WBJEE 2020 Paper 1 (Mathematics) Question Paper with Solutions

Let complex number

$P = \cos \theta + i\sin \theta$

$\therefore$ $\left| P \right| = 1$ and $\theta \in \left[ {0,2\pi } \right)$

Now, ${P^4} = {(\cos \theta + i\sin \theta )^4}$

$\Rightarrow {P^4} = \cos 4\theta + i\sin 4\theta$

Imaginary part of P⁴ = 0

$\therefore$ sin 4$\theta$ = 0

$\Rightarrow 4\theta = n\pi \Rightarrow \theta = {\pi \over 4}$

$0 \le \theta \le 2\pi$

= $0 \le x < 8$

$\Rightarrow$ n = 8

The equation of the plane through the points $(2, - 1, - 3)$ and parallel to the line,

${{x - 1} \over 2} = {{y + 2} \over 3} = {z \over { - 4}}$ and ${x \over 2} = {{y - 1} \over { - 3}} = {{z - 2} \over 2}$

\left| {\matrix{ {x - 2} & {y + 1} & {z + 3} \cr 2 & 3 & { - 4} \cr 2 & { - 3} & 2 \cr } } \right| = 0

$\Rightarrow (x - 2)(6 - 12) - (y + 1)(4 + 8) + (z + 3)( - 6 - 6) = 0$

$\Rightarrow - 6x + 12 - 12y - 12 - 12z - 36 = 0$

$\Rightarrow - 6x - 12y - 12z - 36 = 0$

$\Rightarrow x + 2y + 2z + 6 = 0$

Given, curve

y² = x³

On differentiating both sides w.r.t. x, we get

$2y = {{dy} \over {dx}} = 3{x^2}$

$\Rightarrow {{dy} \over {dx}} = {{3{x^2}} \over {2y}}$

$\Rightarrow {\left( {{{dy} \over {dx}}} \right)_{({m^2},{m^3})}} = {{3({m^4})} \over {2{m^3}}} = {3 \over 2}m$

$\therefore$ Slope of tangent at (m², m³) is ${3 \over 2}m$.

Now slope of normal at (m², m³) to the curve y² = x³ is

${ - {{dx} \over {dy}}}$ = ${{ - 2y} \over {3{x^2}}}$

$\Rightarrow {\left( { - {{dx} \over {dy}}} \right)_{({m^2},{m^3})}} = {{ - 2{m^3}} \over {3{m^4}}} = {{ - 2} \over {3M}}$

Given, slope of tangent = slope of normal

$\therefore$ ${{3m} \over 2} = {{ - 2} \over {3M}}$

$\Rightarrow mM = {{ - 4} \over 9}$

${{6a} \over {5b}} + {{10b} \over {3a}}$

We know that, $AM \ge GM$

$\therefore$ ${{{{6a} \over {5b}} + {{10b} \over {3a}}} \over 2} \ge \sqrt {{{6a} \over {5b}} \times {{10b} \over {3a}}}$

$\Rightarrow {{6a} \over {5b}} + {{10b} \over {3a}} \ge 2 \times 2 = 4$

Hence, least possible value of ${{6a} \over {5b}} + {{10b} \over {3a}}$ is 4.

Given, $P = {{10} \over {100}} = {1 \over {10}}$

$q = 1 - P = 1 - {1 \over {10}} = {9 \over {10}}$

$P(X \ge 1) = 1 - P(X = 0)$

$P(X \ge 1) = 1 - {}^n{C_0}{\left( {{1 \over {10}}} \right)^0}{\left( {{9 \over {10}}} \right)^n}$

$1 - {\left( {{9 \over {10}}} \right)^n} \ge {1 \over 2}$

${\left( {{9 \over {10}}} \right)^n} \le {1 \over 2}$

For n = 7 the least number of round to must so fire.

Let A = \left| {\matrix{ {{a^2}} & {bc} & {{c^2} + ab} \cr {{a^2} + ab} & {{b^2}} & {ca} \cr {ab} & {{b^2} + bc} & {{c^2}} \cr } } \right|

Taking common a, b, c from C₁, C₂ and C₃ respectively.

