Practice Start Test

Trigonometric Ratio and Identities

JEE Mathematics · 51 questions · Page 5 of 6 · Click an option or "Show Solution" to reveal answer

Q41
The value of 2sin (12∘^\circ) −- sin (72∘^\circ) is :
A 5(1−3)4{{\sqrt 5 (1 - \sqrt 3 )} \over 4}
B 1−58{{1 - \sqrt 5 } \over 8}
C 3(1−5)2{{\sqrt 3 (1 - \sqrt 5 )} \over 2}
D 3(1−5)4{{\sqrt 3 (1 - \sqrt 5 )} \over 4}
Correct Answer
Option D
Solution
2sin⁡12∘−sin⁡72∘2\sin 12^\circ - \sin 72^\circ
=sin⁡12∘+(−2cos⁡42∘ . sin⁡30∘)= \sin 12^\circ + ( - 2\cos 42^\circ \,.\,\sin 30^\circ )
=sin⁡12∘−cos⁡42∘= \sin 12^\circ - \cos 42^\circ
=sin⁡12∘−sin⁡48∘= \sin 12^\circ - \sin 48^\circ
=2sin⁡18∘ . cos⁡30∘= 2\sin 18^\circ \,.\,\cos 30^\circ
=−2(5−14) . 32= - 2\left( {{{\sqrt 5 - 1} \over 4}} \right)\,.\,{{\sqrt 3 } \over 2}
=3(1−5)4= {{\sqrt 3 \left( {1 - \sqrt 5 } \right)} \over 4}
Q42
Let f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3π+θ))−2(1−sin⁡22θ)f(\theta ) = 3\left( {{{\sin }^4}\left( {{{3\pi } \over 2} - \theta } \right) + {{\sin }^4}(3\pi + \theta )} \right) - 2(1 - {\sin ^2}2\theta ) and S={θ∈[0,π]:f′(θ)=−32}S = \left\{ {\theta \in [0,\pi ]:f'(\theta ) = - {{\sqrt 3 } \over 2}} \right\}. If 4β=∑θ∈Sθ4\beta = \sum\limits_{\theta \in S} \theta , then f(β)f(\beta ) is equal to
A 98\dfrac{9}{8}
B 32\dfrac{3}{2}
C 54\dfrac{5}{4}
D 118\dfrac{11}{8}
Correct Answer
Option C
Solution

f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3x+θ))−2(1−sin⁡22θ)f(\theta)=3\left(\sin ^{4}\left(\dfrac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 x+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right) S={θ∈[0,π]:f′(θ)=−32}S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\dfrac{\sqrt{3}}{2}\right\} ⇒f(θ)=3(cos⁡4θ+sin⁡4θ)−2cos⁡22θ\Rightarrow \mathrm{f}(\theta)=3\left(\cos ^{4} \theta+\sin ^{4} \theta\right)-2 \cos ^{2} 2 \theta ⇒f(θ)=3(1−12sin⁡22θ)−2cos⁡22θ\Rightarrow \mathrm{f}(\theta)=3\left(1-\dfrac{1}{2} \sin ^{2} 2 \theta\right)-2 \cos ^{2} 2 \theta ⇒f(θ)=3−32sin⁡22θ−2cos⁡2θ\Rightarrow \mathrm{f}(\theta)=3-\dfrac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta =32−12cos⁡22θ=32−12(1+cos⁡4θ2)=\dfrac{3}{2}-\dfrac{1}{2} \cos ^{2} 2 \theta=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{1+\cos 4 \theta}{2}\right) f(θ)=54−cos⁡4θ4f(\theta)=\dfrac{5}{4}-\dfrac{\cos 4 \theta}{4} f′(θ)=sin⁡4θf^{\prime}(\theta)=\sin 4 \theta ⇒f′(θ)=sin⁡4θ=−32\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\dfrac{\sqrt{3}}{2} ⇒4θ=nπ+(−1)nπ3\Rightarrow 4 \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \dfrac{\pi}{3} ⇒θ=nπ4+(−1)nπ12\Rightarrow \theta=\dfrac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \dfrac{\pi}{12}

⇒θ=π12,(π4−π12),(π2+π12),(3π4−π12)⇒4β=π4+π2+3π4=3π2⇒β=3π8⇒f(β)=54−cos⁡3π24=54\begin{aligned} & \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) \\\\ & \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\\\ & \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4} \end{aligned}
Q43
96cos⁡π33cos⁡2π33cos⁡4π33cos⁡8π33cos⁡16π3396\cos {\pi \over {33}}\cos {{2\pi } \over {33}}\cos {{4\pi } \over {33}}\cos {{8\pi } \over {33}}\cos {{16\pi } \over {33}} is equal to :
A 4
B 2
C 1
D 3
Correct Answer
Option D
Solution

