BITSAT 2020 Question Paper with Solutions

For real roots, discriminant,

$D = {b^2} - 4ac \ge 0$

$= {\cos ^2}\beta - 4(\cos \beta - 1)sin\beta \ge 0$

$= {\cos ^2}\beta + 4(1 - \cos \beta )\sin \beta \ge 0$

So, sin$\beta$ should be > 0. [$\because$ ${\cos ^2}\beta \ge 0,\,1 - \cos \beta \ge 0$]

$\Rightarrow \beta \in (0,\pi )$.

We have, $2f(xy) = {(f(x))^x} + {(f(y))^x}$, $x,y \in R$ and $f(1) = a$.

Putting x = 1, y = 2, we get

$2f(2) = {(f(1))^2} + f(2) \Rightarrow f(2) = {a^2}$

Putting x = 1, y = 3, we get

$2f(3) = {(f(1))^3} + f(3) \Rightarrow f(3) = {a^3}$, and so on.

$\therefore$ $f(n) = {a^n}$

$\sum\limits_{k = 1}^n {f(i) = a + {a^2} + .. + {a^n} = {{a({a^n} - 1)} \over {a - 1}}}$.

The given inequality is equivalent to

$1 < |x - 3| + |4 - x|$

i.e., $1 < |x - 3| + |x - 4|$ ...... (i)

For x < 3, we have

$3 - x + 4 - x = 7 - 2x > 1$

i.e., x < 3, which is true in the domain.

For $3 \le x < 4$, Eq. (i) gives

$x - 3 + 4 - x = 1 < 1$ ..... (ii)

For 4 $\le$ x, we have

$2x - 7 > 1 \Rightarrow x > 4$ ...... (iii)

From Eqs. (i) and (iii), we have the solution of the inequality as x < 3 or x > 4.

i.e., $x \in ] - \infty ,3) \cup (4,\infty$ [which is equal to R $-$ [3, 4].

We have, ${{dy} \over {dx}} + {{\tan y} \over x} = {{\tan y\sin y} \over {{x^2}}}$

Now, divide by tan y sin y both sides, we get

$\cot y\cos ec\,y{{dy} \over {dx}} + {{\cos ec\,y} \over x} = {1 \over {{x^2}}}$ ...... (i)

Put cosec y = z

$\therefore$ $- \cos ec\,y\cot y{{dy} \over {dx}} = {{dz} \over {dx}}$

$\therefore$ Eq. (i) reduces to

${{dz} \over {dx}} - {z \over x} = - {1 \over {{x^2}}}$

$IF = {e^{\int { - {1 \over x}dx} }} = {e^{ - \log x}} = {1 \over x}$

Hence solution is

${z \over x} = \int { - {1 \over {{x^3}}}dx = {1 \over {2{x^2}}} + c}$

$\Rightarrow {1 \over {x\sin y}} = {{1 - 2c{x^2}} \over {2{x^2}}}$

$\Rightarrow 2x = \sin y{(1 - 2cx)^2}$

$\Rightarrow 2x = \sin y(1 - 2c{x^2})$

The given equation is

$2(1 + i){x^2} - 4(2 - i)x - 5 - 3i = 0$

$\Rightarrow x = {{4(2 - i) \pm \sqrt {16{{(2 - i)}^2} + 8(1 + i)(5 + 3i)} } \over {4(1 + i)}}$

$= - {i \over {1 + i}}$ or ${{4 - i} \over {1 + i}} = {{ - 1 - i} \over 2}$ or, ${{3 - 5i} \over 2}$

Now, $\left| {{{ - 1 - i} \over 1}} \right| = \sqrt {{1 \over 4} + {1 \over 4}} = \sqrt {{1 \over 2}}$

and $\left| {{{3 - 5i} \over 2}} \right| = \sqrt {{9 \over 4} + {{25} \over 4}} = \sqrt {{{17} \over 2}}$

Also, $\sqrt {{{17} \over 2}} > \sqrt {{1 \over 2}}$

Hence, required root is ${{3 - 5i} \over 2}$.

Possible even digits are 2, 4, 6, 8, 0.

Case I Number has digits 4, 6, 8, 0.

(Here, sum of digits 18, divisible by 3)

Number of arrangements = 3 $\times$ 3!