$\therefore$ A = abc\left| {\matrix{ a & c & {c + a} \cr {a + b} & b & a \cr b & {b + c} & c \cr } } \right|

Apply R₁ $\to$ R₃

A = abc\left| {\matrix{ {2(a + b)} & {2(b + c)} & {2(c + a)} \cr {a + b} & b & a \cr b & {b + c} & c \cr } } \right|

A = 2abc\left| {\matrix{ {a + b} & {b + c} & {c + a} \cr {a + b} & b & a \cr b & {b + c} & c \cr } } \right|

Apply R₁ $\to$ R₁ $-$ R₂

A = 2abc\left| {\matrix{ 0 & c & c \cr {a + b} & b & a \cr b & {b + c} & c \cr } } \right|

A = 2ab{c^2}\left| {\matrix{ 0 & 1 & 1 \cr {a + b} & b & a \cr b & {b + c} & c \cr } } \right|

Apply C₂ $\to$ C₂ $-$ C₃

A = 2ab{c^2}\left| {\matrix{ 0 & 0 & 1 \cr {a + b} & {b - a} & a \cr b & b & c \cr } } \right|

$A = 2ab{c^2}\left| {(a + b)b - (b - a)b} \right|$

$= 2ab{c^2}\left| {ab + {b^2} - {b^2} + ab} \right|$

A = 4a²b²c²

$\therefore$ k = 4

We have,

${\cos ^{ - 1}}\left( {{y \over b}} \right) = \log {\left( {{x \over n}} \right)^n}$

$\Rightarrow$ y = b cos(n log x $-$ n log n)

On differentiating both sides w.r.t.x, we get

${{dy} \over {dx}} = - b\,\sin (n\,\log \,x - n\,\log \,n) \times {n \over x}$

$\Rightarrow x{{dy} \over {dx}} = - nb\,\sin (n\,\log \,x - n\,\log \,n)$

$x{{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} = {{{n^2}b\,\cos (n\,\log \,x - n\,\log \,n)} \over x}$

$\Rightarrow$ ${x^2}{{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}} = - {n^2}y$

$\Rightarrow {x^2}{{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}} + {n^2}y = 0$

Given parabola

y = ax² + bx + c

put x = 1, y = 1, we get

1 = a + b + c ......(i)

y = x is tangent of the parabola at (1, 1).

$\therefore$ ${\left( {{{dy} \over {dx}}} \right)_{(1,1)}} = 2ax + b$

$\Rightarrow$ 1 = 2a + b .......(ii)

Also curve passes through ($-$1, 0)

$\therefore$ 0 = a $-$ b + c ........(iii)

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

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

and $b = {1 \over 2}$

Let

$I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {} \int\limits_{1/(k + \beta )}^{1/(k + \alpha )} {{{dx} \over {1 + x}}}$

$I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {[\log (1 + x)]} _{{1 \over {k + \beta }}}^{{1 \over {k + \alpha }}}$

$I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left[ {\log \left( {1 + {1 \over {k + \alpha }}} \right) - \log \left( {1 + {1 \over {k + \beta }}} \right)} \right]}$

$I = \mathop {\lim }\limits_{n \to \infty } \,\log \left( {{{\alpha + 2} \over {\alpha + 1}}} \right)\left( {{{\beta + 1} \over {\beta + 2}}} \right)\left( {{{\alpha + 3} \over {\alpha + 2}}} \right)\left( {{{\beta + 2} \over {\beta + 3}}} \right)....\left( {{{\alpha + n + 1} \over {\alpha + n}}} \right)\left( {{{\beta + n} \over {\beta + n + 1}}} \right)$

$I = \mathop {\lim }\limits_{n \to \infty } \,\left( {\log \left( {{{\beta + 1} \over {\alpha + 1}}} \right)\left( {{{\alpha + n + 1} \over {\beta + n + 1}}} \right)} \right)$

$I = {\log _e}\left( {{{\beta + 1} \over {\alpha + 1}}} \right)$

Given, equation of latusrectum of parabola is x + y = 8

and equation of tangent at vertex is x + y = 12

Since both are parallel.

$\therefore$ The distance between latusrectum and equation of tangent at vertex is 'a'

$\therefore$ $a = {4 \over {\sqrt {{1^2} + {1^2}} }} = {4 \over {\sqrt 2 }} = 2\sqrt 2$

$\therefore$ Length of latusrectum = $4a = 8\sqrt 2$ unit

The expression ax² + bx + c has the same sign as that of a for all x if D $\le$ 0

$\therefore$ b² $-$ 4ac $\le$ 0

Given, x² + y² = a²

On differentiating both sides w.r.t. x, we get,

$2x + 2y{{dy} \over {dx}} = 0$

$\Rightarrow {{dy} \over {dx}} = {{ - x} \over y}$

$\Rightarrow {\left( {{{dy} \over {dx}}} \right)^2} = {{{x^2}} \over {{y^2}}}$

Let $I = \int\limits_0^a {\sqrt {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} dx}$

$I = \int\limits_0^a {\sqrt {1 + {{{x^2}} \over {{y^2}}}} dx}$

$\Rightarrow$ $I = \int\limits_0^a {\sqrt {{{{x^2} + {y^2}} \over {{y^2}}}} dx}$

$I = \int\limits_0^a {{a \over {\sqrt {{a^2} - {x^2}} }}dx}$ [$\because$ x² + y² = a²]

$\Rightarrow$ $I = a\left[ {{{\sin }^{ - 1}}{x \over a}} \right]_0^a$

$\Rightarrow$ $I = a\left[ {{{\sin }^{ - 1}}{a \over a} - {{\sin }^{ - 1}}0} \right]$

$\Rightarrow$ $I = a\left[ {{{\sin }^{ - 1}}1 - {{\sin }^{ - 1}}0} \right]$

= $a\left[ {{\pi \over 2} - 0} \right] = {\pi \over 2}a$

Given, parabola y² = 4ax


Let the point A on parabola (at², 2at) in $\Delta$OAB

$\cot \,\alpha = {{a{t^2}} \over {2at}} = {t \over 2}$

$t = 2\,\cot \,\alpha$

In $\Delta OAB$

$\sin \,\alpha = {{AB} \over {OA}}$

$\Rightarrow OA = AB\,\cos ec\,\alpha = 2at\,\cos ec\,\alpha$

$\therefore$ $OA = 2a(2\,\cot \,\alpha )\,\cos ec\,\alpha$

$OA = 4a\,\cot \,\alpha \,\cos ec\,\alpha$

$\therefore$ Length of chord = $4a\,\cot \,\alpha \,\cos ec\,\alpha$

Given, the radius of circle = $\sqrt {17}$

With centre on positive sides of X-axis

$\therefore$ Equation of circle is ${(x - a)^2} + {y^2} = 17$

Since, it passes through (0, 1)

$\therefore$ a² + 1 = 17

$\Rightarrow$ a = 4

$\therefore$ Equation of circle is

${x^2} + {y^2} - 8x - 1 = 0$

Given, $f(x) = \sqrt {\left( {{1 \over {\sqrt x }} - \sqrt {x + 1} } \right)}$

f(x) is defined ${1 \over {\sqrt x }} - \sqrt {x + 1} \ge 0$ and x > 0

${x^2} + x - 1 \le 0$, x > 0

$x = {{ - 1 \pm \sqrt 5 } \over 2}$, x > 0

$\therefore$ Domain of f(x) is $\left( {0,{{\sqrt 5 - 1} \over 2}} \right]$.

Let the unit vector in ZOX plane be

$a = x\widehat i + z\widehat k,\left| a \right| = 1$

$a.\alpha = \left| a \right|\left| \alpha \right|\cos 45^\circ$

$\Rightarrow$ $(x\widehat i + z\widehat k).(2\widehat i + 2\widehat j - \widehat k) = 1 \times 3 \times {1 \over {\sqrt 2 }}$

[$\because$ $\alpha = 2\widehat i + 2\widehat j - \widehat k$]

$2x - z = {3 \over {\sqrt 2 }}$

and $a.\beta = \left| a \right|\left| \beta \right|\cos 60^\circ$

$\Rightarrow (x\widehat i + z\widehat k).(\widehat j - \widehat k) = 1 \times \sqrt 2 \times {1 \over 2}$ [$\because$ $\beta = \widehat j - \widehat k$]

$- z = {1 \over {\sqrt 2 }}$

$z = - {1 \over {\sqrt 2 }}$ and $x = {1 \over {\sqrt 2 }}$

$\therefore$ $a = {1 \over {\sqrt 2 }}\widehat i - {1 \over {\sqrt 2 }}\widehat k$

Given, A = $\left( {\matrix{ {3 - t} \cr { - 1} \cr 0 \cr } \matrix{ {} \cr {} \cr {} \cr } \,\matrix{ 1 \cr {3 - t} \cr { - 1} \cr } \matrix{ {} \cr {} \cr {} \cr } \matrix{ 0 \cr 1 \cr 0 \cr } } \right)$

$\left| A \right| = (3 - t)(0 + 1) - 1(0) + 0$

$\Rightarrow \left| A \right| = 3 - t$

Given, $\left| A \right| = 5$

$\therefore$ $3 - t = 5$

$\Rightarrow t = - 2$

We have,

$\phi (x) = f(x) + f(1 - x)$ and $f'(x) < 0$ in [0, 1]

$\phi '(x) = f'(x) - f'(1 - x)$

For monotonic increasing, f'(x)-f'(1-x) $\ge$ 0

$\Rightarrow$ $f'(x) \ge f'(1 - x)$

Since, f'(x) is decreasing.

$\therefore$ $x \le 1 - x \Rightarrow x \le {1 \over 2}$

$\Rightarrow$ $\phi (x)$ is monotonic increasing in $\left[ {0,{1 \over 2}} \right]$ and monotonic decreasing in $\left[ {{1 \over 2},1} \right]$

Let (h, k) be the centre and r be the radius of the variable circle.

Since, variable circle touches the circle x² + y² = a²

$\therefore$ $\sqrt {{h^2} + {k^2}} = a$ ....(i)

Also variable circle touches the circle x² + y² = 4ax

$\therefore$ $\sqrt {{{({h^2} + 2a)}^2} + {k^2}} = 2a + r$ ....(ii)

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

$\sqrt {{{({h^2} + 2a)}^2} + {k^2}} - \sqrt {{h^2} + {k^2}} = a$

$\therefore$ Locus of centre of variable circle is

$\sqrt {{{(x - 2a)}^2} + {y^2}} - \sqrt {{x^2} + {y^2}} = a$

which represents the hyperbola.

Area of region

$\{ (x,y):{x^2} + {y^2} \le 1 \le x + y\}$

$\therefore$ x² + y² = 1 and x + y = 1


Area of shaded region = Area of quadrant $-$ Area of triangle

$= {\pi \over 4}{(1)^2} - {1 \over 2} \times 1 \times 1$

= ${\pi \over 4} - {1 \over 2}$

Given, $P(A' \cap B') = {3 \over 5}$

$\Rightarrow 1 - P(A \cup B) = {3 \over 5}$

$\Rightarrow P(A \cup B) = {2 \over 5}$

$\Rightarrow P(A) + P(B) - P(A \cap B) = {2 \over 5}$

$\Rightarrow P(A) + P(B) - P(A)P(B) = {2 \over 5}$ ...(i)

(since A and B are independent events)

Also, given $P(A \cap B) = {1 \over {20}}$

$\Rightarrow P(A)P(B) = {1 \over {20}}$ ... (ii)

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

$P(A) = {1 \over 4}\,or\,{1 \over 5}$

We have,


$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 0}^n {{1 \over n}} f\left( {{j \over n}} \right)$

By standard formula, we write

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 0}^n {{1 \over n}} f\left( {{j \over n}} \right) = \int\limits_0^1 {f(x)\,} dx$

Where, $\mathop {\lim }\limits_{n \to \infty } \left( {{j \over n}} \right)$ (as j = n) = 1

Given, equation

${x^{{{(\log 3x)}^2}}} - {9 \over 2}\log 3\,(x + 5) = 3\sqrt 3$

$\Rightarrow \left( {{{({{\log }_3}\,x)}^2} - {9 \over 2}{{\log }_3}x + 5} \right){\log _3}x = {3 \over 2}{\log _3}3$

Put log₃ x = t

$\therefore$ $\left( {{t^2} - {9 \over 2}t + 5} \right)t = {3 \over 2}$

$\Rightarrow 2{t^2} - 9{t^2} + 10t - 3 = 0$

$\Rightarrow (t - 3)(t - 1)\left( {t - {1 \over 2}} \right) = 0$

$\Rightarrow t = 3,\,1,\,{1 \over 2}$

$\Rightarrow {\log _3}x = 3,\,1,\,{1 \over 2}$

$\Rightarrow x = {3^3},\,{3^1},\,{3^{{1 \over 2}}}$

Given $\alpha = \widehat i + a\widehat j + {a^2}\widehat k,\,\beta = \widehat i + b\widehat j + {b^2}\widehat k$ and $\,\gamma = \widehat i + c\widehat j + {c^2}\widehat k$

$\therefore$ \left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| \ne 0

Now, \left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0

\left| {\matrix{ a & {{a^2}} & 1 \cr b & {{b^2}} & 1 \cr c & {{c^2}} & 1 \cr } } \right| + \left| {\matrix{ a & {{a^2}} & {{a^3}} \cr b & {{b^2}} & {{b^3}} \cr c & {{c^2}} & {{c^3}} \cr } } \right| = 0

\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right|(1 + abc) = 0

$\Rightarrow$ abc + 1 = 0\left[ {\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| \ne 0} \right]

$\Rightarrow$ abc = $- 1$

We have, ${x^2} + 2xy + a{y^2} = 0$ and $a{x^2} + 2xy + {y^2} = 0$ have common line.

y = x satisfies both lines.

$\therefore$ ${x^2} + 2{x^2} + a{x^2} = 0$

$3{x^2} + a{x^2} = 0 \Rightarrow a = - 3$

Equation of lines are

${x^2} + 2xy - 3{y^2} = 0$ and $- 3{x^2} + 2xy + {y^2} = 0$

$(x + 3y)(x - y) = 0$ and $(3x + y)(x - y) = 0$

The other lines are x + 3y = 0 and 3x + y = 0.

Given, f(x) = sin x + cos ax

sin x is periodic with 2$\pi$.

f(x) is periodic only on sin x and cos x both are periodic.

$\therefore$ cos ax is periodic at ${{2\pi } \over a}$.

LCM of ${2\pi }$ and ${{2\pi } \over a}$ is possible only a is rational number.

Note :
(1) When $n$ is odd then the period of ${\sin ^n}\theta$, ${\cos ^n}\theta$, ${\csc ^n}\theta$, ${\sec ^n}\theta$ = $2\pi$

(2) When $n$ is even then the period of ${\sin ^n}\theta$, ${\cos ^n}\theta$, ${\csc ^n}\theta$, ${\sec ^n}\theta$ = $\pi$

(3) When $n$ is even/odd then the period of ${\tan ^n}\theta$, ${\cot ^n}\theta$ = $\pi$

(3) When $n$ is even/odd then the period of $\left| {{{\sin }^n}\theta } \right|$, $\left| {{{\cos }^n}\theta } \right|$, $\left| {{{\csc }^n}\theta } \right|$, $\left| {{{\sec }^n}\theta } \right|$, $\left| {{{\tan }^n}\theta } \right|$, $\left| {{{\cot }^n}\theta } \right|$ = $\pi$

Intersecting points of both curves are $P( - 2,1)$ and $Q( - 2, - 1)$.


$\therefore$ Area of the bounded region

$= \int_{ - 1}^1 {((1 - 3{y^2}) - ( - 2{y^2}))\,dy}$

$= 2\int_0^1 {(1 - {y^2})\,dy}$

$= 2\left( {y - {{{y^3}} \over 3}} \right)_0^1$

$= 2\left( {1 - {1 \over 3}} \right) = {4 \over 3}$ sq unit