=2 tan^{-1}\left[tan{θ\over2}\right]\)

⇒ S = θ = tan^-1 x

Concept:

The identities of trigonometry are:

• \(\rm \tan a={\sin a\over \cos a}\)
• \(\rm \sin 2a = 2\sin a\cos a\)
• \(\rm \cos 2a=(\cos^2 a- \sin^2 a)= (2\cos^2 a-1)=(1-2\sin^2 a)\)

 

Calculation:

Given 3 sin x + 4 cos x = 5

\(\boldsymbol{⇒ \rm 3(2 \sin {x\over2} \cos {x\over2}) + 4(1 - 2 \sin^2 {x\over2}) = 5}\)

\(⇒ \rm 6 \sin {x\over2}\cos {x\over2}+ 4 - 8 \sin^2 {x\over2}= 5\)

\(\rm ⇒ 6 \sin {x\over2}\cos {x\over2}- 8 \sin^2 {x\over2}= 1\)

\(⇒ \rm 6 \sin {x\over2}\cos {x\over2}- 8 \sin^2{x\over2} = \sin^2{x\over2} + \cos^2{x\over2}\)  (∵ sin2 a + cos2 a = 1)

Dividing the equation by \(\rm\cos^2{x\over2}\)

⇒ \(\rm 6\frac{\sin {x\over2}\cos {x\over2}}{\cos^2 {x\over2}}-8\frac{\sin^2 {x\over2}}{\cos^2 {x\over2}}=\frac{\sin^2 {x\over2}+\cos^2 {x\over2}}{\cos^2 {x\over2}}\)

⇒ \(\rm 6\tan {x\over2}-8\tan^2 {x\over2}=\tan^2 {x\over2}+1\)

⇒ \(\boldsymbol{\rm 6\tan {x\over2}-9\tan^2 {x\over2}=1}\)

Concept:

• The cardinality of a set A is the measure of the "number of elements" of the set. It is denoted by n(A).

For example, a set containing 3 elements has a cardinality of 3.

• If A ⊂ B, then A ∪ B = B.

 

Calculations:

Since, A ⊂ B and B ⊂ C, therefore A ∪ B ∪ C = C.

⇒ n(A ∪ B ∪ C) = n(C).

∴ The cardinality (number of elements) of A ∪ B ∪ C = cardinality (number of elements) of C.