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### JEE Main 2017 (Online) 8th April Morning Slot

Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$-$ 1 and it leaves remainder 6 when divided by x + 1; then :
A
p(2) = 11
B
p(2) = 19
C
p($-$ 2) = 19
D
p($-$ 2) = 11

## Explanation

Let, P(x) = ax2 + bx + c

As, P(0) = 1,

$\therefore\,\,\,$ a(0)2 + b(0) + c = 1

$\Rightarrow $$\,\,\, c = 1 \therefore\,\,\, P(x) = ax2 + bx + 1 If P(x) is divided by x - 1, remainder = 4 \Rightarrow$$\,\,\,$ P$\left( 1 \right) = 4$

$\therefore\,\,\,$ a + b + 1 = 4 . . . . . (1)

If    P(x) is divided by x + 1, remainder = 6

$\Rightarrow $$\,\,\, P(- 1) = 6 \therefore\,\,\, a - b + 1 = 6 . . . .(2) By solving (1) and (2) we get, a = 4, and b = -1 \therefore\,\,\, P(x) = 4x2 - x + 1 P(2) = 4(2)2 - 2 + 1 = 15 P(- 2) = 4 (-2)2 - (- 2) + 1 = 19 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot The sum of all the real values of x satisfying the equation 2(x-1)(x2 + 5x - 50) = 1 is : A 16 B 14 C -4 D - 5 ## Explanation We know, 2x = 1 only when x = 0. Similarly, 2(x-1)(x2 + 5x - 50) = 1 when (x-1)(x2 + 5x - 50) = 0 \Rightarrow (x - 1)(x + 10)(x - 5) = 0 \therefore x = 1, -10, 5 Sum of real values of x = 1 + (-10) + 5 = -4 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) Let S = { x \in R : x \ge 0 and 2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0}. Then S A contains exactly four elements B is an empty set C contains exactly one element D contains exactly two elements ## Explanation Given, 2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 Case 1 : When \sqrt x - 3 \ge 0, then equation becomes 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0 \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0 \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0 \therefore\,\,\, \sqrt x = 0,4 but as \sqrt x - 3 \ge 0 or \sqrt x \ge 3 then \sqrt x \ne 0 \therefore\,\,\, \sqrt x = 4 value is possible. Case 2 : When \sqrt x - 3 < 0 or \sqrt x < 3. There equation becomes - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0 \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0 \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0 \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0 \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0 \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0 \therefore\,\,\, \sqrt x = 2,6 as \sqrt x < 3 so \sqrt x \ne 6 \therefore\,\,\, \sqrt x = 2 is possible. So, total possible value of \sqrt x = 2,4 or for x possible values are 4, 16. \therefore\,\,\, Set S contains exactly two elements 4 and 16. 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot If \lambda \in R is such that the sum of the cubes of the roots of the equation, x2 + (2 - \lambda ) x + (10 - \lambda ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is : A 4\sqrt 2 B 2\sqrt 5 C 2\sqrt 7 D 20 ## Explanation Let \alpha , \beta are the roots of the equation, \therefore \alpha + \beta = \lambda - 2 and \alpha$$\beta$ = 10 $-$ $\lambda$

${\alpha ^3} + {\beta ^3}$ = ($\alpha$ + $\beta$)3 $-$ 3$\alpha$$\beta$ ($\alpha$ + $\beta$)

= ($\lambda$ $-$ 2)3 $-$ 3(10 $-$ $\lambda$)($\lambda$ $-$ 2)

= $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda$ + 52

Let $f(\lambda$) = $\lambda ^3$ $-$ 3$\lambda ^2$ $-$ 24$\lambda$ + 52

$\therefore$ ${{df(\lambda )} \over {d\lambda }}$ = 3$\lambda ^2$ $-$ 6$\lambda$ $-$ 24

$\therefore$ at maximum of minimum ${{df(\lambda )} \over {d\lambda }}$ = 0

$\therefore$ $\lambda ^2$ $-$ 2$\lambda$ $-$ 8 = 0

$\Rightarrow$ ($\lambda$ + 2) ($\lambda$ $-$ 4) = 0

$\Rightarrow$ $\lambda$ = $-$2, 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 2$\lambda$ $-$ 2

When $\lambda$ = $-$2

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = $-$ 6 < 0

$\therefore$ at $\lambda$ = $-$2, f($\lambda$) has maximum value.

When $\lambda$ = 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 6 > 0

$\therefore$ at $\lambda$ = 4, f($\lambda$) has minimum value.

$\therefore$ When $\lambda$ = 4 equation is,

x2 $-$ 2x + 6 = 0

$\therefore$ ($\alpha$ $-$ $\beta$)2 = ($\alpha$ + $\beta$)2 $-$ 4$\alpha \beta$

$\Rightarrow$ x2 $-$ 4 $\times$ 6

= $-$ 20

$\Rightarrow$ ($\alpha$ $-$ $\beta$) = $2\sqrt 5 i$

$\Rightarrow$ $\left| {\alpha - \beta } \right|$ = $2\sqrt 5$ (ans)