Eamcet Mathematics - Inverse Circular Functions Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
\(x\epsilon\left({3\pi\over 2}, 2\pi\right)\)then the value of the expression sin-1 [cos {cos-1 (cos x)} + sin-1 (sin x)], is
- (a)
\(-{\pi\over 2}\)
- (b)
\({\pi\over 2}\)
- (c)
0
- (d)
\(\pi\)
\(x\epsilon\left({3\pi\over 2}, 2\pi\right)\)
Then, cos-1(cosx)=2π-x
and sin-1 (sin x)=x-2π
cos-1 (cos x) + sin-1 (sin x) = 0
Therefore, sin-1 [cos {cos-1 (cos x) + sin-1(sin x)}]
\(= sin^{-1} {cos (0)} = sin^{-1} (1) ={\pi\over 2}\)
The sum of the infinite terms of the series \(tan^{-1}\left(1\over 3\right)+tan^{-1}\left(2\over 9\right)+tan^{-1}\left(4\over 33\right)+......\) is equal to
- (a)
\(\pi\over 6\)
- (b)
\(\pi\over 4\)
- (c)
\(\pi\over 3\)
- (d)
\(\pi\over 2\)
\(T_r=tan^{-1}\left(2^{r-1}\over 1+2^{2r-1}\right)\)
\(=tan^{-1}\left(2^r-2^{r-1}\over 1+2^r.2^{r-1}\right)\)
tan-1 (2r) - tan-1 (2r-1)
\(S_n=\sum_{r=1}^n T_r=\sum_{r=1}^n tan^{-1} (2^r) - tan^{-1} (2^{r-1})\)
= tan-1 (2n) - tan-1 (20)
= tan-1 (2n) - tan-1 (1)
\(= tan^{ -1} (2^n) -{\pi\over 4}\)
Hence, \(S_\infty=tan^{-1}(2^\infty)-{\pi\over 4}\)
\(=tan^{-1}(\infty)-{\pi\over 4}\)
\(={\pi\over 2}-{\pi\over 4}={\pi\over 4}\)
If \(sin^{-1}\sqrt{(x^2+2x+1)}+sec^{c-1}\sqrt{(x^2+2x+1)}={\pi\over 2}\)x ≠ 0, then the value of \(2sec^{-1}\left(x\over 2\right)+sin^{-1}\left(x\over 2\right)\) is equal to
- (a)
\(-{3\pi\over 2}\)
- (b)
\({3\pi\over 2}\)
- (c)
\(-{\pi\over 2}\)
- (d)
\({\pi\over 2}\)
\(sin^{-1}\sqrt{(x^2+2x+1)}+sec^{c-1}\sqrt{(x^2+2x+1)}={\pi\over 2}\)
or \(sin^{-1}|x+1|+cos^{-1}\left(1\over |x+1|\right)={\pi\over 2}\)
or \(|x+1|={1\over|x+1|}(x\neq-1)\)
1 x + 1|2= 1
| x + 1|= 1
x+1=±1
x = 0, - 2
But given x ≠ 0
x =- 2
Then \(2sec^{-1}\left(x\over 2\right)+sin^{-1}\left(x\over 2\right)\)
= 2 sec-1 (-1) + sin-1 (-1)
\(=2\pi-{\pi\over 2}={3\pi\over 2}\)
The greatest of tan1, tan-1 1, sin-11, sin1, cos 1, is
- (a)
sin 1
- (b)
tan 1
- (c)
tan-1 1
- (d)
none of these
1 Radian = 57017'44.8"
\(sin^{-1}1={\pi\over 2}\)is the greatest.
If the mapping f(x)=ax+b,a>0 maps [-1,1] onto [0, 2], then cot [cot-1 7 + cot-1 8 + cot-1 18] is equal to
- (a)
f( - 1)
- (b)
f(0)
- (c)
f(1)
- (d)
f(2)
f(x)=ax+b
f'(x)=a>O
f(x) is an increasing function.
⇒ f(-1) = 0 and f(1) = 2
or , -a + b = 0
and a + b = 2
Then, a= b =1
⇒ f(x)= x + 1
Now, cot [cot-1 7 + cot-1 8 + cot-1 18]
\(=cot\left\{ tan^{-1}\left(1\over 7\right)+tan^{-1}\left(1\over 8\right)+tan^{-1}\left(1\over 18\right) \right\}\)
\(=cot\left\{ tan^{-1}\left({ {1\over 7}+{1\over 8}} \over1-{1\over 7}.{1\over 8} \right)+tan^{-1}\left(1\over 18\right) \right\}\)
\(=cot\left\{ tan^{-1}\left({15\over 55} \right)+tan^{-1}\left(1\over 18\right) \right\}\)
\(=cot\left\{ tan^{-1}\left({3\over 11} \right)+tan^{-1}\left(1\over 18\right) \right\}\)
\(=cot\left\{ tan^{-1}\left({{3\pi\over 11}+{1\over 18}\over1-{3\over 1.1}.{1\over 18}}\right) \right\}\)
\(=cot\left\{ tan^{-1}\left({65\over 195} \right) \right\}\)
\(=cot\left\{ tan^{-1}\left({1\over 3} \right) \right\}\)
= cot (cot-1 3) = 3 = 1 + 2 = 1(2)
The number of solutions for the equation \(2sin^{-1}\sqrt{(x^2-x+1)}+cos^{-1}\sqrt{(x^2-x)}={3\pi\over 2}\)is
- (a)
1
- (b)
2
- (c)
3
- (d)
infinite
\(x^2-x+1=\left(x-{1\over 2}\right)^2+{3\over 4}>0\)
For \(sin^{-1}\sqrt{(x^2-x+1)}\)
0 < X2 - X + 1< 1
or x2 - x > 0
For \(cos^{-1}\sqrt{(x^2-x)}\)
X2 - X < 0
From Eqs. (i) and (ii), we get
X2 - x = 0
x =0,1
The number of solutions of the equation,\(cos^{-1}+m\ cos^{-1}x={n\pi\over 2}\) where m> 0, n > 0, is
- (a)
0
- (b)
1
- (c)
3
- (d)
infinite
For cos-1(1 - x)
-1 < - 1 -x > 1
⇒ 1 > -1+ x> -1
or 2 > x > 0
or x ∈ [0,2] ... (i)
For cos-1 x
⇒ - 1 < x < 1 ... (ii)
From Eqs. (i) and (ii), we get
0 < x < 1
LHS = cos-1(1 - x) + m cos-1 x is defined in [0, 1]
So, LHS > 0, but RHS < 0(x < 0)
Equality holds, if
LHS = 0 and RHS = 0
cos-1(1 - x) + m cos-1 x = 0
⇒ cos-1(1 - x) = 0 and cos-1 x = 0
Which is impossible
ie, no solution.
Hence, number of solution = 0
If \(cos^{-1}\left(x\over 3\right)+cos^{-1}\left(y\over 2\right)={\theta\over 2}\)then the value of \(4x^2-12xy\ cos\left(\theta\over 2\right)+9y^2\)is equal to
- (a)
18(1 + cos θ)
- (b)
18(1- cos θ)
- (c)
36 (1 + cos θ)
- (d)
36 (1 - cos θ)
\(cos^{-1}\left( x\over 3\right)+cos^{-1}\left(y\over 2\right)={\theta\over2}\)
\(\Rightarrow cos\left\{ cos^{-1}\left( x\over 3\right)+cos^{-1}\left(y\over 2\right) \right\}=cos\left(\theta\over 2\right)\)
\(\Rightarrow\ cos \left\{cos^{-1}\left(x\over 3\right) \right\}.co\left\{cos^{-1}\left(y\over 2\right)\right\}\)
\(-sin\left\{cos^{-1}\left(x\over 3\right)\right\}.\left\{cos^{-1}\left(y\over 2\right)\right\}=cos{\theta\over 2}\)
\(⇒{x\over 3}.{y\over 2}-\sqrt{\left(1-{x\over 3}\right)}.\sqrt{1-\left({y\over 3}\right)^2}=\left(cos{\theta\over 2}\right)\)
\(⇒{x^2y^2\over 36}+cos^2\left(\theta\over 2\right)-{2xy\over 6}cos\left(\theta\over 2\right)\)
\(=1-{x^2\over 9}-{y^2\over 4}+{x^2y^2\over 2}\)
\(\Rightarrow\ 4x^2-xy\ cos\left(\theta\over 2\right)+9y^2\)
\(=\left({1-cos^2{\theta\over 2}}\right)\)
\(=36\left(1-{1+cos\theta\over 2}\right)=18(1-cos\theta)\)
\(cos^{-1}\left[cos\left(-{17\over 5}\pi\right)\right]\)is equal to
- (a)
\(-{17\pi\over 15}\)
- (b)
\({17\pi\over 15}\)
- (c)
\({2\pi\over 15}\)
- (d)
\({13\pi\over 15}\)
\(cos^{-1}\left\{cos\left(-{17\pi\over 15}\right)\right\}\)
\(cos\left(-{17\pi\over 15}\right)=cos{17 \pi\over 15}<0\)
Principal value\(=2\pi'-{17\pi\over 15}={13\pi\over 15}\)
\(tan\left\{2tan^{-1}\left(1\over 5\right)-{\pi\over 4}\right\}\)is equal to
- (a)
\({5\over 4}\)
- (b)
\({5\over 16}\)
- (c)
\(-{7\over 17}\)
- (d)
\({7\over 17}\)
\(tan\left\{ 2tan^{-1}\left(1\over 5\right)-{\pi\over 4}\right\}\)
\(=tan\left\{tan^{-1}\left(2\times{1\over 5}\over 1-{1\over 25}\right)-tan^{-1}1\right\}\)
\(=tan\left\{tan^{-1}\left(10\over 24\right)-tan^{-1}1\right\}\)
\(=tan\ tan^{-1}\left\{{10\over 24}-1\over 1+{10\over 24}\right\}\)
\(=-{14\over 34}\)
\(=-{7\over 17}\)
\(x+{1\over x}=2\) the principal value of sin-1 x is
- (a)
\(\pi\over 4\)
- (b)
\(\pi\over 2\)
- (c)
\(\pi\)
- (d)
\(3\pi\over 2\)
\(x+{1\over x}=2\)
\(⇒\left(\sqrt x-{1\over \sqrt x}\right)^2=0\)
\(⇒{x-1\over \sqrt x}=0\)
x=1
So, the principal value of sin-1 x is \(\pi\over 2\)
If cos-1 x + cos-1 y + cos-1 z = 3rr, then x y + y z + z is equal to
- (a)
-3
- (b)
0
- (c)
3
- (d)
-1
0 < cos-1 X < π ,0 < cos-1 < π
0 < cos-1 z < π
But given, cos-1 x + cos-1 y + cos-1 z = 3π
Which is possible only when
cos-1 x = π, cos-1 y = π and cos-1 z = π
x = -1, y = -1 and z =-1
Then, xy + yz + zx =3
The number of real solutions of \(tan^{-1}\sqrt{\{x(x+1)\}}+sin^{-1}\sqrt{\{x^2+x+1\}}={{\pi\over 2}}is\)
- (a)
zero
- (b)
one
- (c)
two
- (d)
infinite
Since, \(tan^{-1}\sqrt{\{x(x+1)\}}+sin^{-1}\sqrt{\{x^2+x+1\}}={{\pi\over 2}}\)
\(\Rightarrow\ cos^{-1}\left(1\over\sqrt{(x^2+x+1)}\right)+sin^{-1}\left(\sqrt3\over 2\right)\sqrt{(x^2+x+1)}={\pi\over 2}\)
Hence, \({1\over \sqrt{(x^2+x+1)}}=\sqrt{(x^2+x+1)}\)
or X2 + x = 0
or x = -1,0
Asolution of the equation tan-1(1 + x) + tan-1(1- x) = \(\pi\over 2\) is
- (a)
x=1
- (b)
x=-1
- (c)
x=0
- (d)
x=ㅠ
tan-1(1 + x) + tan-1(1- x) = \(\pi\over 2\)
\(tan^{-1}\left(1+x+1-x\over 1-(1-x^2)\right)={\pi\over 2}\)
\(tan^{-1}\left(2\over x^2\right)={\pi\over 2}\)
\(cot^{-1}\left(2\over x^2\right)={\pi\over 2}\)
or \({x^2\over 2}cot{\pi\over 2}=0\)
x2=0
x=0
If X2 + y2 + z2 =r2, then \(tan^{-1}\left(xy\over zr\right)+tan^{-1}\left(yz\over xr\right)+tan^{-1}\left(zx\over yr\right)\)is equal
- (a)
π
- (b)
\(π\over 2\)
- (c)
0
- (d)
none of these
\(tan^{-1}\left(xy\over zr\right)+tan^{-1}\left(yz\over xr\right)+tan^{-1}\left(zx\over yr\right)\)
\(=tan\left[{ {xy\over zr}+{yz\over xr}+{zx\over yr}-{xyz\over r^3}} \over{1\left(x^2+y^2+z^2\over r^2\right)}\right]\)
\(=tan^{-1}\infty={\pi\over 2}\)
The values of x satisfying sin-1 x + sin-1 (1- x) = cos-1 x are
- (a)
0
- (b)
\(\frac{1}{2}\)
- (c)
1
- (d)
2
\({ sin }^{ -1 }x+{ sin }^{ -1 }\left( 1-x \right) ={ cos }^{ -1 }x\)
\(\Rightarrow \quad \frac { \pi }{ 2 } -\cos ^{ -1 }{ x } +\frac { \pi }{ 2 } -\cos ^{ -1 }{ \left( 1-x \right) } ={ cos }^{ -1 }x\)
\(\Rightarrow \quad 2\cos ^{ -1 }{ x } =\pi -\cos ^{ -1 }{ \left( 1-x \right) } \)
\(\\ \\ \Rightarrow \quad \cos ^{ -1 }{ \left( { 2x }^{ 2 }-1 \right) } =\cos ^{ -1 }{ \left( x-1 \right) } \)
\(\Rightarrow \quad { 2x }^{ 2 }-1=x-1\)
\(\Rightarrow \quad x\left( 2x-1 \right) =0\)
\(\Rightarrow \quad \therefore \quad x=0,\frac { 1 }{ 2 } \)
If 6 sin-1 (x2- 6x + 8.5) = \(\pi\), then
- (a)
x = 1
- (b)
x = 2
- (c)
x = 3
- (d)
x = 4
\(\because\) 6 sin-1(x2-6x+8.5) = \(\pi\)
\(\therefore \) sin-1(x2-6x+8.5)=\(\frac{\pi}{6}\)
\(\Rightarrow\) x2-6x+8.5 = sin\((\frac{\pi}{6})\)
\(\Rightarrow\) x2-6x+8 = 0
\(\Rightarrow\) (x-4)(x-2) = 0
\(\therefore \) x = 2, 4
Let f(x) = \({ e }^{ { cos }^{ -1s }in\left( x+\frac { \pi }{ 3 } \right) }\), then
- (a)
\(f\left( \frac { 8\pi }{ 9 } \right) ={ e }^{ \frac { 5\pi }{ 18 } }\)
- (b)
\(f\left( \frac { 8\pi }{ 9 } \right) ={ e }^{ \frac { 13\pi }{ 18 } }\)
- (c)
\(f\left( -\frac { 7\pi }{ 4 } \right) ={ e }^{ \frac { \pi }{ 12 } }\)
- (d)
\(f\left( -\frac { 7\pi }{ 4 } \right) ={ e }^{ \frac { 11\pi }{ 12 } }\)
\(f\left( -\frac { 8\pi }{ 9 } \right) ={ e }^{ { cos }^{ -1 }sin\left( \frac { 8\pi }{ 9 } +\frac { \pi }{ 3 } \right) }={ e }^{ { cos }^{ -1 }sin\left( \frac { 11\pi }{ 9 } \right) }\)
