JEE Mathematics Inverse Circular Functions MCQs Set 12

Answer: (c) \(\{\pi\}\)

Explanation:
1. The domain needs \([x^2+\tfrac12]\in\{0,1\}\), so \(x^2<\tfrac32\).
2. If \(x^2<\tfrac12\): \(\sin^{-1}0+\cos^{-1}(-1)=\pi\).
3. If \(\tfrac12\le x^2<\tfrac32\): \(\sin^{-1}1+\cos^{-1}0=\pi\). The range is \(\{\pi\}\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) \(\dfrac12\)

Explanation:
1. Each \(\cos^{-1}\beta_r\le\dfrac\pi2\) for \(\beta_r\ge0\), so every term equals \(\dfrac\pi2\): \(\beta_r=0\) and \(A=0\).
2. Near 0: \((1+x^2)^{1/3}\approx1\) and \((1-2x)^{1/4}\approx1-\dfrac x2\), so the numerator is \(\approx\dfrac x2\).
3. The limit is \(\dfrac{x/2}{x}=\dfrac12\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) \(\left\{\log\dfrac\pi2\right\}\)

Explanation:
1. \(\log[x]\) needs \([x]\ge1\), and \(\sin^{-1}[x]\) needs \([x]\in\{-1,0,1\}\) with \(\sin^{-1}[x]>0\): so \([x]=1\).
2. \(f=\sin^{-1}(0)+\log\dfrac\pi2=\log\dfrac\pi2\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (b) \(15^\circ\)

Explanation:
1. Put \(t=x^2\): \(\dfrac t{t^2+2t+3}=\dfrac1{t+3/t+2}\le\dfrac1{2\sqrt3+2}\).
2. Maximum argument \(\dfrac{2\sqrt3-2}{2\sqrt3+2}=\dfrac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3=\tan15^\circ\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) \(4\)

Explanation:
1. \((x^2+1)^2-4x^2=(x^2-1)^2\), so the equation is \(|\tan^{-1}x|=|x^2-1|\).
2. Both sides are even in \(x\); for \(x\ge0\) the graphs of \(\tan^{-1}x\) and \(|x^2-1|\) meet at two points (one in \((0,1)\), one in \((1,\infty)\)).
3. Hence there are 4 solutions in all.

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) \(\dfrac{20}{n^2+20n+1}\)

Explanation:
1. Each term \(\tan^{-1}\dfrac1{1+k(k+1)}=\tan^{-1}(k+1)-\tan^{-1}k\) for \(k=n,\dots,n+19\).
2. \(S=\tan^{-1}(n+20)-\tan^{-1}n\), so \(\tan S=\dfrac{20}{1+n(n+20)}\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) \(2\)

Explanation:
1. Sum and product of roots are positive; the discriminant condition: \((\tan^{-1}\alpha)^3\ge8(\cot^{-1}\alpha)^3\).
2. \(\tan^{-1}\alpha\ge2\cot^{-1}\alpha=\pi-2\tan^{-1}\alpha\), so \(\tan^{-1}\alpha\ge\dfrac\pi3\) and \(\alpha\ge\sqrt3\).
3. The least integer is 2.

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) \(\tan^{-1}n\)

Explanation:
1. With \(a=n^2-n+1\): \((n^2+1)((n-1)^2+1)-a^2=1\), so \(\sec^2-1=\dfrac1{a^2}\) and the term is \(\tan^{-1}\dfrac1{n^2-n+1}\).
2. \(\dfrac1{1+n(n-1)}\) gives \(\tan^{-1}n-\tan^{-1}(n-1)\); the sum telescopes to \(\tan^{-1}n\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) \(\cos\dfrac\pi8\)

Explanation:
1. With \(a=\sin^{-1}x,\ b=\cos^{-1}x\): \(a^3-b^3+ab(a-b)=(a-b)(a+b)^2\).
2. \(a+b=\dfrac\pi2\), so \(a-b=\dfrac\pi4\) and \(a=\dfrac{3\pi}8\).
3. \(x=\sin\dfrac{3\pi}8=\cos\dfrac\pi8\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) \(\cot^{-1}\dfrac12\)

Explanation:
1. The \(n\)th term is \(\cot^{-1}\left(n^2+\dfrac34\right)=\tan^{-1}\left(n+\dfrac12\right)-\tan^{-1}\left(n-\dfrac12\right)\).
2. \(S_n=\tan^{-1}\left(n+\dfrac12\right)-\tan^{-1}\dfrac12\to\dfrac\pi2-\tan^{-1}\dfrac12=\cot^{-1}\dfrac12\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) \(\dfrac\pi4\)

Explanation:
1. \(\tan^{-1}\dfrac1{n^2+n+1}=\tan^{-1}(n+1)-\tan^{-1}n\) for \(n=1,2,\dots\).
2. The sum is \(\dfrac\pi2-\tan^{-1}1=\dfrac\pi4\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (b) \(x=\dfrac12\)

Explanation:
1. Taking cosine: \(\sqrt3x=\sin(\cos^{-1}x)=\sqrt{1-x^2}\), so \(x^2=\dfrac14\).
2. For \(x=-\dfrac12\): \(\cos^{-1}x>\dfrac\pi2\) and the sum exceeds \(\dfrac\pi2\); hence \(x=\dfrac12\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) \((\alpha+\beta)(\alpha^2+\beta^2)\)

Explanation:
1. Let \(\theta=\tan^{-1}\dfrac\alpha\beta\), \(r=\sqrt{\alpha^2+\beta^2}\): \(\cos\theta=\dfrac\beta r,\ \sin\theta=\dfrac\alpha r\).
2. \(\operatorname{cosec}^2\dfrac\theta2=\dfrac2{1-\cos\theta}\) gives the first term \(\dfrac{\alpha^3r}{r-\beta}\); \(\tan^{-1}\dfrac\beta\alpha=\dfrac\pi2-\theta\) gives \(\sec^2=\dfrac2{1+\sin\theta}\) and the second term \(\dfrac{\beta^3r}{r+\alpha}\).
3. Using \(r^2-\beta^2=\alpha^2\): the first term is \(\alpha r(r+\beta)\); using \(r^2-\alpha^2=\beta^2\): the second is \(\beta r(r-\alpha)\).
4. Sum \(=r[\alpha r+\alpha\beta+\beta r-\alpha\beta]=r^2(\alpha+\beta)=(\alpha^2+\beta^2)(\alpha+\beta)\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) \(\dfrac\pi4\)

Explanation:
1. \(\cot^{-1}(2n^2)=\tan^{-1}\dfrac2{4n^2}=\tan^{-1}(2n+1)-\tan^{-1}(2n-1)\).
2. The sum is \(\dfrac\pi2-\tan^{-1}1=\dfrac\pi4\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) 1

Explanation:
1. \(\dfrac\pi2-\tan^{-1}x=\cot^{-1}x\sim\dfrac1x\).
2. \(\left(\dfrac1x\right)^{1/x}=e^{-\ln x/x}\to e^0=1\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (b) 3

Explanation:
1. \(\cos2\theta=-1\) gives \(\theta=\pm\dfrac\pi2\), i.e. \(\cot(\tan^{-1}u)=\dfrac1u=\pm1\) with \(u=\sec(6\operatorname{cosec}^{-1}x)\).
2. So \(\cos(6\varphi)=\pm1\) with \(\varphi=\operatorname{cosec}^{-1}x\in\left(0,\dfrac\pi2\right]\): \(\varphi=\dfrac\pi6,\dfrac\pi3,\dfrac\pi2\).
3. 3 solutions.

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) 7

Explanation:
1. \(3\cos^{-1}(4x)\ge\pi\) gives \(\cos^{-1}(4x)\ge\dfrac\pi3\), so \(-1\le4x\le\dfrac12\).
2. \(a=-\dfrac14,\ b=\dfrac18\): \(4a+64b=-1+8=7\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) \(\left[\dfrac12,1\right]\)

Explanation:
1. With \(A=\cos^{-1}x\): the bracket equals \(\cos\dfrac\pi3\cos A+\sin\dfrac\pi3\sin A=\cos\left(\dfrac\pi3-A\right)\).
2. So the second term is \(\left|\dfrac\pi3-A\right|\) and the identity \(A+\left|\dfrac\pi3-A\right|=\dfrac\pi3\) holds when \(A\le\dfrac\pi3\), i.e. \(x\ge\dfrac12\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) \(\text{odd values of }n\)

Explanation:
1. \(\tan x=-3\), so \(\tan2x=\dfrac{-6}{1-9}=\dfrac34\) and \(12\tan2x=9\).
2. \(\cos x=(-1)^n\dfrac1{\sqrt{10}}\), so \(\dfrac{\sqrt{10}}{\cos x}=10(-1)^n\).
3. \(9+10(-1)^n+1=0\) requires \(n\) odd.

