question stringlengths 79 9.83k | answer stringlengths 33 9.39k |
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<p>Let $$A = \{ x \in R:|x + 1| < 2\} $$ and $$B = \{ x \in R:|x - 1| \ge 2\} $$. Then which one of the following statements is NOT true?</p>
Options:
[{"identifier": "A", "content": "$$A - B = ( - 1,1)$$"}, {"identifier": "B", "content": "$$B - A = R - ( - 3,1)$$"}, {"identifier": "C", "content": "$$A \\cap B = ( ... | ["B"]
Explanation:
<p>A = ($$-$$3, 1) and B = ($$-$$ $$\infty$$, $$-$$1] $$\cup$$ [3, $$\infty$$)</p>
<p>So, A $$-$$ B = ($$-$$1, 1)</p>
<p>B $$-$$ A = ($$-$$ $$\infty$$, $$-$$3] $$\cup$$ [3, $$\infty$$) = R $$-$$ ($$-$$3, 3)</p>
<p>A $$\cap$$ B = ($$-$$3, $$-$$1]</p>
<p>and A $$\cup$$ B = ($$-$$ $$\infty$$, 1) $$\cup... |
<p>$$
\text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $$<br/><br/>$$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }
$$</p>
<p>Then the number of elements in $$\mathrm{S} \cap \mathrm{T}$$ is :</p>
Options:
[{"identifier": "A", "content": "7"}, {"identif... | ["D"]
Explanation:
<p>$$|{x^2}| - 7|x| + 9 \le 0$$</p>
<p>$$ \Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]$$</p>
<p>As $$x \in Z$$</p>
<p>So, x can be $$ \pm \,2, \pm \,3, \pm \,4, \pm \,5$$</p>
<p>Out of these values of x,</p>
<p>$$x = 3, - 4, - 5$$</p>
<p>satisfy S as w... |
<p>Let $$\lambda \in \mathbb{R}$$ and let the equation E be $$|x{|^2} - 2|x| + |\lambda - 3| = 0$$. Then the largest element in the set S = {$$x+\lambda:x$$ is an integer solution of E} is ______</p>
Options:
[] | 5
Explanation:
$D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0$
<br/><br/>
$|\lambda-3| \leq 1$
<br/><br/>
$-1 \leq \lambda-3 \leq 1$
<br/><br/>
$2 \leq \lambda \leq 4$
<br/><br/>
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
<br/><br/>
$=1 \pm \sqrt{1-|\lambda-3|}$
<br/><br/>
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
... |
The number of real roots of the equation $x|x|-5|x+2|+6=0$, is :
Options:
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["B"]
Explanation:
$$
\begin{aligned}
& x|x|-5|x+2|+6=0 \\\\
& \text { Case-I : } \\\\
& \text { When } x<-2 \text { then } \\\\
& -x^2+5(x+2)+6=0 \\\\
& \Rightarrow x^2-5 x-16=0 \\\\
& \Rightarrow x=\frac{5 \pm \sqrt{25+64}}{2} \\\\
& \therefore x=\frac{5-\sqrt{89}}{2} \text { is accepted }
\end{aligned}
$$
<br/><... |
<p>The set of all $$a \in \mathbb{R}$$ for which the equation $$x|x-1|+|x+2|+a=0$$ has exactly one real root, is :</p>
Options:
[{"identifier": "A", "content": "$$(-\\infty, \\infty)$$"}, {"identifier": "B", "content": "$$(-6, \\infty)$$"}, {"identifier": "C", "content": "$$(-\\infty,-3)$$"}, {"identifier": "D", "cont... | ["A"]
Explanation:
$$
x|x-1|+|x+2|+a=0
$$
<br/><br/>Case I : If $x<-2$ then
<br/><br/>$$
-x^2+x-x-2+a=0
$$
<br/><br/>$$
a=x^2+2
$$
<br/><br/>$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$
<br/><br/>Case II : If $-2 \leq x<1$ then
<br/><br/>$$
\begin{aligned}
& -x^2+x+x+2+a=0 \\\\
& a=x^2-2 x-2
\end{aligned}
$$
<b... |
<p>The number of real solutions of the equation $$x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0$$ is _________.</p>
Options:
[] | 1
Explanation:
<p>The given equation is $$x(x^2+3|x|+5|x-1|+6|x-2|)=0$$, which can be solved by analyzing it in parts. It can be broken down into: $$x=0$$ and $$x^2+3|x|+5|x-1|+6|x-2|=0$$. </p><p>For $$x=0$$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.</p><b>Cas... |
<p>The number of distinct real roots of the equation $$|x+1||x+3|-4|x+2|+5=0$$, is _______</p>