Let

A=96cos⁡π33cos⁡2π33cos⁡4π33cos⁡8π33cos⁡16π33⇒2A=96×2(cos⁡π33cos⁡2π33cos⁡4π33cos⁡8π33cos⁡16π33)⇒2A×sin⁡π33=96×(2sin⁡π33cos⁡π33cos⁡2π33cos⁡4π33⋅cos⁡8π33cos⁡16π33)⇒2A×sin⁡π33=6×sin⁡32π33=6×sin⁡π33⇒2A=6⇒A=3\begin{aligned} & A=96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33} \\\\ & \Rightarrow 2 A=96 \times 2\left(\cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\ & \Rightarrow 2 A \times \sin \frac{\pi}{33} =96 \times\left(2 \sin \frac{\pi}{33} \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}\right) \\\\ & \Rightarrow 2 A \times \sin \frac{\pi}{33}=6 \times \sin \frac{32 \pi}{33}=6 \times \sin \frac{\pi}{33} \\\\ & \Rightarrow 2 A=6 \Rightarrow A=3 \end{aligned}

Thus, the required answer is 3 .

Q44
The value of 36(4cos⁡29∘−1)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right) is :
A 18
B 36
C 54
D 27
Correct Answer
Option B
Solution
4cos⁡2θ−1=4(1−sin⁡2θ)−1=3−4sin⁡2θ=3sin⁡θ−4sin⁡3θsin⁡θ=sin⁡3θsin⁡θ\begin{aligned} & 4 \cos ^2 \theta-1=4\left(1-\sin ^2 \theta\right)-1 \\\\ & =3-4 \sin ^2 \theta \\\\ & =\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta} \\\\ & =\frac{\sin 3 \theta}{\sin \theta} \end{aligned}
36(4cos⁡29∘−1)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)
=36[sin⁡27∘sin⁡9∘×sin⁡81∘sin⁡27∘×sin⁡243∘sin⁡81∘×sin⁡729∘sin⁡243∘]=36[sin⁡729∘sin⁡9∘]=36×1=36\begin{aligned} & =36\left[\frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}\right] \\\\ & =36\left[\frac{\sin 729^{\circ}}{\sin 9^{\circ}}\right]=36 \times 1=36 \end{aligned}
Q45
If tan⁡A=1x(x2+x+1),tan⁡B=xx2+x+1\tan \mathrm{A}=\dfrac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\dfrac{\sqrt{x}}{\sqrt{x^2+x+1}} and tan⁡C=(x−3+x−2+x−1)1/2,0<A,B,C<π2\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\dfrac{\pi}{2}, then A+B\mathrm{A}+\mathrm{B} is equal to :
A C\mathrm{C}
B π−C\pi-C
C 2π−C2 \pi-C
D π2−C\dfrac{\pi}{2}-\mathrm{C}
Correct Answer
Option A
Solution

To find the sum of two angles in terms of tangent, we can use the tangent addition formula:

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}

Let's compute tan⁡(A+B)\tan (A + B) using the given tan⁡A\tan A and tan⁡B\tan B:

tan⁡A=1x(x2+x+1)\tan A = \frac{1}{\sqrt{x(x^2+x+1)}}
tan⁡B=xx2+x+1\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}

Now, we can apply the addition formula:

tan⁡(A+B)=1x(x2+x+1)+xx2+x+11−1x(x2+x+1)⋅xx2+x+1\tan (A + B) = \frac{\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \frac{1}{\sqrt{x(x^2+x+1)}} \cdot \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}
tan⁡(A+B)=1+x2x(x2+x+1)1−1(x2+x+1)\tan (A + B) = \frac{\frac{1 + \sqrt{x^2}}{\sqrt{x(x^2+x+1)}}}{1 - \frac{1}{(x^2+x+1)}}
tan⁡(A+B)=x2+1x(x2+x+1)(x2+x+1)−1(x2+x+1)\tan (A + B) = \frac{\frac{\sqrt{x^2} + 1}{\sqrt{x(x^2+x+1)}}}{\frac{(x^2+x+1) - 1}{(x^2+x+1)}}
tan⁡(A+B)=(x+1)(x2+x+1)(x2+x)x\tan (A + B) = \frac{(x + 1)\sqrt{(x^2+x+1)}}{(x^2 + x){\sqrt{x}}}
tan⁡(A+B)=(x+1)(x2+x+1)x(x+1)x\tan (A + B) = \frac{{(x + 1)}\sqrt{(x^2+x+1)}}{x{(x + 1)}{\sqrt{x}}}
tan⁡(A+B)=x2+x+1x3/2\tan (A + B) = \frac{\sqrt{x^2+x+1}}{x^{3/2}}

