WBJEE 2017 Paper 1 (Mathematics) Question Paper with Solutions

We have,

${\tan ^{ - 1}}\left( {{{x - 1} \over {x - 2}}} \right) + {\tan ^{ - 1}}\left( {{{x + 1} \over {x + 2}}} \right) = {\pi \over 4}$

$\Rightarrow {\tan ^{ - 1}}\left[ {{{{{x - 1} \over {x - 2}} + {{x + 1} \over {x + 2}}} \over {1 - {{x - 1} \over {x - 2}}.{{x + 1} \over {x + 2}}}}} \right] = {\pi \over 4}$

$\Rightarrow {{(x - 1)(x + 2) + (x - 2)(x + 1)} \over {(x - 2)(x + 2) - (x - 1)(x + 1)}} = \tan {\pi \over 4}$

$\Rightarrow {{{x^2} + x - 2 + {x^2} - x - 2} \over {{x^2} - 4 - {x^2} + 1}} = 1$

$\Rightarrow {{2{x^2} - 4} \over { - 3}} = 1$

$\Rightarrow 2{x^2} - 4 = - 3$

$\Rightarrow 2{x^2} = 1$

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

$\Rightarrow x = \pm {1 \over {\sqrt 2 }}$

We have, \eqalign{ & 8x - 3y - 5z = 0 \cr & 5x - 8y + 3z = 0 \cr & 3x + 5y - 8z = 0 \cr}

$\therefore$ D = \left| {\matrix{ 8 & { - 3} & { - 5} \cr 5 & { - 8} & 3 \cr 3 & 5 & { - 8} \cr } } \right|

$= 8(64 - 15) + 3( - 40 - 9) - 5(25 + 24)$

$= 8 \times 49 + 3 \times ( - 49) - 5 \times 49$

$= 0$

$\therefore$ The system has infinitely many non-zero solutions.

All the orthogonal matrix are non-singular matrix.

$\therefore$ Q is proper subset of P.

We have,

$\overline x = {{{x_1} + {x_2} + ... + {x_q} + ... + {x_n}} \over n}$

$\Rightarrow \sum x = n\overline x$ ...(i)

If x_q is replaced by ${{x_q}'}$ then new total will be

$\sum {x'} = \sum x - {x_q} - {x_q}'$

$\therefore$ New mean will be

$\overline x ' = {{\sum {x'} } \over n}$

$\overline x ' = {{\sum x - {x_q} + {x_q}'} \over n}$

$\overline x ' = {{n\overline x - {x_q} + {x_q}'} \over n}$ [from Eq. (i)]

$(p + 1)(p + 2)(p + 3)...(p + q)$ is the product of q consecutive natural number $(p,q \in N)$. The product of q consecutive natural number is always divisible by q!.

Let, $f(x) = \sin x(\sin x + \cos x)$

$= {\sin ^2}x + \sin x\cos x$

$= {{1 - {{\cos }^2}2x} \over 2} + {{2\sin x\cos x} \over 2}$

$= {1 \over 2} - {1 \over 2}\cos 2x + {1 \over 2}\sin 2x$

$= {1 \over 2} + {1 \over 2}(\sin 2x - \cos 2x)$

Now, $- \sqrt {{{(1)}^2} + {{(1)}^2}} \le \sin 2x - \cos 2x \le \sqrt {{{(1)}^2} + {{(1)}^2}}$

[$\because$ $\sqrt {{a^2} + {b^2}} \le a\sin x + b\cos x \le \sqrt {{a^2} + {b^2}}$]

$\Rightarrow - \sqrt 2 \le \sin 2x - \cos 2x \le \sqrt 2$

$\Rightarrow - {{\sqrt 2 } \over 2} \le {{\sin 2x - \cos 2x} \over 2} \le {{\sqrt 2 } \over 2}$

$\Rightarrow {1 \over 2} - {{\sqrt 2 } \over 2} \le {1 \over 2} + {{\sin 2x - \cos 2x} \over 2} \le {1 \over 2} + {{\sqrt 2 } \over 2}$

$\Rightarrow {{1 - \sqrt 2 } \over 2} \le f(x) \le {{1 + \sqrt 2 } \over 2}$

$\therefore$ ${{1 - \sqrt 2 } \over 2} \le k \le {{1 + \sqrt 2 } \over 2}$

We have,

|A|\, = \left| {\matrix{ 1 & {\cos \theta } & 0 \cr { - \cos \theta } & 1 & {\cos \theta } \cr { - 1} & { - \cos \theta } & 1 \cr } } \right|

$= 1[1 - ( - \cos \theta )(\cos \theta )] - cos\theta [ - cos\theta + cos\theta ] + 0(co{s^2}\theta + 1)$

$= 1 + {\cos ^2}\theta$

Now, we know that

$- 1 \le \cos \theta \le 1$

$\Rightarrow 0 \le {\cos ^2}\theta \le 1$

$\Rightarrow 1 \le 1 + {\cos ^2}\theta \le 2$

$\Rightarrow 1 \le \,|A|\, \le 2$

$\therefore$ $|A|\, \in [1,2]$

In a non-leap year, total number of days is 365. Out of them, there are 52 weeks and 1 day extra. Thus, a non-leap year always has 52 Sunday. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

Out of these 7 cases, we have Sunday in one case.

$\therefore$ Total number of outcomes = 7

Number of favourable outcomes = 1

Hence, required probability = ${1 \over 7}$

Total number of 5-digit numbers having all the digits distinct = ${}^{10}{P_5} - {}^9{P_4}$

$= {{10!} \over {5!}} - {{9!} \over {5!}} = {{10 \times 9!} \over {5!}} - {{9!} \over {5!}}$

$= {{9!} \over {5!}}(10 - 1) = {{9!} \over {(9 - 4)!}}(9)$

$= 9 \times {}^9{P_4}$

Since, $f:R \to R$ is injective and $f(x)f(y) = f(x + y)$, $\forall x,y \in R$.

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

Again, f(x), f(y), f(z) are in GP.

$\Rightarrow {(f(y))^2} = f(x)\,.\,f(y)$

$\Rightarrow {a^{2y}} = {a^x}\,.\,{a^z}$

$\Rightarrow {a^{2y}} = {a^{x + z}}$

$\Rightarrow 2y = x + z$

$\therefore$ x, y, z are in AP.

For every real number x, x² $\ge$ 0

$\therefore$ (x, x) $\in$ P

Hence, P is reflexive.

Now, let (x, y) $\in$ P

$\Rightarrow$ xy $\ge$ 0

$\Rightarrow$ yx $\ge$ 0

$\Rightarrow$ (y, x) $\in$ P

Hence, P is symmetric.

Again, ($-$ 1, 0) $\in$ P and (0, 2) $\in$ P.

But ($-$ 1, 2) $\notin$ P

as ($-$ 1) (2) = $-$ 2 < 0

$\therefore$ P is not transitive.

${(1 + x + {x^2})^9} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{18}}{x^{18}}$

Put x = $-$ 1, we get

${(1 - 1 + 1)^9} = {a_0} - {a_1} + {a_2} + ... + {a_{18}}$

$\Rightarrow 1 = {a_0} - {a_1} + {a_2} + ... + {a_{18}}$ ....(i)

Put x = 1, we get

${(1 - 1 + 1)^9} = {a_0} - {a_1} + {a_2} + ... + {a_{18}}$

$\Rightarrow {3^9} = {a_0} + {a_1} + {a_2} + ... + {a_{18}}$ ....(ii)

On adding Eqs. (i) and (ii), we get

${3^9} + 1 = 2({a_0} + {a_2} + ... + {a_{18}})$

$\Rightarrow {a_0} + {a_2} + {a_4} + ... + {a_{18}} = {{{3^9} + 1} \over 2}$

$= {{19683 + 1} \over 2}$

$= {{19684} \over 2}$

= 9842, which is even number.

We have,

x$\rho$y $\Rightarrow$ x $-$ y is zero or irrational.

Now, x $-$ x = 0

$\Rightarrow$ (x, x) $\in$ $\rho$

$\therefore$ $\rho$ is reflexive.

Again, if x $-$ y is either zero or irrational, then y $-$ x will also be either zero or irrational.

$\Rightarrow$ (x, y) $\in$ $\rho$

$\Rightarrow$ (y, x) $\in$ $\rho$

$\therefore$ $\rho$ is symmetric.

Again, (2, $\sqrt 3$) $\in$ $\rho$ and ($\sqrt 3$, 4) $\in$ $\rho$

But (2, 4) $\notin$ $\rho$

$\therefore$ $\rho$ is not transitive.

We have,

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

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

$\therefore$ $IF = {e^{\int {{{{x^2} + 1} \over {x({x^2} - 1)}}dx} }}$

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

Let ${{{x^2} + 1} \over {x(x - 1)(x + 1)}} = {A \over x} + {B \over {x - 1}} + {C \over {x + 1}}$

$\Rightarrow {x^2} + 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)$

Put x = 0,

$\therefore$ 1 = $-$ A

$\Rightarrow$ A = $-$ 1

Put x = 1,

$\therefore$ 2 = 2B

$\Rightarrow$ B = 1

Put x = $-$ 1,

$\therefore$ 2 = 2C

$\Rightarrow$ C = 1

$\therefore$ ${{{x^2} + 1} \over {x(x - 1)(x + 1)}} = {{ - 1} \over x} + {1 \over {x - 1}} + {1 \over {x + 1}}$

$\therefore$ $IF = {e^{\int {\left( {{{ - 1} \over x} + {1 \over {x - 1}} + {1 \over {x + 1}}} \right)} \,dx}}$

$= {e^{[ - \log x + \log (x - 1) + \log (x + 1)]}}$

$= {e^{\log \left( {{{{x^2} - 1} \over x}} \right)}} = {{{x^2} - 1} \over x}$

$= x - {1 \over x}$

Let $x = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$

Then, $x \times \widehat i = - \beta \widehat k + \gamma \widehat j$

$x\widehat j = \widehat k - \gamma \widehat i$

$x \times \widehat k = - \alpha \widehat j + \beta \widehat i$

Now, ${(x \times \widehat i)^2} = (x \times \widehat i)\,.\,(x \times \widehat i)$

$= ( - \beta \widehat k + \gamma \widehat j)\,.\,( - \beta \widehat k + \gamma \widehat j)$

$= {\beta ^2} + {\gamma ^2}$

Similarly, ${(x \times \widehat j)^2} = {\alpha ^2} + {\gamma ^2}$

and ${(x \times \widehat k)^2} = {\alpha ^2} + {\beta ^2}$

$\therefore$ ${(x \times \widehat i)^2} + {(x \times \widehat j)^2} + {(x \times \widehat k)^2}$

$= {\beta ^2} + {\gamma ^2} + {\alpha ^2} + {\gamma ^2} + {\alpha ^2} + {\beta ^2}$

$= 2({\alpha ^2} + {\beta ^2} + {\gamma ^2}) = 2|x{|^2}$

Let $y = \mathop {\lim }\limits_{x \to 0} {(\sin x)^{2\tan x}}$

$\Rightarrow \log y = \mathop {\lim }\limits_{x \to 0} \,log{(\sin x)^{2\tan x}}$

$= 2\mathop {\lim }\limits_{x \to 0} \tan x\log \sin x$

$= 2\mathop {\lim }\limits_{x \to 0} {{\log \sin x} \over {\cot x}}$

$= 2\mathop {\lim }\limits_{x \to 0} {{{1 \over {\sin x}}.\cos x} \over { - \cos e{c^2}x}}$

$= 2\mathop {\lim }\limits_{x \to 0} ( - \sin x\cos x)$

$= 2 \times 0 = 0$

$\therefore$ $\log y = 0 \Rightarrow y = {e^0} = 1$

Given equations are $2{x^2} + 3xy + 4{y^2} + x + 18y + 25 = 0$ .... (i)

$2{x^2} + 3xy + 4{y^2} + 1 = 0$ ... (ii)

Let the origin be transferred to (p, q) axes being parallel to the previous axes; then the equation (i) becomes.

$2{(x' + p)^2} + 3(x' + p)(y' + q) + 4{(y' + q)^2} + (x' + p) + 18(y' + q) + 25 = 0$

$\Rightarrow 2{x^2} + 2{p^2} + 4x'p + 3x'y' + 3x'q + 3py' + 3pq + 4y{'^2} + 4{q^2} + 8y'q + x' + p + 18y' + 18q + 25 = 0$

$\Rightarrow 2x{'^2} + 4y{'^2} + 3x'y' + (4p + 3q + 1)x'1 + (3p + 8q + 18)y' + 2{p^2} + 3pq + 4{q^2} + p + 25 = 0$

From Eq. (ii) coefficient of x' and y' must be zero.

$\therefore$ $4p + 3q + 1 = 0$ .... (iii)

$3p + 8q + 18 = 0$ .... (iv)

By solving Eqs. (iii) and (iv), we get

p = 2, q = $-$ 3

$\mathop {\lim }\limits_{x \to 0} {{2f(x) - 3f(2x) + f(4x)} \over {{x^2}}}$

$= \mathop {\lim }\limits_{x \to 0} {{2f'(x) - 3f'(2x).2 + f'(4x).4} \over {2x}}$

$= \mathop {\lim }\limits_{x \to 0} {{f'(x) - 3f'(2x) + 2f'(4x)} \over x}$

$= \mathop {\lim }\limits_{x \to 0} {{f''(x) - 3f''(2x).2 + 2f''(4x).4} \over 1}$

$= \mathop {\lim }\limits_{x \to 0} f''(x) - 6f''(2x) + 8f''(4x)$

$= f''(0) - 6f''(0) + 8f''(0)$

$= k - 6k + 8k$ [$\because$ $f''(0) = k$]

$= 3k$

Let function f(x) = x² (x $-$ 1)

$\Rightarrow$ f'(x) = 3x² $-$ 2x

and f''(x) = 6x $-$ 2

Now, f(0) = f(1) = f'(0) = 0

Then, according to question,

at x = ${1 \over 3}$, f''(x) = 0

(i.e. c = 1/3 for some C$\in$R)

Let the coordinates of C be ($\alpha$, $\beta$).

$\therefore$ Coordinates of centroid

$= \left( {{{2 - 2 + \alpha } \over 3},{{ - 3 + 1 + \beta } \over 3}} \right)$

$= \left( {{\alpha \over 3},{{\beta - 2} \over 3}} \right)$

Since, centroid lie on 2x + 3y = 1.

$\therefore$ ${{2\alpha } \over 3} + 3\left( {{{\beta - 2} \over 3}} \right) = 1$

$\Rightarrow {{2\alpha } \over 3} + {{3\beta - 6} \over 3} = 1$

$\Rightarrow 2\alpha + 3\beta - 6 = 3 \Rightarrow 2\alpha + 3\beta = 9$

$\therefore$ Locus of point C will be 2x + 3y = 9.

We know that,

det (AB) = det (A) det (B)

$\therefore$ det (AB) = 0

$\Rightarrow$ det (A) . det (B) = 0

\Rightarrow \left| {\matrix{ {x + 2} & {3x} \cr 3 & {x + 2} \cr } } \right|\left| {\matrix{ x & 0 \cr 5 & {x + 2} \cr } } \right| = 0

$\Rightarrow \{ {(x + 2)^2} - 9x\} \{ x(x + 2) - 0\} = 0$

$\Rightarrow ({x^2} + 4x + 4 - 9x)x(x + 2) = 0$

$\Rightarrow x(x + 2)({x^2} - 5x + 4) = 0$

$\Rightarrow x(x + 2)(x - 1)(x - 4) = 0$

$\Rightarrow x = 0, - 2,1,4$

Let a_n be the general term of a GP whose first term is a and common ratio is r.