= \({ e }^{ { cos }^{ -1\quad }cos\frac { 13\pi }{ 18 } }=e^{ \frac { 13\pi }{ 18 } }\)
and \(f\left( -\frac { 7\pi }{ 4 } \right) ={ e }^{ { cos }^{ -1 }sin\left( -\frac { 7\pi }{ 4 } +\frac { \pi }{ 3 } \right) }\)
= \({ e }^{ { cos }^{ -1 }sin\left( -\frac { 17\pi }{ 12 } \right) }\)
= \({ e }^{ { cos }^{ -1 }sin\frac { \pi }{ 12 } }=e^{ \frac { \pi }{ 12 } }\)
If \(\sum_{i=i}^{2n}sin^{-1}x_i=n\pi\ then\ \sum_{i=1}^{2n}\)is equal to
- (a)
n
- (b)
2n
- (c)
\(n(n+1)\over 2\)
- (d)
none of these
\(\sum_{i=i}^{2n}sin^{-1}x_i=n\pi\ =\ \sum_{i=1}^{2n}\left(\pi\over 2\right)\)
Which is possible only when
\(sin^{-1}x_i={\pi\over 2}∀i\)
\(x_i=1∀i\)
then \(\sum_{i=1}^{2n}x_i=\sum_{i=1}^{2n}1=2n\)
The inequality sin-1 (sin 5) > X2 - 4x holds, if
- (a)
\(x=2-\sqrt{(9-2\pi)}\)
- (b)
\(x=2+\sqrt{(9-2\pi)}\)
- (c)
\(x\epsilon (2-\sqrt{(9-2\pi)}, 2+\sqrt{(9-2\pi)}\)
- (d)
\(x>2+\sqrt{(9-2\pi)}\)
sin-1 (sin 5) > x2 - 4x
sin-1 {sin (5 - 2π)} > X2 - 4x
⇒ 5 - 2π > X2 - 4x
⇒ 9-2π > (x-2)2
⇒ (x - 2)2 < 9 - 2π
\(⇒ -\sqrt{(9-2\pi)}< x-2<\sqrt{(9-2\pi)}\)
\(⇒2-\sqrt{(9-2\pi)}< x <+\sqrt{(9-2\pi)}\)
\(x\epsilon(2-\sqrt{(9-2\pi)}2+\sqrt{(9-2\pi)})\)
The value of tan2(sec-12) + cot2 (cosec-13) is
- (a)
13
- (b)
15
- (c)
11
- (d)
none of these
tan2(sec-12) + cot2(cosec-13)
= sec2(sec-12) - 1+ cosec2(cosec-13) - 1
= (see sec-12)2 -1 + (cosec cosec-1 3)2 - 1
=22-1+32-1
=4-1+9-1=11
3The equation sin-1 x = 2 sin-1 a has a solution for
- (a)
all real values of a
- (b)
a < 1
- (c)
\(-{1\over \sqrt2}\le a\le {1\over \sqrt2}\)
- (d)
-1< a< 1
Since, sin-1 x = 2 sin-1 a
\(⇒-{\pi\over 2}\le sin^{-1}x\le{\pi\over 2}\)
\(⇒-{\pi\over 2}\le sin^{-1}a\le{\pi\over 2}\)
\(⇒-{\pi\over 2}\le sin^{-1}a\le{\pi\over 4}\)
\(⇒sin\left(-{\pi\over 4}\right)\le a\le sin\left({\pi\over 4}\right)\)
\(⇒-{1\over \sqrt2}\le a\le{1\over \sqrt2}\)
If \(\alpha \le \sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } +\tan ^{ -1 }{ x } \le \beta \), then
- (a)
\(\alpha =0\)
- (b)
\(\beta =\pi /2\)
- (c)
\(\alpha =0\pi /4\)
- (d)
\(\beta =\pi \)
We have, \(\sin ^{ -1 }{ x } +\cos ^{ -1 }{ x } +\tan ^{ -1 }{ x } =\frac { \pi }{ 2 } +\tan ^{ -1 }{ x } \)
\(\because \quad \quad -\frac { \pi }{ 2 } \le \tan ^{ -1 }{ x } \le \frac { \pi }{ 2 } \)
\(\Rightarrow \quad 0\le \frac { \pi }{ 2 } +\tan ^{ -1 }{ x } \le \pi \)
\(\therefore \quad \quad \alpha =0,\quad \beta =\pi \)
The solution of the equation sin[2cos-1{cot(2tan-1x)}] = 0 are
- (a)
\(\pm 1\)
- (b)
\(1\pm \sqrt { 2 } \)
- (c)
\(-1\pm \sqrt { 2 } \)
- (d)
none of these
\(\therefore \quad sin\left[ 2{ cos }^{ -1 }\left\{ cot\left( 2{ tan }^{ -1 }x \right) \right\} \right] =0\)
\(\Rightarrow \quad 2{ cos }^{ -1 }\left\{ cot\left( 2{ tan }^{ -1 }x \right) \right\} =n\pi ,\ n\epsilon I\)
\(\Rightarrow \quad { cos }^{ -1 }\left( cot0\left\{ 2{ tan }^{ -1 }x \right\} \right) =\frac { n\pi }{ 2 } =0,\frac { \pi }{ 2 } ,\pi \) \(\left( \because \quad 0\le { cos }^{ -1 }x\le \pi \right) \)
\(\Rightarrow \quad cot\left( 2{ tan }^{ -1 }x \right) =1,\ 0,-1\)
\(\Rightarrow \quad cot\left\{ { tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \right\} =1,0,-1\)
\(\Rightarrow \quad cot\left\{ { tan }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 2{ x } } \right) \right\} =1,0,-1\)
\(\Rightarrow \quad \frac { 1-{ x }^{ 2 } }{ 2{ x } } =1,0,-1\)
\(\Rightarrow \quad \ 1-{ x }^{ 2 }=2x,0,-2x\)
or \({ x }^{ 2 }+2x-1=0,\ { x }^{ 2 }-1=0,\ { x }^{ 2 }-2x-1=0\)
or \(x=-1\pm \sqrt { 2 } ,\ x=\pm 1,\ x=1\pm \sqrt { 2 } \)
\(\alpha ,\beta \ and\ \gamma \) are the angles given by \(\alpha =2{ tan }^{ -1 }\left( \sqrt { 2 } -1 \right) ,\ \beta =3{ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) +{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \ and\ \gamma ={ cos }^{ -1 }\left( \frac { 1 }{ 3 } \right) \), then
- (a)
\(\alpha >\beta \)
- (b)
\(\beta >\gamma \)
- (c)
\(\gamma >\alpha \)
- (d)
none of these
\(\because \quad \quad \alpha =2{ tan }^{ -1 }\left( \sqrt { 2 } -1 \right) =2.22\frac { { 1 }^{ \circ } }{ 2 } ={ 45 }^{ \circ }\)
\(\Rightarrow \quad cos\quad \alpha =\frac { 1 }{ \sqrt { 2 } } \)
\(\beta =3{ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) +{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \)
\(=3.\left( \frac { \pi }{ 4 } \right) -\frac { \pi }{ 6 } ={ 105 }^{ \circ }\)
\(\Rightarrow \quad cos\beta =-\frac { \left( \sqrt { 3 } -1 \right) }{ 2\sqrt { 2 } } \)
and \(\gamma ={ cos }^{ -1 }\left( \frac { 1 }{ 3 } \right) \)
\(\Rightarrow \quad cos\quad \gamma =\frac { 1 }{ 3 } \)
\(\therefore \quad \quad \beta >\alpha \)
\(cos\beta <cos\gamma \Rightarrow \beta >\gamma \)
\(cos\alpha >cos\gamma \Rightarrow \alpha <\gamma \)
Indicate the relation which is true
- (a)
tan |tan-1x| = |x|
- (b)
cot |cot-1x| = |x|
- (c)
tan-1 |tan x| = |x|
- (d)
sin |sin-1 x| = |x|
tan |tan-1 x| = |tan tan-1 x| = |x| (non-periodic)
cot |cot-1 x| = |cot cot-1 x| = |x| (non-periodic)
sin |sin-1 x| = |sin sin-1 x| = |x| (non-periodic)
\({ cos }^{ -1 }\left( \sqrt { \frac { a-x }{ a-b } } \right) ={ sin }^{ -1 }\left( \sqrt { \frac { x-b }{ a-b } } \right) \) is possible, if
- (a)
a > x > b
- (b)
a < x < b
- (c)
a = x = b
- (d)
a>b and x, takes any value
LHS and RHS of the given equation are defined, if \(\frac { x-b }{ a-b } >0\ and\ \frac { x-b }{ a-b } >0\). Either a>x>b or a< x<b. For a=x=b, LHS and RHS are not defined.
The value of cos-1 (cas 12) - sin-1(sin 12) is
- (a)
0
- (b)
ㅠ
- (c)
8ㅠ-24
- (d)
none of these
Let P = cos-1 (cos 12) - sin-1(sin 12)
cos 12 > 0 and sin 12 < 0
P = cos-1 cos (4π -12) - sin-1 sin (12 - 4π)
= (4π -12)- (12- 4π)= 8π - 24
\(\theta =\tan ^{ -1 }{ \left( 2\quad { tan }^{ 2 }\theta \right) } -{ tan }^{ -1 }\left\{ \left( \frac { 1 }{ 3 } \right) tan\quad \theta \right\} ,\)if
- (a)
\(tan\ \theta =-2\)
- (b)
\(tan\ \theta =0\)
- (c)
\(tan\ \theta =1\)
- (d)
\(tan\ \theta =2\)
\(\theta =\tan ^{ -1 }{ \left( 2\ { tan }^{ 2 }\theta \right) } -{ tan }^{ -1 }\left\{ \left( \frac { 1 }{ 3 } \right) tan\ \theta \right\} \)
\(\Rightarrow \ tan\ \theta =\frac { 2\tan ^{ 2 }{ \theta } -\left( \frac { 1 }{ 3 } \right) tan\ \theta }{ 1+\left( \frac { 2 }{ 3 } \right) \tan ^{ 3 }{ \theta } } \)
\(\Rightarrow \ tan\ \theta =\left[ \frac { 2\ tan\ \theta -\left( \frac { 1 }{ 3 } \right) }{ 1+\left( \frac { 2 }{ 3 } \right) \tan ^{ 3 }{ \theta } } -1 \right] =0\)
Which is true, if tan \(\theta =0\)
or \(\frac { 2\ tan\ \theta -\left( \frac { 1 }{ 3 } \right) }{ 1+\left( \frac { 2 }{ 3 } \right) \tan ^{ 3 }{ \theta } } =1\)
\(\Rightarrow \ \ \tan ^{ 3 }{ \theta } -3\ tan\ \theta +2=0\)
\(\\ \Rightarrow \quad \left( tan\theta -1 \right) ^{ 2 }\left( tan\theta +2 \right) =0\)
\(\theta =1\)
\(tan\ \theta =-2\)
If cosec-1x = sin-1\(\left( \frac { 1 }{ x } \right) \), then x may be
- (a)
1
- (b)
\(-\frac { 1 }{ 2 } \)
- (c)
\(\frac { 3 }{ 2 } \)
- (d)
\(-\frac { 3 }{ 2 } \)
Since, cosec-1 x is defined if \(x\le -1\ or\ x\ge 1\)
\(\therefore\) cosec-1 x = sin-1 \(\left( \frac { 1 }{ x } \right) \)is not true for \(x=-\frac { 1 }{ 2 } \)
\(2{ cot }^{ -1 }7+{ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \) is equal to
- (a)
\({ cot }^{ -1 }\left( \frac { 44 }{ 117 } \right) \)
- (b)
\({ cosec }^{ -1 }\left( \frac { 125 }{ 117 } \right) \)
- (c)
\({ tan }^{ -1 }\left( \frac { 4 }{ 117 } \right) \)
- (d)
\({ cos }^{ -1 }\left( \frac { 44 }{ 125 } \right) \)
\(2{ cot }^{ -1 }7\ ={ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \)
\(=\ { cos }^{ -1 }\left( \frac { 1-\frac { 1 }{ 49 } }{ 1+\frac { 1 }{ 49 } } \right) ={ cos }^{ -1 }\frac { 24 }{ 25 } \)
Now, \(2{ cot }^{ -1 }7+{ cot }^{ -1 }\frac { 3 }{ 5 } \)
\(=\ { cot }^{ -1 }\frac { 24 }{ 25 } +{ cos }^{ -1 }\frac { 3 }{ 5 } \)
\(=\ { cos }^{ -1 }\left( \frac { 24 }{ 25 } .\frac { 3 }{ 5 } -\frac { 7 }{ 25 } .\frac { 4 }{ 5 } \right) ={ cos }^{ -1 }\frac { 44 }{ 125 } \)
Since, \(\frac { 44 }{ 125 } >0\)
\(\therefore \quad 0<{ cos }^{ -1 }\frac { 44 }{ 125 } <\frac { \pi }{ 2 } \)
Let \({ cos }^{ -1 }\frac { 44 }{ 125 } =\theta ,\quad cos\quad \theta =\frac { 44 }{ 125 } \)
\(\therefore \quad \ cosec\ \theta \ =\ \frac { 44 }{ 125 } \ or\ \theta =cosec^{ -1 }\frac { 125 }{ 117 } \)
Also, \(cot\ \theta =\frac { 44 }{ 117 } \)
\(\therefore \quad \theta ={ cot }^{ -1 }\frac { 44 }{ 117 } \)
If the equation sin-1(x2+x+1)+cos-1(\(\lambda\) x+1) = \(\frac { \pi }{ 2 } \) has exactly two solutions, then \(\lambda \) cannot have the integral value
- (a)
-1
- (b)
0
- (c)
1
- (d)
2
The given equation holds, if x2+x+1= \(\lambda \) x+1 and -1\(\le \) x2 + x + 1 \(\le \) 1
\(\Rightarrow\) x (x+1-\(\lambda \)) = 0 and -1 \(\le \) x \(\le \) 0
\(\Rightarrow\) x = 0 or \(\le \) -1 and -1 \(\le \) x \(\le \) 0
\(\therefore \) x = 0 is one solution and for another different solution -1\(\le \lambda \)-1< 0
\(\Rightarrow\) 0\(\le \lambda \) < 1, so only integral value \(\lambda \) can have is 0.
If a, b are positive quantities and, if \(a_1={a+b\over 2},b_1=\sqrt{a_1,b}\) \(a_2={a_1+b_1\over1},b_2\sqrt{a_1b_1}\) and so on, then
- (a)
\(a_\infty={sqrt{(b^2-a^2)}\over cos^{-1}\left(a\over b\right)}\)
- (b)
\(b_\infty={sqrt{(b^2-a^2)}\over cos^{-1}\left(a\over b\right)}\)
- (c)
\(b_\infty={sqrt{(a^2-b^2)}\over cos^{-1}\left(b\over a\right)}\)
- (d)
none of these
Let a=b cosΦ
Then, \(a_1={a+b\over 2}={b(1+cos\pi)\over 2}\)
\(=b\ cos^2{\phi\over 2}\)
\(⇒b_1=\sqrt{a_1b}=b\ cos{\phi\over 2}\)
Now, \(a_2={a_1+b_2\over 2}\)
\(=b\ cos{\phi\over 2}cos^2{\phi\over 4}\)
\(b_2=b\ cos{\phi\over 2}cos{\phi\over 4}\)
Similarly \(b_3=b\ cos\left(\phi\over 2\right)cos\left(\phi\over 2^2\right)cos\left(\phi\over 2^3\right)\)
and so on
Now
\(b_\infty=\underset{n\rightarrow\infty}{lim}b\ cos\left(\phi\over 2\right)cos\left(\phi\over 2^@\right)\)
\(cos\left(\phi\over 2^3\right).....cos\left(\phi\over 2^n\right)\)
\(=\underset{n\rightarrow\infty}{lim}{b\ sin\phi\over 2^nsin\left(\phi\over 2^n\right)}\)
\(=\underset{n\rightarrow\infty}{lim}{b\ sin\phi\over\phi {2^nsin\left(\phi\over 2^n\right)\over \left(\phi\over 2^n\right)}}\)
\(={{b\ sin\phi}\over \phi\left\{ \underset{n\rightarrow\infty}{lim}\ sin{\left(\phi\over 2^n\right)\over \left(\phi\over 2^n\right)}\right\}}\)
\(={b\ sin\phi\over \phi}=b{\sqrt{(1-cos^2)}\over\left(a\over b\right)}\)
\(={\sqrt{(b^2-a^2)}\over cos^{-1}\left(a\over b\right)}\)
The value of \(tan^{-1}\left(c_1x-y\over c_1y+x\right)+tan^{-1}\left(c_2-c_2\over 1+c_2c_2\right)+tan^{_1}\left(c_3-c_2\over 1+c_3c_2\right)+...+\left(1\over c_n\right)\)is equal
- (a)
\(tan^{-1}\left(y\over x\right)\)
- (b)
\(tan^{-1}\left(x\over y\right)\)
- (c)
\(-tan^{-1}\left(x\over y\right)\)
- (d)
none of these
we have
\(tan^{-1}\left(c_1x-y\over c_1y+x\right)+tan^{-1}\left(c_2-c_2\over 1+c_2c_2\right)+tan^{_1}\left(c_3-c_2\over 1+c_3c_2\right)+...+\left(1\over c_n\right)\)
\(=tan^{1}\left( {x\over y}-{1\over c_1}\over 1+{x\over y}{1\over c_1}\right)+tan^{-1}\left({1\over c_1}-{1\over vc_2}\over 1+{1\over c_1c_2}\right)+tan^{-1}\left({1\over c_2}-{1\over c_3}\over 1+{1\over c_2c_3}\right)+...+tan^{-1}\left(1\over c_n\right)\)
\(=tan^{-1}\left(x\over y\right)-tan^{-1}\left(1\over c_1\right)+tan^{-1}\left(1\over c_1\right)-tan^{-1}\left(1\over c_2\right)+tan^{-1}\left(1\over c_2\right)-tan^{-1}\left(1\over c_3\right)+...-tan^{-1}\left(1\over c_n\right)+tan{-1}\left(1\over c_n\right)=tan^{-1}\left(x\over y\right)\)
The value \(\left\{\left(cos^{-1}\left(-{2\over 7}\right)-{\pi\over 2}\right)\right\}\)is
- (a)
\(2\over 3\sqrt5\)
- (b)
\(2\over 3\)
- (c)
\(1\over \sqrt5\)
- (d)
\(4\over \sqrt5\)
\(\left\{\left(cos^{-1}\left(-{2\over 7}\right)-{\pi\over 2}\right)\right\}\)
\(=tan\left\{\pi-cos^{-1}\left(2\over 7\right)-{\pi\over 2}\right\}\)
\(=tan\left\{{\pi\over2}-cos^{-1}\left(2\over 7\right)\right\}=tan\left\{sin^{-1}-{2\over 7}\right\}\)
\(=tan\ tan^{-1}\left(2\over 3\sqrt5\right)={2\over 3\sqrt5}\)
The value(s) of x satisfying the equation \({ sin }^{ -1 }\left| sin\ x \right| =\sqrt { sin^{ -1 }\left| sin\ x \right| } \) is/are given by (n is any integer)
- (a)
\(n\pi -1\)
- (b)
\(n\pi\)
- (c)
\(n\pi +1\)
- (d)
\(2n\pi +1\)
The solution of y =\(\sqrt { y } \) is y = 0 or y = 1
If sin-1|sin x| = 1
\(\Rightarrow\) x = 1 or \(\pi-1\) [in the interval \(\left( 0,\pi \right) \)]
But y = sin-1 |sin x| is periodic with periodic \(\pi\), so x = n\(\pi\) + 1 or n\(\pi\) - 1.
Again, if sin-1|sin x| = 0
\(\Rightarrow\) x = n\(\pi\)
If tan-1 y = 4 tan-1 x, then y is infinite, if
- (a)
\({ x }^{ 2 }=3+2\sqrt { 2 } \)
- (b)
\({ x }^{ 2 }=3-2\sqrt { 2 } \)
- (c)
\({ x }^{4 }=6x^2-1\)
- (d)
\({ x }^{ 4 }=6x^2+1\)
If we put x = tan-1 \(\theta \), the given equality becomes tan-1 y = 4\(\theta \)
\(\Rightarrow \quad y\ =\ tan\ 4\theta =\frac { 2\ tan\ 2\theta }{ 1-tan^{ 2 }2\theta } \)
\(=\ \frac { 2\left[ \frac { 2\quad tan\quad \theta }{ 1-{ tan }^{ 2 }\theta } \right] }{ 1-\left( \frac { 2\quad tan\quad \theta }{ 1-{ tan }^{ 2 }\theta } \right) ^{ 2 } } \)
\(=\ \frac { 2\times 2x\left( 1-x^{ 2 } \right) }{ \left( 1-x^{ 2 } \right) ^{ 2 }-4{ x }^{ 2 } } =\frac { 4x\left( 1-{ x }^{ 2 } \right) }{ 1-{ 6x }^{ 2 }+{ x }^{ 4 } } \)
So, that y is infinite, if x4-6x2+1 = 0
\(\Rightarrow \quad \quad { x }^{ 2 }=\frac { 6\pm \sqrt { 36-4 } }{ 2 } =3\pm 2\sqrt { 2 } \)
If cos-1x = tan-1 x, then
- (a)
\({ x }^{ 2 }=\left( \sqrt { 5 } -1 \right) /2\)
- (b)
\({ x }^{ 2 }=\left( \sqrt { 5 } +1 \right) /2\)
- (c)
\(sin\left( { cos }^{ -1 }x \right) =\left( \sqrt { 5 } -1 \right) /2\)
- (d)
\(\\ \\ tan\left( { cos }^{ -1 }x \right) =\left( \sqrt { 5 } -1 \right) /2\)
\({ cos }^{ -1 }x\ =\ { tan }^{ -1 }x\)
\(\Rightarrow \ \ x\ =\ cos\theta \ =\ tan\theta \)
\(\Rightarrow \quad { cos }^{ 2 }\theta =sin\theta \)
\(\Rightarrow \quad { sin }^{ 2 }\theta \ +\ sin\ \theta -1=0\)
\(\Rightarrow \ sin\ \theta \ =\frac { -1\pm \sqrt { 1+4 } }{ 2 } \)
\(\Rightarrow \ sin\ \theta \ =\frac { \sqrt { 5 } -1 }{ 2 } \)
\(\\ \\ \Rightarrow \quad { x }^{ 2 }=\ cos^{ 2 }\theta =\frac { \sqrt { 5 } -1 }{ 2 } \)
So, (a) is correct.
and sin(cos-1 x) = \(sin\ \theta \ =\ \left( \frac { 1 }{ 2 } \right) \left( \sqrt { 5 } -1 \right) \)
So, (c) is correct.
and tan(cos-1 x) = \(tan\ \theta \ \neq \ \left( \frac { 1 }{ 2 } \right) \left( \sqrt { 5 } -1 \right) \)
So, (d) is not correct.
Hence, (a) and (c) are correct.
The number of the positive integral solutions of \(tan^{-1}x+cos^{-1}\left(y\over \sqrt{(1+y^2)}\right)=sin^{-1}\left(3\over\sqrt{10}\right)\) is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Since, \(tan^{-1}x+cos^{-1}\left(y\over \sqrt{(1+y^2)}\right)=sin^{-1}\left(3\over \sqrt{10}\right)\)
\(tan^{-1}x+tan^{-1}\left(1\over y\right)=tan^{-1}3\)
or \(tan^{-1}\left(1\over y\right)=tan^{-1}3-tan^{-1}x\)
\(=tan^{-1}\left(3-x\over 1+3x\right)\)
or \(y={1+3x\over 3-x}=-3+{10\over 3-x}\)
For positive integer y, x must be 1 and 2
Y (at x =1) =2
and y (at x =2) =7
Hence, solutions are (x, y) =(1, 2), (2, 7)
If a1 a2, a3, ... , an is an AP with common difference d, then \(tan\left[tan^{-1}\left(d\over 1+a_1a_2\right)+tan^{-1}\left(d\over 1+a_2a_3\right)+...+tan^{-1}\left(d\over 1+a_{n-1}a_n\right)\right]\)
- (a)
\((n-1)d\over a_1+a_n\)
- (b)
\((n-1)d\over1+ a_1a_n\)
- (c)
\(nd\over1+ a_1a_n\)
- (d)
\(a_n-a_1\over a_n+a_1\)
a2 - a1 = a3 - a2 = ...=an-an-1 =d
\(tan\left[tan^{-1}\left(d\over 1+a_1a_2\right)+tan^{-1}\left(d\over 1+a_2a_3\right)+...+tan^{-1}\left(d\over 1+a_{n-1}a_n\right)\right]\)
\(=tan\left[\sum_{r=2}^{n}tan^{-1}\left(d\over 1+a_{r-1}a_r\right)\right]\)
\(=tan\left[\sum_{r=2}^{n}tan^{-1}\left(a_r-a_{r-1}\over 1+a_{r-1}a_r\right)\right]\)
\(=tan\left[\sum_{r=2}^{n}(tan^{-1}a_r-tan^{-1}a_{r-1})\right]\)
=tan (tan-1 an - tan-1 a1)
\(=tan\left(tan^{-1}\left(a_n-a_1\over 1+a_1a_n\right)\right)\)
\(={a_n-a_1\over 1+a_1a_n}={a_1+(n-1d-a_1\over 1+a_1a_n}\)
\(={(n-1)d\over 1+a_1a_n}\)
Let f : A\(\rightarrow \)B be a function defined by y = f(x) such that f is both one-one (Injective) and onto (surjective)(ie, bijective), then there exists a unique function g: B\(\rightarrow \)A such that \(f\left( x \right) =y\Leftrightarrow g\left( y \right) =x,\ \forall x\epsilon A\ y\epsilon B\), then g is said to be inverse of f. Thus, g = f-1: B\(\rightarrow \)A = \(\left[ \left\{ f\left( x \right) ,x \right\} :\left\{ x,\ f(x) \right\} \epsilon { f }^{ -1 } \right] \).If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that functon. On the basis of above information, answer the following question: If x takes negative permissible value, then sin-1 x is equal to
- (a)
\({ cos }^{ -1 }\sqrt { { \left( 1-{ x }^{ 2 } \right) } } \)
- (b)
\({ cos }^{ -1 }\left( \frac { \sqrt { { 1-{ x }^{ 2 } } } }{ x } \right) \)
- (c)
\(\pi -{ cos }^{ -1 }\sqrt { \left( 1-{ x }^{ 2 } \right) } \)
- (d)
\(-\pi +{ cot }^{ -1 }\left( \frac { \sqrt { { 1-{ x }^{ 2 } } } }{ x } \right) \)
\(\sin ^{ -1 }{ x } =\begin{cases} { cos }^{ -1 }\sqrt { \left( { 1-x }^{ 2 } \right) } \ \ \ \ ,\ \quad \quad 0\le x\le 1 \\ -cos^{ -1 }\sqrt { \left( { 1-x }^{ 2 } \right) } \ \ ,\quad -1\le x<0 \end{cases}\)
Let f : A\(\rightarrow \)B be a function defined by y = f(x) such that f is both one-one (Injective) and onto (surjective)(ie, bijective), then there exists a unique function g: B\(\rightarrow \)A such that \(f\left( x \right) =y\Leftrightarrow g\left( y \right) =x,\ \forall x\epsilon A\ y\epsilon B\), then g is said to be inverse of f. Thus, g = f-1: B\(\rightarrow \)A = \(\left[ \left\{ f\left( x \right) ,x \right\} :\left\{ x,\ f(x) \right\} \epsilon { f }^{ -1 } \right] \).If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that function. On the basis of above information, answer the following question: If \(\frac { 3\pi }{ 2 } \le x\le \frac { 5\pi }{ 2 } \) then sin-1(sin x) is equal to
- (a)
x
- (b)
-x
- (c)
\(2\pi -x\)
- (d)
\(x-2\pi \)
For \(\frac { 3\pi }{ 2 } \le x\le \frac { 5\pi }{ 2 } \)
\({ sin }^{ -1 }\left( sin\ x \right) =x-2\pi \)
Let f : A\(\rightarrow \)B be a function defined by y = f(x) such that f is both one-one (Injective) and onto (surjective)(ie, bijective), then there exists a unique function g: B\(\rightarrow \)A such that \(f\left( x \right) =y\Leftrightarrow g\left( y \right) =x,\ \forall x\epsilon A\ y\epsilon B\), then g is said to be inverse of f. Thus, g = f-1: B\(\rightarrow \)A = \(\left[ \left\{ f\left( x \right) ,x \right\} :\left\{ x,\ f(x) \right\} \epsilon { f }^{ -1 } \right] \).If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that functon. On the basis of above information, answer the following question: If x>1, then the value of 2 tan-1x + sin-1 \(\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \) is
- (a)
\(\frac { \pi }{ 4 } \)
- (b)
\(\frac { \pi }{ 2 } \)
- (c)
\(\pi\)
- (d)
\(\frac { 3\pi }{ 2 } \)
\(\because \quad \quad 2{ tan }^{ -1 }x=\begin{cases} { sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right) \quad \quad \ ,\quad -1\le x\le 1 \\ \pi -{ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right) \ \ ,\quad x>1 \\ -\pi -{ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,\quad x<-1 \end{cases}\)
\(\therefore \) For x > 1
\(2{ tan }^{ -1 }x=\pi -sin^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) =\pi \)
\(\Rightarrow \quad 2{ tan }^{ -1 }x+{ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) \)
Let f : A\(\rightarrow \)B be a function defined by y = f(x) such that f is both one-one (Injective) and onto (surjective)(ie, bijective), then there exists a unique function g: B\(\rightarrow \)A such that \(f\left( x \right) =y\Leftrightarrow g\left( y \right) =x,\ \forall x\epsilon A\ y\epsilon B\), then g is said to be inverse of f. Thus, g = f-1: B\(\rightarrow \)A = \(\left[ \left\{ f\left( x \right) ,x \right\} :\left\{ x,\ f(x) \right\} \epsilon { f }^{ -1 } \right] \).If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that function. On the basis of above information, answer the following question: If \(0\le x\le 1\), then the least and greatest values of tan-1 \(\left( \frac { 1+x }{ 1-{ x } } \right) \) are
- (a)
\(-\frac { \pi }{ 4 } ,\frac { \pi }{ 4 } \)
- (b)
\(0,\frac { \pi }{ 4 } \)
- (c)
\(\frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \)
- (d)
\(0,\pi \)
\(\because \quad { tan }^{ -1 }\left( \frac { 1+x }{ 1-x } \right) ={ tan }^{ -1 }1+{ tan }^{ -1 }x=\frac { \pi }{ 4 } +{ tan }^{ -1 }x\)
Given, \(0\le x\le 1\)
\(\Rightarrow \quad { tan }^{ -1 }0\le { tan }^{ -1 }x\le { tan }^{ -1 }1\)
\(\Rightarrow \quad 0\le { tan }^{ -1 }x\le \frac { \pi }{ 4 } \)
\(\Rightarrow \quad \frac { \pi }{ 4 } +0\le \frac { \pi }{ 4 } +{ tan }^{ -1 }x\le \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \)
\(\Rightarrow \quad \ \frac { \pi }{ 4 } \le { tan }^{ -1 }\left( \frac { 1+x }{ 1-x } \right) \frac { \pi }{ 2 } \)
Hence, least and greatest values of tan-1 \(\left( \frac { 1+x }{ 1-x } \right) \)are \(\frac { \pi }{ 4 } \) and \(\frac { \pi }{ 2 } .\)
\(\sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left( \frac { { x }_{ r }-{ x }_{ r-1 } }{ 1+{ x }_{ r-1 }{ x }_{ r } } \right) =\sum _{ r=1 }^{ n }{ \left( { tan }^{ -1 }{ x }_{ r }-{ tan }^{ -1 }{ x }_{ r-1 } \right) } ={ tan }^{ -1 }{ x }_{ n }-{ tan }^{ -1 }{ x }_{ 0 },\forall n\epsilon N\)
On the basis of above information, answer the following questions:
The value of \({ cosec }^{ -1 }\sqrt { 5 } +{ cosec }^{ -1 }\sqrt { 65 } +{ cosec }^{ -1 }\sqrt { \left( 325 \right) } +.....\) to \(\infty \) is
- (a)
\(\pi \)
- (b)
\(\frac { 3\pi }{ 4 } \)
- (c)
\(\frac { \pi }{2 } \)
- (d)
\(\frac { \pi }{ 4 } \)
\(\because \quad \quad { cosec }^{ -1 }\sqrt { 5 } +{ cosec }^{ -1 }\sqrt { 65 } +{ cosec }^{ -1 }\sqrt { \left( 325 \right) } +.....\infty \)
\(=\quad { cot }^{ -1 }2+{ cot }^{ -1 }8+{ cot }^{ -1 }18+....\infty \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { cot }^{ -1 }\left( { 2r }^{ 2 } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { tan }^{ -1 }\left( \frac { 2 }{ { 4r }^{ 2 } } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { tan }^{ -1 }\left( \frac { 2 }{ { 4r }^{ 2 } } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { tan }^{ -1 }\left( \frac { \left( 2r+1 \right) -\left( 2r-1 \right) }{ 1+\left( 2r+1 \right) \left( 2r-1 \right) } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( { tan }^{ -1 }\left( 2r+1 \right) -{ tan }^{ -1 }\left( 2r-1 \right) \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \left\{ \left( { tan }^{ -1 }\left( 2n+1 \right) -{ tan }^{ -1 }1 \right) \right\} } \)
\(=\quad { tan }^{ -1 }\infty -{ tan }^{ -1 }1\)
\(=\quad \frac { \pi }{ 2 } -\frac { \pi }{ 4 } =\frac { \pi }{ 4 } \)
\(\sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left( \frac { { x }_{ r }-{ x }_{ r-1 } }{ 1+{ x }_{ r-1 }{ x }_{ r } } \right) =\sum _{ r=1 }^{ n }{ \left( { tan }^{ -1 }{ x }_{ r }-{ tan }^{ -1 }{ x }_{ r-1 } \right) } ={ tan }^{ -1 }{ x }_{ n }-{ tan }^{ -1 }{ x }_{ 0 },\forall n\epsilon N\)
On the basis of above information, answer the following questions:
The sum to infinite terms of the series \({ cot }^{ -1 }\left( { 2 }^{ 2 }+\frac { 1 }{ 2 } \right) +{ cot }^{ -1 }\left( { 2 }^{ 3 }+\frac { 1 }{ 2^{ 2 } } \right) +{ cot }^{ -1 }\left( { 2 }^{ 4 }+\frac { 1 }{ 2^{ 3 } } \right) +.....\)is
- (a)
\(\frac { \pi }{ 4 } \)
- (b)
\(\frac { \pi }{ 2 } \)
- (c)
cot-1 2
- (d)
-cot-1 2
\( \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { cot }^{ -1 }\left( { 2 }^{ r+1 }+\frac { 1 }{ { 2 }^{ r } } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { tan }^{ -1 }\left( \frac { 2 }{ { 1+2^{ r }.2 }^{ r+1 } } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ { tan }^{ -1 }\left( \frac { { 2 }^{ r+1 }-{ 2 }^{ r } }{ 1+2^{ r }.{ 2 }^{ r+1 } } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( { tan }^{ -1 }{ 2 }^{ r+1 }-{ tan }^{ -1 }{ 2 }^{ r } \right) } } \)
\(=\quad \lim _{ n\rightarrow \infty }{ \left[ { tan }^{ -1 }{ 2 }^{ n+1 }-{ tan }^{ -1 }{ 2 } \right] } \)
\(=\quad { tan }^{ -1 }\infty -{ tan }^{ -1 }2\)
\(=\quad \frac { \pi }{ 2 } -{ tan }^{ -1 }2\)
\(=\quad { cot }^{ -1 }2\)
Principal values for inverse circular functions:
x<0 | x\(\ge \)0 |
\(-\frac { \pi }{ 2 } \le \ { sin }^{ -1 }x<0\) | \(0\le { sin }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <\ cos^{ -1 }x\le \pi \) | \(0\le { cos }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } <{ tan }^{ -1 }x<0\) | \(0\le { tan }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ cot }^{ -1 }x<\pi \) | \(0\le { cot }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ sec }^{ -1 }x\le \pi \) | \(0\le { sec }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } \le { cosec }^{ -1 }x<0\) | \(0<{ cosec }^{ -1 }x\le \frac { \pi }{ 2 } \) |
Ex. \({ sec }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } ,\ { tan }^{ -1 }\left( -\sqrt { 3 } \right) =-\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } \)
On the basis of above information, answer the following questions:
The principal value of \(\sin ^{ -1 }{ \left( sin\frac { 4\pi }{ 3 } \right) +\cos ^{ -1 }{ \left( cos\frac { 4\pi }{ 3 } \right) } } \) is
- (a)
\(\frac { 8\pi }{ 3 } \)
- (b)
\(\frac { 4\pi }{ 3 } \)
- (c)
\(\frac { 2\pi }{ 3 } \)
- (d)
\(\frac { \pi }{ 3 } \)
\(\because \quad \quad cos\left( \frac { 4\pi }{ 3 } \right) =cos\left( \pi +\frac { \pi }{ 3 } \right) =-cos\frac { \pi }{ 3 } =-\frac { 1 }{ 2 } <0\)
\(\therefore \quad \quad \frac { \pi }{ 2 } <{ cos }^{ -1 }\left( cos\frac { 4\pi }{ 3 } \right) \le \pi \)
\(\Rightarrow \quad \quad { cos }^{ -1 }\left( cos\frac { 4\pi }{ 3 } \right) ={ cos }^{ -1 }cos\left( 2\frac { 2\pi }{ 3 } \right) ={ cos }^{ -1 }cos\frac { 2\pi }{ 3 } =\frac { 2\pi }{ 3 } \)
and \(sin\left( \frac { 4\pi }{ 3 } \right) =sin\left( \pi +\frac { \pi }{ 3 } \right) =-sin\frac { \pi }{ 3 } =\frac { -\sqrt { 3 } }{ 2 } <0\)
\(\therefore \quad \quad -\frac { \pi }{ 2 } \le { sin }^{ -1 }\left( sin\frac { 4\pi }{ 3 } \right) <0\)
\(\therefore \quad { sin }^{ -1 }\left( sin\frac { 4\pi }{ 3 } \right) ={ sin }^{ -1 }\left\{ sin\left( \pi -\frac { 4\pi }{ 3 } \right) \right\} =-\frac { \pi }{ 3 } \)
Hence, \({ sin }^{ -1 }\left( sin\frac { 4\pi }{ 3 } \right) +{ cos }^{ -1 }\left( cos\frac { 4\pi }{ 3 } \right) =-\frac { \pi }{ 3 } +\frac { 2\pi }{ 3 } =\frac { \pi }{ 3 } \)
Principal values for inverse circular functions:
x<0 | x\(\ge \)0 |
\(-\frac { \pi }{ 2 } \le \ { sin }^{ -1 }x<0\) | \(0\le { sin }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <\ cos^{ -1 }x\le \pi \) | \(0\le { cos }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } <{ tan }^{ -1 }x<0\) | \(0\le { tan }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ cot }^{ -1 }x<\pi \) | \(0\le { cot }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ sec }^{ -1 }x\le \pi \) | \(0\le { sec }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } \le { cosec }^{ -1 }x<0\) | \(0<{ cosec }^{ -1 }x\le \frac { \pi }{ 2 } \) |
Ex. \({ sec }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } ,\ { tan }^{ -1 }\left( -\sqrt { 3 } \right) =-\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } \)
On the basis of above information, answer the following questions:
The principal value of \(\sin ^{ -1 }{ \left( sin\ 5 \right) -\cos ^{ -1 }{ \left( cos\ 5 \right) } } \) is
- (a)
0
- (b)
\(2\pi -10\)
- (c)
\(-\pi \)
- (d)
\(3\pi -10\)
\(\\ \because \quad 5=5\times { 57 }^{ \circ }\approx { 57 }^{ \circ }\)
\(\therefore\) sin 5 < 0 and cos 5 < 0
\(\Rightarrow\) sin-1(sin 5) = sin-1{sin(\(\pi\) - 5)} = \(\pi\) - 5
and cos-1(cos 5) = cos-1{cos(2\(\pi\) - 5)} = 2\(\pi\) - 5
\(\therefore\) sin-1(sin 5)-cos-1(cos 5) = (\(\pi\) - 5) - (2\(\pi\) -5) = - \(\pi\)
Principal values for inverse circular functions:
x<0 | x\(\ge \)0 |
\(-\frac { \pi }{ 2 } \le \ { sin }^{ -1 }x<0\) | \(0\le { sin }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <\ cos^{ -1 }x\le \pi \) | \(0\le { cos }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } <{ tan }^{ -1 }x<0\) | \(0\le { tan }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ cot }^{ -1 }x<\pi \) | \(0\le { cot }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ sec }^{ -1 }x\le \pi \) | \(0\le { sec }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } \le { cosec }^{ -1 }x<0\) | \(0<{ cosec }^{ -1 }x\le \frac { \pi }{ 2 } \) |
Ex. \({ sec }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } ,\ { tan }^{ -1 }\left( -\sqrt { 3 } \right) =-\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } \)
On the basis of above information, answer the following questions:
The principal value of \({ tan }^{ -1 }\left( tan\left( -\frac { 3\pi }{ 4 } \right) \right) +{ cot }^{ -1 }cot\left( -\frac { 3\pi }{ 4 } \right) \) is
- (a)
\(\frac { \pi }{ 2 } \)
- (b)
\(\pi\)
- (c)
\(\frac { -3\pi }{ 2 } \)
- (d)
0
\(\because \quad \quad \tan ^{ -1 }{ tan } \left( -\frac { 3\pi }{ 4 } \right) =\tan ^{ -1 }{ tan } \left( -\frac { 3\pi }{ 4 } +\pi \right) =-\frac { 3\pi }{ 4 } +\pi =\frac { \pi }{ 4 } \)
and \({ cot }^{ -1 }cot\left( -\frac { 3\pi }{ 4 } \right) ={ cot }^{ -1 }cot\left( \pi -\frac { 3\pi }{ 4 } \right) =\pi -\frac { 3\pi }{ 4 } =\frac { \pi }{ 4 } \)
\(\therefore \quad \quad \tan ^{ -1 }{ tan } \left( -\frac { 3\pi }{ 4 } \right) +{ cot }^{ -1 }\left\{ cot\left( -\frac { 3\pi }{ 4 } \right) \right\} =\frac { \pi }{ 4 } +\frac { \pi }{ 4 } =\frac { \pi }{ 2 } \)
Principal values for inverse circular functions:
x<0 | x\(\ge \)0 |
\(-\frac { \pi }{ 2 } \le \ { sin }^{ -1 }x<0\) | \(0\le { sin }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <\ cos^{ -1 }x\le \pi \) | \(0\le { cos }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } <{ tan }^{ -1 }x<0\) | \(0\le { tan }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ cot }^{ -1 }x<\pi \) | \(0\le { cot }^{ -1 }x\le \frac { \pi }{ 2 } \) |
\(\frac { \pi }{ 2 } <{ sec }^{ -1 }x\le \pi \) | \(0\le { sec }^{ -1 }x<\frac { \pi }{ 2 } \) |
\(-\frac { \pi }{ 2 } \le { cosec }^{ -1 }x<0\) | \(0<{ cosec }^{ -1 }x\le \frac { \pi }{ 2 } \) |
Ex. \({ sec }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } ,\ { tan }^{ -1 }\left( -\sqrt { 3 } \right) =-\frac { \pi }{ 3 } not\frac { 2\pi }{ 3 } \)
On the basis of above information, answer the following questions:
The value of sin-1 [cos{cos-1(cos x)+sin-1(sin x)}], where \(x\epsilon \left( \frac { \pi }{ 2 } ,\pi \right) \) is
- (a)
\(\frac { \pi }{ 2 } \)
- (b)
\(-\pi\)
- (c)
\(\pi\)
- (d)
\(-\frac { \pi }{ 2 } \)
\(\because \quad \ { sin }^{ -1 }\left[ cos\left\{ { cos }^{ -1 }\left( cos\ x \right) +{ sin }^{ -1 }\left( sin\ x \right) \right\} \right] \)
\(=\ { sin }^{ -1 }\left\{ cos\left( x+\pi -x \right) \right\} \ as\ x\epsilon \left( \frac { \pi }{ 2 } ,\pi \right) \)
\(=\ { sin }^{ -1 }\left( cos\ \pi \right) \)
\(=\ { sin }^{ -1 }\left( -1 \right) =-\frac { \pi }{ 2 } \)