Let us first simplify tan⁡C\tan \mathrm{C}:

tan⁡C=x−3+x−2+x−1=1x3+1x2+1x=1+x+x2x3=x2+x+1x3/2.\tan \mathrm{C} = \sqrt{x^{-3}+x^{-2}+x^{-1}} = \sqrt{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{x}} = \sqrt{\frac{1+x+x^2}{x^3}} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}.

We see that:

tan⁡(A+B)=tan⁡C\tan (A + B) = \tan C

Since tangent is positive and all the angles AA, BB, and CC are in the first quadrant, we can say that:

A+B=CA + B = C

Hence, the answer is: Option A : CC

Q46
The number of solutions, of the equation esin⁡x−2e−sin⁡x=2e^{\sin x}-2 e^{-\sin x}=2, is :
A 0
B 1
C 2
D more than 2
Correct Answer
Option A
Solution

Take

esin⁡x=t(t>0)e^{\sin x}=t(t>0)
⇒t−2t=2⇒t2−2t=2⇒t2−2t−2=0⇒t2−2t+1=3⇒(t−1)2=3⇒t=1±3⇒t=1±1.73⇒t=2.73 or −0.73 (rejected as t>0)⇒esin⁡x=2.73⇒log⁡eesin⁡x=log⁡e2.73⇒sin⁡x=log⁡e2.73>1\begin{aligned} & \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\ & \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\ & \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\ & \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\ & \Rightarrow(\mathrm{t}-1)^2=3 \\ & \Rightarrow \mathrm{t}=1 \pm \sqrt{3} \\ & \Rightarrow \mathrm{t}=1 \pm 1.73 \\ & \Rightarrow \mathrm{t}=2.73 \text{ or }-0.73 \text{ (rejected as } \mathrm{t}>0) \\ & \Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73 \\ & \Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73 \\ & \Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1 \end{aligned}

So no solution.

Q47
For α,β∈(0,π/2)\alpha, \beta \in(0, \pi / 2), let 3sin⁡(α+β)=2sin⁡(α−β)3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta) and a real number kk be such that tan⁡α=ktan⁡β\tan \alpha=k \tan \beta. Then, the value of kk is equal to
A 5
B −-2/3
C −-5
D 2/3
Correct Answer
Option C
Solution

To find the value of

kk

, the given conditions are:

3sin⁡(α+β)=2sin⁡(α−β)3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)

And

tan⁡α=ktan⁡β\tan \alpha = k \tan \beta

For the first equation, using the sum and difference formulas for sine, we can rewrite the equation as:

3(sin⁡αcos⁡β+cos⁡αsin⁡β)=2(sin⁡αcos⁡β−cos⁡αsin⁡β)3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta)

Simplifying this, we get:

3sin⁡αcos⁡β+3cos⁡αsin⁡β=2sin⁡αcos⁡β−2cos⁡αsin⁡β3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta

Rearranging the terms, we obtain:

5sin⁡βcos⁡α=−sin⁡αcos⁡β5 \sin \beta \cos \alpha = - \sin \alpha \cos \beta

Dividing both sides by

sin⁡αcos⁡β\sin \alpha \cos \beta

, we get:

5sin⁡βcos⁡αsin⁡αcos⁡β=−1\frac{5 \sin \beta \cos \alpha}{\sin \alpha \cos \beta} = -1

Which simplifies to:

5tan⁡β=−tan⁡α5 \tan \beta = - \tan \alpha

So, taking the reciprocal, we have:

tan⁡α=−5tan⁡β\tan \alpha = -5 \tan \beta

Therefore, by comparing this equation with the given

tan⁡α=ktan⁡β\tan \alpha = k \tan \beta

, we find that

k=−5k = -5

. Thus, the value of

kk

is

−5-5

.