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) \(x^4=6x^2-1\)

Explanation:
1. \(y\) is infinite when \(4\tan^{-1}x=\dfrac\pi2\) (or \(\dfrac{3\pi}2\)): \(\tan^{-1}x=\dfrac\pi8\) or \(\dfrac{3\pi}8\).
2. \(x=\sqrt2-1\) or \(\sqrt2+1\) (and negatives), so \(x^2=3\mp2\sqrt2\).
3. These are the roots of \(t^2-6t+1=0\), i.e. \(x^4=6x^2-1\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) 0

Explanation:
1. \(\sin^{-1}u+\cos^{-1}v=\dfrac\pi2\) needs \(u=v\) with \(|u|\le1\): \(x^2+x+1=ax+1\), so \(x=0\) or \(x=a-1\).
2. \(|x^2+x+1|\le1\) needs \(-1\le x\le0\), so \(x=a-1\in[-1,0)\) gives \(0\le a<1\).
3. The only integer is \(a=0\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) \(-\dfrac14\)

Explanation:
1. Taking tangents: \(\dfrac{4x-x^3}{2-3x^2}=3x\) gives \(8x^3-2x=0\).
2. Roots \(0,\ \pm\dfrac12\) (each checked in the original equation).
3. \(\alpha\beta+\beta\gamma+\gamma\alpha=-\dfrac14\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) 1

Explanation:
1. \(x^2+x+1\le1\) and \(x^2+x\ge0\) force \(x^2+x=0\): \(x=0\) or \(x=-1\).
2. Both satisfy the equation (\(2\cdot\dfrac\pi2+\dfrac\pi2=\dfrac{3\pi}2\)). Sum \(=-1\), so \(p=1\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (b) 1

Explanation:
1. \(\cos^{-1}\left(-\dfrac{\sqrt3}2\right)=\dfrac{5\pi}6\).
2. \(-\cos\left(\dfrac{5\pi}6+\dfrac\pi6\right)=-\cos\pi=1\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (c) 3

Explanation:
1. \(\operatorname{cosec}^{-1}\dfrac54=\sin^{-1}\dfrac45\).
2. \(\sin^{-1}\dfrac x5=\dfrac\pi2-\sin^{-1}\dfrac45=\cos^{-1}\dfrac45=\sin^{-1}\dfrac35\), so \(x=3\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) 4

Explanation:
1. \(\cos^{-1}x=a+\cos^{-1}\dfrac y2\) gives \(x=\dfrac y2\cos a-\sin a\sqrt{1-\dfrac{y^2}4}\).
2. Rearrange and square: \(x^2-xy\cos a+\dfrac{y^2}4\cos^2a=\sin^2a-\dfrac{y^2}4\sin^2a\), i.e. \(4x^2-4xy\cos a+y^2=4\sin^2a\); \(k=4\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) 1

Explanation:
1. The largest values are \(\dfrac{5\pi}2\) (at \(x=1\)) and \(3\pi\) (at \(y=-1\)).
2. Their sum is exactly \(\dfrac{11\pi}2\), so only \((1,-1)\) works.

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (b) \(-\dfrac12\)

Explanation:
1. \(\sin(\cot^{-1}(1+x))=\dfrac1{\sqrt{1+(1+x)^2}}\) and \(\cos(\tan^{-1}x)=\dfrac1{\sqrt{1+x^2}}\).
2. Equal when \((1+x)^2=x^2\): \(x=-\dfrac12\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (d) 1

Explanation:
1. The series are \(\ln(1+x)\) and \(\ln(1+x^2)\).
2. \(\sin^{-1}u+\cos^{-1}v=\dfrac\pi2\) needs \(u=v\): \(1+x=1+x^2\), so \(x=1\) (as \(x\ne0\)).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (a) \(\dfrac{25}{23}\)

Explanation:
1. \(1+\sum_{k=1}^n2k=1+n(n+1)\), so each term is \(\tan^{-1}(n+1)-\tan^{-1}n\).
2. The sum is \(\tan^{-1}24-\tan^{-1}1=\tan^{-1}\dfrac{23}{25}\), and the cotangent is \(\dfrac{25}{23}\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.

Answer: (b) -2

Explanation:
1. Each \(\cos^{-1}\le\pi\), so the sum is at most \(2\pi^2\) with equality only when both equal \(\pi\).
2. \(x=y=-1\) and \(x^5+y^5=-2\).

Teacher's Note:
1. Always check the principal value ranges of the inverse functions before simplifying.
2. Convert sums of inverse tangents using tan(A + B) and look for telescoping terms.