Match the statements given in Column - I with the values given in Column- II.\,\,\,\,\,\,\,\,\,\,\,\, Column- I (A) \,\,\,\, If \overrightarrow a = \widehat j + \sqrt 3 \widehat k,\overrightarrow b = - \widehat j + \sqrt 3 \widehat k and \overrightarrow c = 2\sqrt 3 \widehat k form a triangle, then the internal angle of the triangle between \overrightarrow a and \overrightarrow b is (B) \,\,\,\, If \int\limits_a^b ( f( x ) - 3x )dx = a^2 - b^2, then the value of f ( \pi \over 6 ) is (C) \,\,\,\, The value of \pi ^2 \over \ell n3\int\limits_7/6^5/6 \sec ( \pi x )dx is (D) \,\,\,\, The maximum value of | Arg( 1 \over 1 - z ) | for | z | = 1,\,z e 1 is given by \,\,\,\,\,\,\,\,\,\,\,\, Column- II (p) \,\,\,\,\pi \over 6 (q) \,\,\,\,2\pi \over 3 (r) \,\,\,\,\pi \over 3 (s) \,\,\,\,\pi (t) \,\,\,\,\pi \over 2

( A ) \to q;\,\,( B ) \to p;\,\,( C ) \to s;\,\,( D ) \to t

( A ) \to q;\,\,( B ) \to p;\,\,( C ) \to t;\,\,( D ) \to s

( A ) \to p;\,\,( B ) \to q;\,\,( C ) \to s;\,\,( D ) \to t

( A ) \to q;\,\,( B ) \to s;\,\,( C ) \to p;\,\,( D ) \to t

Vector Algebra PYQ — JEE Advanced 2011 · Shift 2 · Q30 with Solution

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Correct answer: A

We match each item of Column-I with Column-II.

1. Part (A)

Given $\vec a=\hat j+\sqrt3\hat k,\qquad \vec b=-\hat j+\sqrt3\hat k,\qquad \vec c=2\sqrt3\hat k$ and these form a triangle.

Notice $\vec a+\vec b=(\hat j-\hat j)+(\sqrt3+\sqrt3)\hat k=2\sqrt3\hat k=\vec c$ So in the triangle, the two sides meeting at the included angle can be taken as $\vec a$ and $\vec b$ .

Now, $\vec a\cdot \vec b=(1)(-1)+(\sqrt3)(\sqrt3)=-1+3=2$ Also, $|\vec a|=\sqrt{1+3}=2,\qquad |\vec b|=\sqrt{1+3}=2$ Hence $\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\frac{2}{4}=\frac12$ Therefore $\theta=\frac{\pi}{3}$ But this is the angle between the vectors as placed tail-to-tail. In the triangle, the internal angle between the sides corresponding to $\vec a$ and $\vec b$ is the supplementary angle: $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ So, $(A)\to q$

2. Part (B)

Given $\int_a^b (f(x)-3x)\,dx=a^2-b^2$ for arbitrary $a,b$ .

Rewrite right side: $a^2-b^2=-(b^2-a^2)=\int_a^b (-2x)\,dx$ Thus $\int_a^b (f(x)-3x)\,dx=\int_a^b (-2x)\,dx$ Since this holds for all $a,b$ , $f(x)-3x=-2x$ So $f(x)=x$ Hence $f\left(\frac{\pi}{6}\right)=\frac{\pi}{6}$ Therefore, $(B)\to p$

3. Part (C)

We need $\frac{\pi^2}{\ln 3}\int_{7/6}^{5/6}\sec(\pi x)\,dx$

First, $\int \sec(\pi x)\,dx=\frac1\pi \ln|\sec(\pi x)+\tan(\pi x)|+C$ Therefore $\int_{7/6}^{5/6}\sec(\pi x)\,dx=\frac1\pi\left[\ln|\sec(\pi x)+\tan(\pi x)|\right]_{7/6}^{5/6}$ Now evaluate:

For $x=\frac56$ , $\pi x=\frac{5\pi}{6},\quad \sec\frac{5\pi}{6}=-\frac{2}{\sqrt3},\quad \tan\frac{5\pi}{6}=-\frac1{\sqrt3}$ So $\sec\frac{5\pi}{6}+\tan\frac{5\pi}{6}=-\sqrt3$ Hence absolute value is $\sqrt3$ .

For $x=\frac76$ , $\pi x=\frac{7\pi}{6},\quad \sec\frac{7\pi}{6}=-\frac{2}{\sqrt3},\quad \tan\frac{7\pi}{6}=\frac1{\sqrt3}$ So $\sec\frac{7\pi}{6}+\tan\frac{7\pi}{6}=-\frac1{\sqrt3}$ Hence absolute value is $\frac1{\sqrt3}$ .

Thus

=\frac1\pi\ln 3$$ Therefore $$\frac{\pi^2}{\ln 3}\int_{7/6}^{5/6}\sec(\pi x)\,dx =\frac{\pi^2}{\ln 3}\cdot \frac{\ln 3}{\pi}=\pi$$ So, $$(C)\to s$$ --- ## 4. Part (D) We need the maximum value of $$\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right|$$ for $$|z|=1,\ z\ne 1$$ Let $$z=e^{i\theta},\quad \theta\ne 0\pmod{2\pi}$$ Then $$1-z=1-e^{i\theta}=e^{i\theta/2}\left(e^{-i\theta/2}-e^{i\theta/2}\right) =-2i\,e^{i\theta/2}\sin\frac\theta2$$ So $$\Arg(1-z)=\Arg\left(-2i\,e^{i\theta/2}\sin\frac\theta2\right)$$ This implies geometrically that as $z$ moves on the unit circle, the point $1-z$ lies on a circle through the origin, and $$\Arg\left(\frac1{1-z}\right)=-\Arg(1-z)$$ Hence we need the largest possible magnitude of argument of $1-z$. Now $1-z$ traces the circle centered at $1$ with radius $1$, excluding the origin. From the origin, the arguments of points on this circle range between $-\frac\pi2$ and $\frac\pi2$, and values arbitrarily close to $\pm\frac\pi2$ are attained as $z\to 1$. Thus the maximum value is $$\frac\pi2$$ So, $$(D)\to t$$ --- ## 5. Final matching We obtained:

(A)\to q,\qquad (B)\to p,\qquad (C)\to s,\qquad (D)\to t

This corresponds to **Option A**. --- ## 6. Comparison with stored answer Stored correct answer: **A** Our derived answer: **A** So the stored answer agrees.

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1. Part (A)

Given

\vec a=\hat j+\sqrt3\hat k,\qquad \vec b=-\hat j+\sqrt3\hat k,\qquad \vec c=2\sqrt3\hat k

and these form a triangle.

Notice

\vec a+\vec b=(\hat j-\hat j)+(\sqrt3+\sqrt3)\hat k=2\sqrt3\hat k=\vec c

So in the triangle, the two sides meeting at the included angle can be taken as

\vec a

and

\vec b

Now,

\vec a\cdot \vec b=(1)(-1)+(\sqrt3)(\sqrt3)=-1+3=2

Also,

|\vec a|=\sqrt{1+3}=2,\qquad |\vec b|=\sqrt{1+3}=2

Hence

\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\frac{2}{4}=\frac12

Therefore

\theta=\frac{\pi}{3}

But this is the angle between the vectors as placed tail-to-tail. In the triangle, the internal angle between the sides corresponding to

\vec a

and

\vec b

is the supplementary angle:

\pi-\frac{\pi}{3}=\frac{2\pi}{3}

So,

(A)\to q

2. Part (B)

Given

\int_a^b (f(x)-3x)\,dx=a^2-b^2

for arbitrary

a,b

Rewrite right side:

a^2-b^2=-(b^2-a^2)=\int_a^b (-2x)\,dx

Thus

\int_a^b (f(x)-3x)\,dx=\int_a^b (-2x)\,dx

Since this holds for all

a,b

f(x)-3x=-2x

f(x)=x

Hence

f\left(\frac{\pi}{6}\right)=\frac{\pi}{6}

Therefore,

(B)\to p

3. Part (C)

We need

\frac{\pi^2}{\ln 3}\int_{7/6}^{5/6}\sec(\pi x)\,dx

First,

\int \sec(\pi x)\,dx=\frac1\pi \ln|\sec(\pi x)+\tan(\pi x)|+C

Therefore

\int_{7/6}^{5/6}\sec(\pi x)\,dx=\frac1\pi\left[\ln|\sec(\pi x)+\tan(\pi x)|\right]_{7/6}^{5/6}

Now evaluate:

For

x=\frac56

\pi x=\frac{5\pi}{6},\quad \sec\frac{5\pi}{6}=-\frac{2}{\sqrt3},\quad \tan\frac{5\pi}{6}=-\frac1{\sqrt3}

\sec\frac{5\pi}{6}+\tan\frac{5\pi}{6}=-\sqrt3

Hence absolute value is

\sqrt3

For

x=\frac76

\pi x=\frac{7\pi}{6},\quad \sec\frac{7\pi}{6}=-\frac{2}{\sqrt3},\quad \tan\frac{7\pi}{6}=\frac1{\sqrt3}

\sec\frac{7\pi}{6}+\tan\frac{7\pi}{6}=-\frac1{\sqrt3}

Hence absolute value is

\frac1{\sqrt3}

Thus

=\frac1\pi\ln 3$$ Therefore $$\frac{\pi^2}{\ln 3}\int_{7/6}^{5/6}\sec(\pi x)\,dx =\frac{\pi^2}{\ln 3}\cdot \frac{\ln 3}{\pi}=\pi$$ So, $$(C)\to s$$ --- ## 4. Part (D) We need the maximum value of $$\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right|$$ for $$|z|=1,\ z\ne 1$$ Let $$z=e^{i\theta},\quad \theta\ne 0\pmod{2\pi}$$ Then $$1-z=1-e^{i\theta}=e^{i\theta/2}\left(e^{-i\theta/2}-e^{i\theta/2}\right) =-2i\,e^{i\theta/2}\sin\frac\theta2$$ So $$\Arg(1-z)=\Arg\left(-2i\,e^{i\theta/2}\sin\frac\theta2\right)$$ This implies geometrically that as $z$ moves on the unit circle, the point $1-z$ lies on a circle through the origin, and $$\Arg\left(\frac1{1-z}\right)=-\Arg(1-z)$$ Hence we need the largest possible magnitude of argument of $1-z$. Now $1-z$ traces the circle centered at $1$ with radius $1$, excluding the origin. From the origin, the arguments of points on this circle range between $-\frac\pi2$ and $\frac\pi2$, and values arbitrarily close to $\pm\frac\pi2$ are attained as $z\to 1$. Thus the maximum value is $$\frac\pi2$$ So, $$(D)\to t$$ --- ## 5. Final matching We obtained:

(A)\to q,\qquad (B)\to p,\qquad (C)\to s,\qquad (D)\to t

This corresponds to **Option A**. --- ## 6. Comparison with stored answer Stored correct answer: **A** Our derived answer: **A** So the stored answer agrees.

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2. Part (B)

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