Q48
If for θ∈[−π3,0]\theta \in\left[-\dfrac{\pi}{3}, 0\right], the points (x,y)=(3tan⁡(θ+π3),2tan⁡(θ+π6))(x, y)=\left(3 \tan \left(\theta+\dfrac{\pi}{3}\right), 2 \tan \left(\theta+\dfrac{\pi}{6}\right)\right) lie on xy+αx+βy+γ=0x y+\alpha x+\beta y+\gamma=0, then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is equal to :
A 75
B 96
C 80
D 72
Correct Answer
Option A
Solution
 Let ϕ=θ+π3⇒θ=ϕ−π6x=3tan⁡(θ+π3)=3tan⁡(θ+π6)y=2tan⁡ϕtan⁡(ϕ+π6)=tan⁡ϕ+131−tan⁡ϕ⋅13x3=y2+131−y2⋅13\begin{aligned} &\text{ Let } \phi=\theta+\frac{\pi}{3} \Rightarrow \theta=\phi-\frac{\pi}{6}\\ &\begin{aligned} & x=3 \tan \left(\theta+\frac{\pi}{3}\right)=3 \tan \left(\theta+\frac{\pi}{6}\right) \\ & y=2 \tan \phi \\ & \tan \left(\phi+\frac{\pi}{6}\right)=\frac{\tan \phi+\frac{1}{\sqrt{3}}}{1-\tan \phi \cdot \frac{1}{\sqrt{3}}} \\ & \frac{x}{3}=\frac{\frac{y}{2}+\frac{1}{\sqrt{3}}}{1-\frac{y}{2} \cdot \frac{1}{\sqrt{3}}} \end{aligned} \end{aligned}
⇒x=3(y3+2)23−yxy+αx+βy+r=03(y3+223−y)+α(3(y3+2)(23−y))+βy+r=0=(33−β)y2+(6+33α+23β−y)y+(6α+23y)=0\begin{aligned} & \Rightarrow \quad x=\frac{3(y \sqrt{3}+2)}{2 \sqrt{3}-y} \\ & x y+\alpha x+\beta y+r=0 \\ & 3\left(\frac{y \sqrt{3}+2}{2 \sqrt{3}-y}\right)+\alpha\left(3 \frac{(y \sqrt{3}+2)}{(2 \sqrt{3}-y)}\right)+\beta y+r=0 \\ & =(3 \sqrt{3}-\beta) y^2+(6+3 \sqrt{3} \alpha+2 \sqrt{3} \beta-y) y \\ & \quad+(6 \alpha+2 \sqrt{3} y)=0 \end{aligned}

For this identity to hold for all θ\theta, coefficients must be 0

∴β=33γ=−α36+33α+(23)(33)+α3=0⇒α=−23⇒β=6α2+β2+γ2=75\begin{aligned} & \therefore \quad \beta=3 \sqrt{3} \\ & \gamma=-\alpha \sqrt{3} \\ & 6+3 \sqrt{3} \alpha+(2 \sqrt{3})(3 \sqrt{3})+\alpha \sqrt{3}=0 \\ & \Rightarrow \alpha=-2 \sqrt{3} \\ & \Rightarrow \beta=6 \\ & \alpha^2+\beta^2+\gamma^2=75 \end{aligned}
Q49
If 10sin⁡4θ+15cos⁡4θ=610 \sin ^4 \theta+15 \cos ^4 \theta=6, then the value of 27cosec⁡6θ+8sec⁡6θ16sec⁡8θ\dfrac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} is
A 25\dfrac{2}{5}
B 35\dfrac{3}{5}
C 15\dfrac{1}{5}
D 34\dfrac{3}{4}
Correct Answer
Option A
Solution
10sin⁡4θ+15cos⁡4θ=6⇒10sin⁡4θ+10cos⁡4θ+5cos⁡4θ=6⇒10[(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ]+5cos⁡4θ=6⇒10−20(1−cos⁡2θ)cos⁡2θ+5cos⁡4θ=6 Let cos⁡2θ=x10−20(x−x2)+5x2=6\begin{aligned} &\begin{aligned} & 10 \sin ^4 \theta+15 \cos ^4 \theta=6 \\ \Rightarrow & 10 \sin ^4 \theta+10 \cos ^4 \theta+5 \cos ^4 \theta=6 \\ \Rightarrow & 10\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6 \\ \Rightarrow & 10-20\left(1-\cos ^2 \theta\right) \cos ^2 \theta+5 \cos ^4 \theta=6 \end{aligned}\\ &\text{ Let } \cos ^2 \theta=x\\ &10-20\left(x-x^2\right)+5 x^2=6 \end{aligned}
⇒25x2−20x+4=0(5x−2)2=0⇒x=25⇒cos⁡2θ=25⇒sin⁡2θ=35,sec⁡2θ=52,cosec⁡2θ=53\begin{aligned} \Rightarrow & 25 x^2-20 x+4=0 \\ & (5 x-2)^2=0 \Rightarrow x=\frac{2}{5} \\ \Rightarrow & \cos ^2 \theta=\frac{2}{5} \Rightarrow \sin ^2 \theta=\frac{3}{5}, \\ & \sec ^2 \theta=\frac{5}{2}, \operatorname{cosec}^2 \theta=\frac{5}{3} \end{aligned}
27cosec⁡6θ+8sec⁡6θ16sec⁡8θ=27(53)3+8(52)316(52)4=53+5354=2.5354=25\begin{aligned} \frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} & =\frac{27\left(\frac{5}{3}\right)^3+8\left(\frac{5}{2}\right)^3}{16\left(\frac{5}{2}\right)^4} \\ & =\frac{5^3+5^3}{5^4}=\frac{2.5^3}{5^4}=\frac{2}{5} \end{aligned}
Q50
If u=a2cos⁡2θ+b2sin⁡2θ+a2sin⁡2θ+b2cos⁡2θu = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } then the difference between the maximum and minimum values of u2{u^2} is given by :
A (a−b)2{\left( {a - b} \right)^2}
B 2a2+b22\sqrt {{a^2} + {b^2}}
C (a+b)2{\left( {a + b} \right)^2}
D 2(a2+b2)2\left( {{a^2} + {b^2}} \right)
Correct Answer
Option A
Solution

