WBJEE 2019 Paper 1 (Mathematics) Question Paper with Solutions

Given equation of ellipse is

$3{x^2} + 4{y^2} = 48$

$\Rightarrow {{3{x^2}} \over {48}} + {{4{y^2}} \over {48}} = 1 \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over {12}} = 1$

Here, $a = 4$ and $b = 2\sqrt 3$

$\therefore$ $e = \sqrt {1 - {{{b^2}} \over {{a^2}}}}$

$= \sqrt {1 - {{12} \over {16}}} = \sqrt {{4 \over {16}}} = \sqrt {{1 \over 4}} = {1 \over 2}$

$\therefore$ Coordinates of P are = (2, 3)

$\because$ $(4\cos \theta ,2\sqrt 3 \sin \theta ) = (2,3)$ $\because$ $\left[ {p = \left( {ae,{{{b^2}} \over a}} \right)} \right]$

By comparing, we get

$4\cos \theta = 2$ and $2\sqrt 3 \sin \theta = 3$

$\Rightarrow \cos \theta = {1 \over 2}$ and $\sin \theta = {{\sqrt 3 } \over 2}$

$\Rightarrow \cos \theta = \cos {\pi \over 3}$ and $\sin \theta = \sin {\pi \over 3}$

$\Rightarrow \theta = {\pi \over 3}$ and $\theta = {\pi \over 3}$

$\therefore$ eccentric angle of P is ${\pi \over 3}$.

We have,

y = x + 1 and y = cos x


$\therefore$ Required area = $\int_{ - 1}^0 {(x + 1)dx + \int_0^{\pi /2} {\cos xdx} }$

$= \left[ {{{{x^2}} \over 2} + x} \right]_1^0 + [\sin x]_0^{\pi /2}$

$= \left[ {(0) - \left( {{1 \over 2} - 1} \right)} \right] + [1 - 0]$

$= {1 \over 2} + 1 = {3 \over 2}$ sq unit.

Let V be the volume of spherical balloon of radius r.

Then, $V = {4 \over 3}\pi {r^3}$

$\Rightarrow \log V = \log \left( {{{4\pi } \over 3}} \right) + 3\log r \Rightarrow {1 \over V}{{dV} \over {dr}} = 0 + {3 \over r}$

$\Rightarrow {1 \over V}{{dV} \over {dr}} = {3 \over r} \Rightarrow {1 \over V}{{\Delta V} \over {\Delta r}} = {3 \over r}$

$\Rightarrow {{\Delta V} \over V} = 3{{\Delta r} \over r}$

$\Rightarrow {{\Delta V} \over V} \times 100 = 3{{\Delta r} \over r} \times 100$

$\Rightarrow {{\Delta V} \over V} \times 100 = 3 \times 0.1 = 0.3$

$\therefore$ Percentage increase in volume is 0.3%.

Given that f : [1, 3] $\to$ R be a continuous and differentiable in (1, 3).

and f'(x) = | f(x) |² + 4

By applying LMVT, there exist at least one point c$\in$(1, 3) such that

${{f(3) - f(1)} \over {3 - 1}} = f'(c)$ $\because$ $\left[ {f'(c) = {{f(b) - f(a)} \over {b - a}}} \right]$

$\Rightarrow {{f(3) - f(1)} \over 2} = f'(c)$

$\Rightarrow {{f(3) - f(1)} \over 2} = [f|c{|^2} + 4]$

($\because$ $f'(c) = |f(c){|^2} + 4$)

$\Rightarrow f(3) - f(1) = 2\,.\,|f(c){|^2} + 8$

$\Rightarrow f(3) - f(1) \ge 8$

It is clear from the given options that (B) and (C) the correct.

We have,

x₁, x₂ be the roots of equation ${x^2} - 3x + a = 0$

$\therefore$ ${x_1} + {x_2} = 3$ and ${x_1}{x_2} = a$

Also, x₃, x₄ be the roots of equation ${x^2} - 12x + b = 0$

$\therefore$ ${x_3} + {x_4} = 12$ and ${x_3}{x_4} = b$

Again, x₁, x₂, x₃, x₄ are in GP

$\therefore$ ${x_1} = A,{x_2} = AR,{x_3} = A{R^2},{x_4} = A{R^3}$

Now, ${x_1} + {x_2} = 3$

$\Rightarrow A(1 + R) = 3$ ....(i)

and ${x_3} + {x_4} = 12$

$\Rightarrow A{R^2}(1 + R) = 12$ .... (ii)

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

${R^2} = 4 \Rightarrow R = \pm 2$

When, R = 2, A = 1 and when, R = $-$ 2, A = $-$ 3

$\therefore$ Numbers are either 1, 2, 4, 8 or $-$3, 6, $-$12, 24

But ${x_1} < {x_2} < {x_3} < {x_4}$,

So, ${x_1} = 1,{x_2} = 2,{x_3} = 4,{x_4} = 8$

$\therefore$ $ab = {x_1}{x_2}{x_3}{x_4}$

$= 1 \times 2 \times 4 \times 8 = 64$

We observe the following properties :

Reflexivity : Let a be an arbitrary element of R, then a $\in$ R

$\Rightarrow$ 1 + a . a = 1 + a² > 0 [$\because$ a² > 0 for all a $\in$ R]

$\Rightarrow$ (a, a) $\in$R₁ [By def. of R₁]

Thus, (a, a) $\in$R₁ for all a$\in$R. So, R₁ is reflexive on R.

Symmetry : Let (a, b) $\in$ R

Then, (a, b) $\in$ R₁

$\Rightarrow$ 1 + ab > 0 [$\because$ ab = b for all]

$\Rightarrow$ 1 + ba > 0 $\Rightarrow$ (b, a) $\in$ R₁

Thus, (a, b) $\in$ R₁ $\ne$ (b, a) $\in$ R, for all a, b $\in$R

So, R₁ is symmetric on R.

Transitivity : we observe that $\left( {1,{1 \over 2}} \right) \in {R_1}$ and $\left( {{1 \over 2}, - 1} \right) \in {R_1}$ but (1, $-$1) $\notin$ R₁ because

1 + 1 $\times$ ($-$1) = 0 > 0

So, R₁ is not transitive on R.

Equation of the ellipse is

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$


Foci are S(ae, 0), T($-$ae, 0).

B(0, b) is the end of the minor axis.

STB is an equilateral triangle.

SB = ST

$\Rightarrow$ SB² = ST²

$\Rightarrow$ a²e² + b² = 4a²e²

$\Rightarrow$ b² = 3a²e²

$\Rightarrow$ a²(1 $-$ e²) = 3a²e²

$\Rightarrow$ 1 $-$ e² = 3e²

$\Rightarrow$ 4e² = 1

$\Rightarrow$ e² = ${1 \over 4}$

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

Eccentricity of the ellipse, e = ${1 \over 2}$

Let $L = \mathop {\lim }\limits_{x \to {0^ + }} {({e^x} + x)^{1/x}}$

$\Rightarrow \log L = \mathop {\lim }\limits_{x \to {0^ + }} {{\log ({e^x} + x)} \over x}$

$\Rightarrow \log L = \mathop {\lim }\limits_{x \to {0^ + }} {{{1 \over {({e^x} + x)}}.\,{e^x} + 1} \over 1}$ [using L' Hospital rule]

$\Rightarrow \log L = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^x} + 1} \over {{e^x} + x}}$

$\Rightarrow \log L = 2$

$\Rightarrow L = {e^2}$

Given,

$|\widehat \alpha |\, = \,| \widehat \beta |\, = \,| \widehat\gamma |\, = 1$

and $\widehat \alpha \, \times \,(\widehat \beta \times \widehat \gamma ) = {1 \over 2}(\widehat \beta + \widehat \gamma )$

Now, $\widehat \alpha \, \times \,(\widehat \beta \times \widehat \gamma ) = {1 \over 2}(\widehat \beta + \widehat \gamma )$

$\Rightarrow (\widehat \alpha \,.\,\widehat \gamma )\widehat \beta - (\widehat \alpha \,.\,\widehat \beta )\widehat \gamma = {1 \over 2}\widehat \beta + {1 \over 2}\widehat \gamma$

On comparing we get,

$- (\widehat \alpha \,.\,\widehat \beta ) = {1 \over 2}$

$\Rightarrow |\widehat \alpha ||\widehat \beta |\cos \theta = - {1 \over 2}$

$\Rightarrow \cos \theta = - {1 \over 2}$ [$\because$ $|\widehat \alpha |\, = \,|\widehat \beta |\, = \,1$]

$\Rightarrow \theta = 120^\circ$ or ${{2\pi } \over 3}$

${(x + y)^2}{{dy} \over {dx}} = {a^2},a \ne 0$

Let $x + y = t$

$1 + {{dy} \over {dx}} = {{dt} \over {dx}}$

$\Rightarrow {{dy} \over {dx}} = {{dt} \over {dx}} - 1$

Therefore, ${t^2}\left( {{{dt} \over {dx}} - 1} \right) = {a^2}$

$\Rightarrow {t^2}{{dt} \over {dx}} - {t^2} = {a^2}$

$\Rightarrow {t^2}{{dt} \over {dx}} = {a^2} + {t^2}$

$\Rightarrow {{{t^2}} \over {{a^2} + {t^2}}}dt = dx$ [variable separation]

$\Rightarrow \int {{{{t^2}} \over {({t^2} + {a^2})}}} dt = \int {dx}$ [integration]

$\Rightarrow \int {{{{t^2} + {a^2} - {a^2}} \over {({t^2} + {a^2})}}} dt = x + C'$ [doing a² adding and subtracting]

$\Rightarrow \int {\left( {1 - {{{a^2}} \over {{t^2} + {a^2}}}} \right)dt = x + C'}$

$\Rightarrow \int {dt - \int {{{{a^2}} \over {{t^2} + {a^2}}}dt = x + C'} }$

$\Rightarrow t - {a^2}\int {{{dt} \over {{t^2} + {a^2}}} = x + C'}$

$\Rightarrow t - {a^2}\left[ {{1 \over a}{{\tan }^{ - 1}}\left( {{t \over a}} \right)} \right] = x + C'$

$\Rightarrow (x + y) - a{\tan ^{ - 1}}\left( {{{x + y} \over a}} \right) = x + C'$ [on putting the value of t]

$\Rightarrow {{y - C'} \over a} = {\tan ^{ - 1}}\left( {{{x + y} \over a}} \right)$

$\Rightarrow {{x + y} \over a} = \tan \left( {{{y - C'} \over a}} \right)$

$\Rightarrow {{x + y} \over a} = \tan \left( {{{y + C} \over a}} \right)$ [ where, C = $-$C' is an arbitrary constant]

$\Rightarrow \tan \left( {{{y + C} \over a}} \right) = {{x + y} \over a}$

Since, adj (M') = (adj M)'

= adj (M') $-$ (adj M)'

[$\because$ adj (A') = (adj A)']

= 0, a null matrix

We have,

${{dx} \over {dt}} = 1$ and $y = {{10} \over x}$

Now, $y = {{10} \over x}$

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

$\Rightarrow {{dy} \over {dt}} = {{ - 10} \over {{x^2}}} \times {{dx} \over {dt}}$

$= {{ - 10} \over {{{(5)}^2}}} \times 1$

$= {{ - 10} \over {25}} = {{ - 2} \over 5}$ $\left[ {x = 10,\,{{dx} \over {dt}} = 1} \right]$

$\therefore$ y decreases at the rate of ${2 \over 5}$ unit per second.

Given,

h{f(x)} = x

differentiating w.r.t.x, we get

h'{f(x)} . f'(x) = 1

$\Rightarrow$ $f'(x) = {1 \over {h'\{ f(x)\} }}$

$\Rightarrow f'(x) = {1 \over {{1 \over {1 + \log \{ f(x)\} }}}}$

$\because$ $\left[ {h'(x) = {1 \over {1 + \log x}}} \right]$ (given)

$\Rightarrow f'(x) = 1 + \log (f(x))$

Let $(\alpha ,\beta )$ be the given point, let Q(x, y) be the foot of the perpendicular, and let O be the origin. The line can have any direction

$\angle PQO = 90^\circ$

Point Q lies on the circle having diameter OP. The locus of point Q.

$(x - 0)(x - \alpha ) + (y - 0)(y - \beta ) = 0$

$\Rightarrow {x^2} - x\alpha + {y^2} - y\beta = 0$

$\Rightarrow {x^2} + {y^2} - \alpha x - \beta y = 0$

Let $f(x) = {x^2} + 2x + 3$

$a = f{(x)_{\min }} = {{ - D} \over {4a}} = {{ - (4 - 12)} \over 4} = {8 \over 4} = 2$

and $b = \mathop {\lim }\limits_{\theta \to 0} {{1 - \cos \theta } \over {{\theta ^2}}} = \mathop {\lim }\limits_{\theta \to 0} {{1 - 1 + 2{{\sin }^2}\theta /2} \over {{\theta ^2}}}$

$= \mathop {\lim }\limits_{\theta \to 0} {{2{{\sin }^2}\theta /2} \over {{{(\theta /2)}^2}\,.\,4}} = \mathop {\lim }\limits_{\theta \to 0} {1 \over 2}\,.\,\left[ {{{{{\sin }^2}\theta /2} \over {{{(\theta /2)}^2}}}} \right]$

$= {1 \over 2}\,.\,\mathop {\lim }\limits_{\theta \to 0} {{{{\sin }^2}\theta /2} \over {{{(\theta /2)}^2}}}$

$= {1 \over 2}\,.\,1 = {1 \over 2}$

$b = {1 \over 2}$

Now, $\sum\limits_{r = 0}^n {{a^r}{b^{n - r}}}$

$= \sum\limits_{r = 0}^n {{{(2)}^r}{{\left( {{1 \over 2}} \right)}^{n - r}}}$

$= \sum\limits_{r = 0}^n {{2^r}\,.\,{2^{(r - n)}}}$

$= \sum\limits_{r = 0}^n {{2^{2r - n}}}$

$= {2^{ - n}}\sum\limits_{r = 0}^n {{2^{2r}}}$

$= {2^{ - n}}[1 + {2^2} + {2^4} + {2^6} + ... + {2^{2n}}]$

$= {2^{ - n}}\left[ {{{1\,.\,{{({2^2})}^{n + 1}} - 1} \over {{2^2} - 1}}} \right]$ $\because$ $\left\{ {{S_n} = {{a({r^n} - 1)} \over {r - 1}}} \right\}$

$= {2^{ - n}}\left[ {{{{4^{n + 1}} - 1} \over 3}} \right]$

$= {{{4^{n + 1}} - 1} \over {3\,.\,{2^n}}}$

Let $I = \int {{2^{{2^x}}}.\,{2^x}dx}$

Put $t = {2^x}$

$\Rightarrow dt = {2^x}\log 2dx$

$\therefore$ $I = \int {{{{2^t}} \over {\log 2}}dt}$

$= {{{2^t}} \over {{{(\log 2)}^2}}} + C = {{{2^{{2^x}}}} \over {{{(\log 2)}^2}}} + C$

$\therefore$ $A = {1 \over {{{(\log 2)}^2}}}$

Given, ${\left( {{3^{{1 \over 8}}} + {5^{{1 \over 4}}}} \right)^{84}}$

here, ${T_{r + 1}} = {}^{84}{C_r}{\left( {{3^{{1 \over 8}}}} \right)^{84 - r}}{\left( {{5^{{1 \over 4}}}} \right)^r}$

$= {}^{84}{C_r}\left( {{3^{{{84 - r} \over 8}}}} \right)\,.\,{5^{{r \over 4}}}$

Since, ${T_{r + 1}}$ is rational for r = 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84

$\therefore$ no. of rational terms = 11

Here, total number of terms = 85

$\therefore$ Number of irrational terms = 85 $-$ 11 = 74

Let $f(x) = a{x^2} + bx + c$

$\Rightarrow f(1) = a + b + c < 0$

Again, f(x) has imaginary zeroes. So, a < 0.

Also, f(0) = c. Since f(x) is downward parabola. So, c < 0.

We have,

7²ⁿ + 16n $-$ 1

Put n = 1, we get

7² + 16 $-$ 1

= 49 + 15

= 64

which is divisible by 64.

We have,

$(\cos \theta + i\sin \theta )(\cos 2\theta + i\sin 2\theta )...(\cos n\theta + i\sin n\theta ) = 1$

$\Rightarrow {e^{i\theta }}\,.\,{e^{i(2\theta )}}\,.\,{e^{i(3\theta )}}\,.\,...\,{e^{i(n\theta )}} = 1$

$\Rightarrow {e^{i\theta (1 + 2 + 3 + ... + n)}} = 1$

$\Rightarrow {e^{{{i\,n(n + 1)\theta } \over 1}}} = 1$

$\Rightarrow \cos \left( {{{n(n + 1)} \over 2}\theta } \right) + i\sin \left( {{{n(n + 1)} \over 2}} \right)\theta = 1 + 0i$

$\Rightarrow \cos \left( {{{n(n + 1)} \over 2}\theta } \right) = 1$

$\Rightarrow {{n(n + 1)} \over 2}\theta = 2k\pi$

$\Rightarrow \theta = {{4k} \over {n(n + 1)}}\pi$

Let the equation of line be ${x \over a} + {y \over b} = 1$

Since, the line passing through a fixed point $({x_1},{y_1})$.


$\therefore$ ${{{x_1}} \over a} + {{{y_1}} \over b} = 1$

Since, OAPB is a rectangle, therefore the coordinate of P will be (a, b).

Hence, locus of P is

${{{x_1}} \over x} + {{{y_1}} \over y} = 1$

Let $I = \int {\cos \left\{ {2{{\tan }^{ - 1}}\sqrt {{{1 - x} \over {1 + x}}} } \right\}} dx$

Put $x = \cos 2\theta \Rightarrow dx = - 2\sin 2\theta d\theta$

$\therefore$ $I = \int {\cos \left\{ {2{{\tan }^1}\sqrt {{{1 - \cos 2\theta } \over {1 + \cos 2\theta }}} } \right\}} ( - 2\sin 2\theta )d\theta$

$= - 2\int {\cos \left\{ {2{{\tan }^{ - 1}}\sqrt {{{2{{\sin }^2}\theta } \over {2{{\cos }^2}\theta }}} } \right\}} \sin 2\theta d\theta$

$= - 2\int {\cos (2{{\tan }^{ - 1}}(\tan \theta )\} \sin 2\theta d\theta }$

$= - 2\int {\cos 2\theta \sin 2\theta d\theta }$

$= \int {\cos 2\theta d(\cos 2\theta )}$

$= {{{{\cos }^2}2\theta } \over 2} + C$

$= {{{x^2}} \over 2} + C$

Which is an equation of parabola.

The position vectors of the points A, B, C and D are $3\widehat i - 2\widehat j - \widehat k$, $2\widehat i - 3\widehat j + 2\widehat k$, $5\widehat i - \widehat j + 2\widehat k$ and $4\widehat i - \widehat j - \lambda \widehat k$ respectively.

$BA = (3\widehat i - 2\widehat j - \widehat k) - (2\widehat i - 3\widehat j + 2\widehat k) = \widehat i + \widehat j - 3\widehat k$

$CA = (3\widehat i - 2\widehat j - \widehat k) - (5\widehat i - \widehat j + 2\widehat k) = - 2\widehat i - \widehat j - 3\widehat k$

$DA = (3\widehat i - 2\widehat j - \widehat k) - (4\widehat i - \widehat j + \lambda \widehat k)$

$= - \widehat i - \widehat j - (1 + \lambda )\widehat k$

These are coplanar, so

\left| {\matrix{ 1 & 1 & { - 3} \cr { - 2} & { - 1} & { - 3} \cr { - 1} & { - 1} & { - (1 + \lambda )} \cr } } \right| = 0

= \left| {\matrix{ { - 1} & { - 1} & 3 \cr 2 & 1 & 3 \cr 1 & 1 & {1 + \lambda } \cr } } \right| = 0

$- 1[1 + \lambda - 3] + 1[2 + 2\lambda - 3] + 3[2 - 1] = 0$

$\Rightarrow - 1[\lambda - 2] + 1[2\lambda - 1] + 3 = 0$

$\Rightarrow - \lambda + 2 + 2\lambda - 1 + 3 = 0$

$\Rightarrow \lambda + 4 = 0$

$\Rightarrow \lambda = - 4$

Let the length of conjugate axis = b

and the length of transvers axis = a

Given that, b > a

Now, b² > a²

$\therefore$ Eccentricity (e) = $\sqrt {1 + {{{b^2}} \over {{a^2}}}}$

${e^2} = {\left( {\sqrt {1 + {{{b^2}} \over {{a^2}}}} } \right)^2} > 2$

$\therefore$ $e > \sqrt 2$

Since, product of two non-null matrix can be a null matrix.

Therefore, may be A $\ne$ O₃, B $\ne$ O₃.

$\therefore$ arg(z) $-$ arg($-$ z) = $\theta$ $-$ ($-$ $\pi$ + $\theta$) = $\pi$