[1st place can be filled using 4, 6, 8]

= 3 $\times$ 6 = 18

Case II Number has digits 2, 4, 6, 0

(Here, sum of digits is 12, divisible by 3)

1st place cannot be filled by 0.

Number of arrangements = 3 $\times$ 3! = 18

$\therefore$ Number of numbers = 18 + 18 = 36

Let f(x) = x¹⁰⁰

g(x) = x² $-$ 3x + 2

q(x) = quotient

r(x) = remainder = (ax + b)

$\therefore$ f(x) = g(x) q(x) + r(x)

x¹⁰⁰ = (x² $-$ 3x + 2) q(x) + ax + b

Put x = 1 $\to$ 1 = a + b ..... (i)

Put x = 2 $\Rightarrow$ 2¹⁰⁰ = 2a + b ..... (ii)

From Eqs. (i) and (ii), we get

a = 2¹⁰⁰ $-$ 1, b = 2 $-$ 2¹⁰⁰

Remainder = (2¹⁰⁰ $-$ 1)x + (2 $-$ 2¹⁰⁰)

= (2^{99 + 1} $-$ 1)x $-$ (2⁹⁹ $-$ 1)

$\therefore$ k = 99

We have, $Z = {{7 + i} \over {3 + 4i}}$

$\Rightarrow Z = \left( {{{7 + i} \over {3 + 4i}}} \right)\left( {{{3 - 4i} \over {3 - 4i}}} \right) = {{21 + 4 - 28i + 3i} \over {9 + 16}} = 1 - i$

$\therefore$ ${Z^{14}} = {(1 - i)^{14}} = {[{(1 - i)^2}]^7} = {(1 - 1 - 2i)^7}$

$= {( - 2)^7}{(i)^7} = {2^7}i$

We have,

Vector r is coplanar with vector a and b

$\therefore$ $r = \lambda (b \times (a \times b))$

$\Rightarrow r = \lambda ((b\,.\,b)a - (b\,.\,a)b)$

$\Rightarrow r = \lambda (|b{|^2}a - (a\,.\,b)b)$

$\Rightarrow r = \lambda [5a + b]$ [$\because$ $|b| = \sqrt {{2^2} + {1^2}} = \sqrt 5$]

and $a\,.\,b = ( - \widehat i + \widehat j + \widehat k)\,.\,(2\widehat i + \widehat k) = - 1$

Given, $r\,.\,a = 7$

$\therefore$ $\lambda (5a + b)\,.\,a = 7$

$\Rightarrow \lambda [5|a{|^2} + a\,.\,b] = 7$

$\Rightarrow \lambda [15 - 1] = 7$ [$\because$ $|a{|^2} = 1 + 1 + 1 = 3$]

$\Rightarrow \lambda = {1 \over 2}$

$\therefore$ $r = {1 \over 2}(5a + b)$

$r = {1 \over 2}( - 5\widehat i + 5\widehat j + 5\widehat k + 2\widehat i + \widehat k)$

$r = {{ - 3} \over 2}\widehat i + {5 \over 2}\widehat j + {6 \over 2}\widehat k$

z component of $r = {6 \over 2} = 3$

We have,

$(x - 1)(x - 2)(x - 3)....(x - 10)$

Coefficient of x⁸ = Sum of the terms taken two at a time i.e.,

$\sum\limits_{1 \le i < j \le 10}^{10} {{x_i}\,.\,{x_j} = {1 \over 2}\left[ {{{\left( {\sum\limits_{i = 1}^{10} {{x_i}} } \right)}^2} - \left( {\sum\limits_{i = 1}^{10} {{x_i}^2} } \right)} \right]}$

$= {1 \over 2}[{(1 + 2 + 3 + ... + 10)^2} - {({1^2} + {2^2} + {3^2} + ... + 10)^2}]$

$= {1 \over 2}\left[ {{{\left( {{{10 \times 11} \over 2}} \right)}^2} - \left( {{{10 \times 11 \times 21} \over 6}} \right)} \right]$

$= {1 \over 2}({(55)^2} - 385) = {1 \over 2}(3025 - 385) = {{2640} \over 2} = 1320$

Since x, y, z are three consecutive integers

$\therefore$ 2y = x + z

$\Rightarrow$ (4y²) = (x + z)²

= (x $-$ z)² + 4xz

= ($-$2)² + 4xz [$\because$ x $-$ z = $-$2]

$\Rightarrow$ 4y² = 4 + 4xz

$\Rightarrow$ y² = 1 + xz

Now,

${1 \over 2}{\log _e}x + {1 \over 2}{\log _e}z + {1 \over {2xz + 1}} + {1 \over 3}{\left( {{1 \over {2xz + 1}}} \right)^3} + ...$

$= {1 \over 2}\left[ {{{\log }_e}(xz) + 2\left( {{1 \over {2zx + 1}} + {1 \over 3}\,.\,{1 \over {{{(2zx + 1)}^3}}} + ...} \right)} \right]$

$= {1 \over 2}\left[ {{{\log }_e}(xz) + {{\log }_e}\left( {{{1 + {1 \over {2zx + 1}}} \over {1 - {1 \over {2zx + 1}}}}} \right)} \right]$

$= {1 \over 2}\left[ {{{\log }_e}(xz) + {{\log }_e}\left( {{{1 + xz} \over {xz}}} \right)} \right]$

$= {1 \over 2}[{\log _e}(1 + xz)] = {1 \over 2}[{\log _e}{(y)^2}] = {\log _e}(y)$

Graph of circle whose centre is origin and radius is $\pi$ units and y = sin x is

Area of shaded region = ${1 \over 2}$ Area of circle

= ${1 \over 2}$($\pi$r²) = ${1 \over 2}$$\pi$($\pi$²) [$\because$ r = $\pi$]

= ${{{\pi ^3}} \over 2}$

Given equation of parabola is

x = 4y $-$ y²

$\Rightarrow$ y² $-$ 4y + 4 = $-$ x + 4

$\Rightarrow$ (y $-$ 2)² = $-$ (x $-$ 4)

a = ${{1} \over 4}$

Focus = $\left( {{{15} \over 4},2} \right)$

Parabola cuts Y-axis of (0, 0) and (0, 4)

Now, equation of tangent at (0, 0) is

x = 4y ...... (i)

and equation of tangent at (0, 4) is

x + 4y = 16 .... (ii)

Solving Eqs. (i) and (ii), we get

x = 8, y = 2

$\therefore$ Intersection point of tangent is P(8, 2).

Distance from point P to focus S

Distance = $\sqrt {{{\left( {8 - {{15} \over 4}} \right)}^2} + {{(2 - 2)}^2}}$

$= 8 - {{15} \over 4} = {{17} \over 4}$

We have,

Slope of tangent of the curve is ${{{{(x + 1)}^2} + y - 3} \over {x + 1}}$

$\therefore$ ${{dy} \over {dx}} = {{{{(x + 1)}^2} + y - 3} \over {x + 1}}$

$\Rightarrow {{dy} \over {dx}} = {{{{(x + 1)}^2} - 3} \over {x + 1}} + {y \over {x + 1}}$

$\Rightarrow {{dy} \over {dx}} - {y \over {x + 1}} = {{{{(x + 1)}^2} - 3} \over {x + 1}}$

This is a form of linear differential equations

$\therefore$ $IF = {e^{\int {{{ - 1} \over {x + 1}}dx} }} = {e^{\log {1 \over {1 + x}}}} = {1 \over {1 + x}}$

Solution of given differential equation is

${y \over {x + 1}} = \int {{{{{(x + 1)}^2} - 3} \over {{{(x + 1)}^2}}}dx}$

${y \over {x + 1}} = \int {\left( {1 - {3 \over {{{(x + 1)}^2}}}} \right)dx}$

${y \over {x + 1}} = x + {3 \over {x + 1}} + c$

$\because$ Curve passes through (2, 0)

$\therefore$ $0 = 2 + {3 \over 3} + c$

$\Rightarrow c = - 3$

Hence, equation of curve is

${y \over {x + 1}} = x + {3 \over {x + 1}} - 3$

$y = {x^2} + x + 3 - 3x - 3$

$y = {x^2} - 2x$

We have, $f(x) = {x \over {\sqrt {1 + {x^2}} }}$

$\Rightarrow f(f(x)) = {{f(x)} \over {\sqrt {1 + {{(f(x))}^2}} }} = {{{x \over {\sqrt {1 + {x^2}} }}} \over {\sqrt {1 + {{{x^2}} \over {1 + {x^2}}}} }} = {x \over {\sqrt {1 + 2{x^2}} }}$