Given

u=a2cos⁡2θ+b2sin⁡2θu = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta }
+a2sin⁡2θ+b2cos⁡2θ+ \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }

∴\therefore

u2=a2cos⁡2θ+b2sin⁡2θ+a2sin⁡2θ+b2cos⁡2θ{u^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta
+2(a2cos⁡2θ+b2sin⁡2θ)×(a2sin⁡2θ+b2cos⁡2θ)+ 2\sqrt {\left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)} \times \sqrt {\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)}
⇒u2=\Rightarrow {u^2} =
a2+b2{a^2} + {b^2}

++

2(a4+b4)cos⁡2θsin⁡2θ+a2b2(cos⁡4θ+sin⁡4θ)2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)}
⇒u2=\Rightarrow {u^2} =
a2+b2{a^2} + {b^2}

++

2(a4+b4)cos⁡2θsin⁡2θ+a2b2(1−2cos⁡2θsin⁡2θ)2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)}
⇒u2=\Rightarrow {u^2} =
a2+b2{a^2} + {b^2}

++

2(a4+b4−2a2b2)cos⁡2θsin⁡2θ+a2b22\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}}
⇒u2=\Rightarrow {u^2} =
a2+b2{a^2} + {b^2}

++

2(a2−b2)2(2cos⁡θsin⁡θ)24+a2b22\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}}
⇒u2=\Rightarrow {u^2} =
a2+b2{a^2} + {b^2}

++

2(a2−b2)2sin⁡22θ4+a2b22\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}}

We know

0≤sin⁡22θ≤10 \le {\sin ^2}2\theta \le 1

∴\therefore

0≤(a2−b2)2sin⁡22θ4≤(a2−b2)240 \le {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4}

⇒\Rightarrow

a2b2≤{a^2}{b^2} \le
(a2−b2)2sin⁡22θ4+a2b2{\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}
≤(a2−b2)24+a2b2\le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}

∴\therefore Min value of

u2=a2+b2{u^2} = {a^2} + {b^2}
+2a2b2+ 2\sqrt {{a^2}{b^2}}

=

(a+b)2{\left( {a + b} \right)^2}

and Max value of

u2=a2+b2{u^2} = {a^2} + {b^2}
+2(a2−b2)24+a2b2+ 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}}
=a2+b2= {a^2} + {b^2}
+2(a2−b2)2+4a2b24+ 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2} + 4{a^2}{b^2}} \over 4}}
=a2+b2= {a^2} + {b^2}
+2(a2+b2)24+ 2\sqrt {{{{{\left( {{a^2} + {b^2}} \right)}^2}} \over 4}}
=a2+b2= {a^2} + {b^2}
+ a2+b2+\, {a^2} + {b^2}
=2(a2+b2){ = 2\left( {{a^2} + {b^2}} \right)}

Max of

u2{u^2}

- Min of

u2{u^2}

=

2(a2+b2){2\left( {{a^2} + {b^2}} \right)}

-

(a+b)2{\left( {a + b} \right)^2}

=

2(a2+b2){2\left( {{a^2} + {b^2}} \right)}

-

(a2+b2+2ab){\left( {{a^2} + {b^2} + 2ab} \right)}

=

=2(a2+b2−2ab)4\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}}

=

(a−b)2{\left( {a - b} \right)^2}
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →