1
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
2
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.
3
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)
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Morning Slot

If tanA and tanB are the roots of the quadratic equation, 3x2 $$-$$ 10x $$-$$ 25 = 0, then the value of 3 sin2(A + B) $$-$$ 10 sin(A + B).cos(A + B) $$-$$ 25 cos2(A + B) is :
A
$$-$$ 10
B
10
C
$$-$$ 25
D
25

Explanation

As tan A and tan B are the roots of 3x2 $$-$$ 10x $$-$$ 25 = 0,

So, tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$

= $${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$$ = $${{10/3} \over {28/3}}$$ = $${5 \over {14}}$$

Now, cos2 (A + B) = $$-$$ 1 + 2 cos2 (A + B)

= $${{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,$$ $$ \Rightarrow $$ cos2(A + B) = $${{196} \over {221}}$$

$$\therefore\,\,\,$$ 3sin2(A + B) $$-$$ 10sin(A + B)cos(A + B) $$-$$ 25 cos2(A + B)

= cos2(A + B) [ 3tan2(A + B) $$-$$ 10tan(A + B) $$-$$ 25]

= $${{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}$$

= $$-$$ $${{5525} \over {196}}$$ $$ \times $$ $${{196} \over {221}}$$ = $$-$$ 25

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