Similarly, $f(f(f(x))) = {x \over {\sqrt {1 + 3{x^2}} }}$

$\underbrace {fofofo.....of(x)}_{x\,times} = {x \over {\sqrt {1 + n{x^2}} }} = {x \over {\sqrt {1 + \left( {\sum\limits_{r = 1}^n 1 } \right){x^2}} }}$

We know that,

n(A $\cup$ B) = n(A) + n(B) $-$ n(A $\cap$ B)

$\therefore$ P = 1500 $-$ n(A $\cap$ B)

$\Rightarrow$ n(A $\cap$ B) = 1500 $-$ P

Clearly, 1 $\le$n(A $\cap$ B) $\le$ 500

$\Rightarrow$ 1 $\le$ 1500 $-$ P $\le$ 500

$\Rightarrow$ $-$1499 $\le$ $-$P $\le$ $-$1000

$\Rightarrow$ 1000 $\le$ P $\le$ 1499

We have,

$Z = r{e^{i\theta }}$

$Z = r(\cos \theta + i\sin \theta )$

${e^{iz}} = {e^{ir(\cos \theta + i\sin \theta )}}$

$\Rightarrow {e^{iz}} = {e^{ir\cos \theta + {i^2}r\sin \theta }}$

$\Rightarrow {e^{iz}} = {e^{ir\cos \theta - r\sin \theta }}$

$\Rightarrow {e^{iz}} = {e^{ - r\sin \theta }}\,.\,{e^{ir\cos \theta }}$

$\therefore$ $\arg ({e^{iz}}) = r\cos \theta$

Since, $g = \sqrt {ab}$. Also, a, p, q and b are in AP.

So, common difference d is ${{b - a} \over 3}$.

$\therefore$ $p = a + d = a + {{b - a} \over 3} = {{2a + b} \over 3}$

$q = b - d = b - {{b - a} \over 3} = {{a + 2b} \over 3}$

Now, $(2p - q)(p - 2q)$

$= {{(4a + 2b - a - 2b)} \over 3}\,.\,{{(2a + b - 2a - 4b)} \over 3}$

$= - ab = - {g^2}$

We have P(3, 7, 1) and R(2, 5, 7)

DR's of PR = ($-$1, $-$2, 6)

Angle between line PQ and plane is

$\sin \theta = {{( - 1)(3) + ( - 2) \times 2 + 6 \times 11} \over {\sqrt {1 + 4 + 36} \sqrt {9 + 4 + 121} }}$

$\sin \theta = {{59} \over {\sqrt {41} \sqrt {134} }}$

$PM = \left| {{{3(3) + 2(7) + 11( \times 1) - 9} \over {\sqrt {9 + 4 + 121} }}} \right|$

$= \left| {{{9 + 14 + 11 - 9} \over {\sqrt {134} }}} \right| = {{25} \over {\sqrt {134} }}$

$\sin \theta = {{PM} \over {PQ}}$

$\Rightarrow PQ = {{PM} \over {\sin \theta }} = {{25} \over {\sqrt {134} }} \times {{\sqrt {41} \times \sqrt {134} } \over {59}} = {{25\sqrt {41} } \over {59}}$

We have given,

| a | = 1, | b | = 4, a . b = 2

Now, c = (2a $\times$ b) $-$ 3b

Multiplying by b both sides, we get

$b\,.\,c = (2a \times b)\,.\,b - 3|b{|^2}$

$\Rightarrow b\,.\,c = 0 - 3{(4)^2} = - 48$

and $|c{|^2}\, = \,|(2a \times b)\,.\,3b{|^2}$

$|c{|^2}\, = \,{(2a \times b)^2} + 9|b{|^2} = 4{(a \times b)^2} + 9{(4)^2}$

$\Rightarrow |c{|^2}\, = 4(|a{|^2}\,.\,|b{|^2} - {(a\,.\,b)^2}) + 144$

$= 4(16 - 4) + 144 = 48 + 144 = 192$

$\therefore$ $b\,.\,c = |b||c|cos\theta$

$\Rightarrow cos\theta = {{b\,.\,c} \over {|b||c|}} = {{ - 48} \over {4 \times \sqrt {192} }} = - {{\sqrt 3 } \over 2}$

$\theta = \pi - {\pi \over 6} = {{5\pi } \over 6}$

Let AB = h = height of vertical pole

OA = x = Horizontal distance from O to pole

In $\Delta$OAB

$\tan 45^\circ = {{AB} \over {OA}} = 1 = {{20} \over x} \Rightarrow x = 20$ m

In $\Delta$OPQ

$\tan 30^\circ = {{PQ} \over {OP}} \Rightarrow {1 \over {\sqrt 3 }} = {{20} \over {x + y}}$

$\Rightarrow x + y = 20\sqrt 3$

$\Rightarrow y = 20\sqrt 3 - x = 20\sqrt 3 - 20 = 20(\sqrt 3 - 1)$

$\therefore$ Speed of bird is $20(\sqrt 3 - 1)$ m/s.