Now according to the question,

${a_n} = {a_{n + 1}} + {a_{n + 2}}$

$\Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}}$

$\Rightarrow {r^{n + 1}} = {r^n} + {r^{n + 1}}$

$\Rightarrow 1 = {{{r^n}} \over {{r^{n - 1}}}} + {{{r^{n + 1}}} \over {{r^{n - 1}}}}$

$\Rightarrow 1 = r + {r^2}$

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

$\Rightarrow r = {{ - 1 \pm \sqrt {{{(1)}^2} - 4(1)( - 1)} } \over {2(1)}}$

$= {{ - 1 \pm \sqrt {1 + 4} } \over 2} = {{ - 1 \pm \sqrt 5 } \over 2}$

Since, GP consists only positive terms

$\therefore$ $r = {{\sqrt 5 - 1} \over 2}$

Let $I = \int {\cos (\log x)dx}$

Put log x = t

$\Rightarrow x = {e^t}$

$\therefore$ $dx = {e^t}dt$

$\therefore$ $I = \int {{e^t}\cos t\,dt}$

$= {{{e^t}} \over {{1^2} + {1^2}}}[\cos t + \sin t] + C$

$\because$ $\left[ {\int {{e^{ax}}} \cos bx\,dx = {{{e^{ax}}} \over {{a^2} + {b^2}}}[a\cos bx + b\sin bx] + C} \right]$

$\Rightarrow I = {{{e^t}} \over 2}[\cos t + \sin t] + C$

$= {x \over 2}[\cos (\log x) + \sin (\log x] + C$

$\therefore$ $f(x) = {x \over 2}[\cos (\log x) + \sin (\log x)]$

Let (h, k) be any point on the line $7x - 9y + 10 = 0$, then $7h - 9k + 10 = 0$

$\Rightarrow 7h = 9k - 10$

$\Rightarrow h = {{9k - 10} \over 7}$ .... (i)

Now, perpendicular distance from point (h, k) to the line $3x + 4y = 5$ is ${d_1}$

${d_1} = {{3h + 4k - 5} \over {\sqrt {{3^2} + {4^2}} }}$

$\Rightarrow {d_1} = {{3h + 4k - 5} \over 5}$ .... (ii)

and perpendicular distance from (h, k) to the line $12x + 5y = 7$ is ${d_2}$

$\therefore$ ${d_2} = {{12h + 5k - 7} \over {\sqrt {{{12}^2} + {5^2}} }}$

$\Rightarrow {d_2} = {{12h + 5k - 7} \over {13}}$ ..... (iii)

Now, ${d_1} - {d_2} = {{3h + 4k - 5} \over 5} - {{12h + 5k - 7} \over {13}}$

$\Rightarrow {d_1} - {d_2}$

$= {{13(3h + 4k - 5) - 5(12h + 5k - 7)} \over {65}}$

$= {{39h + 52k - 65 - 60h - 25k + 35} \over {65}}$

$= {{ - 21h + 27k - 30} \over {65}}$

$= {{ - 21\left( {{{9k - 10} \over 7}} \right) + 27k - 30} \over {65}}$ [from Eq. (i)]

$= {{ - 27k + 30 + 27k - 30} \over {65}} = 0$

$\Rightarrow {d_1} - {d_2} = 0 \Rightarrow {d_1} = {d_2}$

Slope of SB, ${m_1} = {{b - 0} \over {0 - ae}} = - {b \over {ae}}$ and slope of S'B, ${m_2} = {{b - 0} \over {0 - ( - ae)}} = {b \over {ae}}$

Since, $\angle SBS'$ is a right angle.

$\therefore$ ${m_1}{m_2} = - 1$

$\Rightarrow {{ - b} \over {ae}} \times {b \over {ae}} = - 1$

$\Rightarrow {b^2} = {a^2}{e^2}$

$\Rightarrow {{{b^2}} \over {{a^2}}} = {e^2}$

$\Rightarrow 1 - {e^2} = {e^2}$

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

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

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

$\Rightarrow e = {1 \over {\sqrt 2 }}$ or $e = - {1 \over {\sqrt 2 }}$