Given that, $A = \{ x:{x^2} = 1\} = \{ - 1,1\}$ and $B = \{ x:{x^4} = 1\} = \{ - 1,1, - i,i\}$

Now, $A - B = \phi$ and $B - A = \{ - i,i\}$

$\therefore$ $A\Delta B = (A - B) \cup (B - A) = \{ - i,i\}$

With four 6-sided dics, there are possible combinations that can be rolled. Out of these, only 80 sum to 10 as the table shows,

$6 \times 6 \times 6 \times 6=6^4=1296$

Let f(x) = Ax² + Bx+ C

$\therefore$ f(1) = A + B + C

and f($-$1) = A $-$ B + C

$\because$ f(1) = f($-$1)

$\Rightarrow$ A + B + C = A $-$ B + C

$\Rightarrow$ 2B = 0 $\Rightarrow$ B = 0

$\therefore$ f(x) = Ax² + C

$\Rightarrow$ f'(x) = 2Ax

$\Rightarrow$ f'(a) = 2Aa

f'(b) = 2Ab and f'(c) = 2Ac

Also, a, b, c are in AP

So, 2Aa, 2Ab and 2Ac are in AP.

$\Rightarrow$ f'(a), f'(b) and f'(c) are also in AP.

A line passing through (1, 2) and parallel to the line 3x $-$ y = 10 is

3(x $-$ 1) $-$ (y $-$ 2) = 0

$\Rightarrow$ 3x $-$ 3 $-$ y + 2 = 0 $\Rightarrow$ 3x $-$ y $-$ 1 = 0

The intersection point of line

x + y + 5 = 0 ..... (i)

and 3x $-$ y $-$ 1 = 0 ..... (ii)

are ($-$1, $-$4)

$\therefore$ Distance from the line x + y + 5 to the point (1, 2) measured parallel to the line 3x $-$ y = 10 is

$\sqrt {{{( - 1 - 1)}^2} + {{( - 4 - 2)}^2}}$

$\sqrt {4 + 36} = \sqrt {40} = 2\sqrt {10}$

We have,

A = \left[ {\matrix{ 2 & 1 \cr 1 & 2 \cr } } \right]

We know that, $\left| {A \to \lambda I} \right| = 0$

\left| {\left[ {\matrix{ 2 & 1 \cr 1 & 2 \cr } } \right] - \left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]} \right| = 0

= \left| {\matrix{ {2 - \lambda } & 1 \cr 1 & {2 - \lambda } \cr } } \right| = 0

$\Rightarrow {(2 - \lambda )^2} - 1 = 0$

$\Rightarrow 4 - 4\lambda + {\lambda ^2} - 1 = 0$

$\Rightarrow {\lambda ^2} - 4\lambda + 3 = 0$

$\therefore$ ${A^2} - 4A + 3I = 0$

Multiply by A^$-$1, we get

$A - 4I + 3{A^{ - 1}} = 0$

$\Rightarrow {A^{ - 1}} = {{ - 1} \over 3}A + {4 \over 3}I$

$\therefore$ $\alpha = - {1 \over 3}$ and $\beta = {4 \over 3}$

$\alpha + \beta = - {1 \over 3} + {4 \over 3} = {3 \over 3} = 1$

We have,

$I = \int {{{8{x^{43}} + 13{x^{38}}} \over {{{({x^{13}} + {x^5} + 1)}^4}}}dx}$

$= \int {{{8{x^{43}} + 13{x^{38}}} \over {{x^{52}}\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}}dx}$

$= \int {{{8\left( {{1 \over {{x^9}}}} \right) + 13\left( {{1 \over {{x^{14}}}}} \right)} \over {{{\left( {1 + {1 \over {{x^8}}} + {1 \over {{x^{13}}}}} \right)}^4}}}dx}$