Options:
[] | 2
Explanation:
<p>Let's analyze the equation $ |x+1||x+3|-4|x+2|+5=0 $ based on different intervals of $ x $.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4nlbkg/60bbafa8-f162-4063-8d33-6f852bb91330/ef943d00-10fb-11ef-aaa0-17ca36a32505/file-1lw4nlbkh.png?format=png" data-orsrc="https://ap... |
<p>The number of distinct real roots of the equation $$|x||x+2|-5|x+1|-1=0$$ is __________.</p>
Options:
[] | 3
Explanation:
<p>$$|x| \quad|x+2|-5|x+1|-1=0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lwgkotns/48dc3e18-87b2-4fd8-a3b9-f30ede35fcf3/fd635a80-1789-11ef-b185-b7bdfa18e39b/file-jaoe38c1lwgkotnt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lwgkotns... |
<p>The number of real solutions of the equation $$x|x+5|+2|x+7|-2=0$$ is __________.</p>
Options:
[] | 3
Explanation:
<p>$$x|x+5|+2|x+7|-2=0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71afe90ed53/c1552330-16a2-11ef-afc3-d53e649859cd/file-1lwexk63o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71a... |
If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals
Options:
[{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["D"]
Explanation:
<p>Let n and (n + 1) be the roots of x<sup>2</sup> $$-$$ bx + c = 0.</p>
<p>Then, n + (n + 1) = b and n(n + 1) = c</p>
<p>$$\therefore$$ b<sup>2</sup> $$-$$ 4c = (2n + 1)<sup>2</sup> $$-$$ 4n(n + 1)</p>
<p>= 4n<sup>2</sup> + 4n + 1 $$-$$ 4n<sup>2</sup> $$-$$ 4n = 1</p> |
If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals
Options:
[{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$1$$"}] | ["D"]
Explanation:
Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots
<br><br>Then $$\alpha + \alpha + 1 = b = $$ sum of -
<br><br>roots $$\alpha \left( {\alpha + 1} \right) = c$$
<br><br>$$=$$ product of roots
<br><br>$$\therefore$$ $${b^2} - 4c $$
<br><br>$$ = {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alph... |
If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is :
Options:
[{"identifier": "A", "content": "less than $$4ab$$ "}, {"identifier": "B", "content": "greater than $$-4ab$$"}, {"identifier": "C", "content": "less than $$-4a... | ["B"]
Explanation:
Given that roots of the equation
<br><br>$$b{x^2} + cx + a = 0$$ are imaginary
<br><br>$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$
<br><br>$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$
<br><br>As $... |
The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has:
Options:
[{"identifier": "A", "content": "infinite number of real roots "}, {"identifier": "B", "content": "no real roots "}, {"identifier": "C", "content": "exactly one real root "}, {"identifier": "D", "content": "exactly four real roots "}] | ["B"]
Explanation:
Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$
<br><br>Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,
<br><br>we get $${t^2} - 4t - 1 = 0$$
<br><br>$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } ... |
The sum of all the real values of x satisfying the equation
<br/>2<sup>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50)</sup> = 1 is :
Options:
[{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$ 5"}] | ["C"]
Explanation:
We know, 2<sup>x</sup> = 1 only when x = 0.
<br><br>Similarly, 2<sup>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50)</sup> = 1 when
<br><br>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50) = 0
<br><br>$$ \Rightarrow $$ (x - 1)(x + 10)(x - 5) = 0
<br><br>$$ \therefore $$ x = 1, -10, 5
<br><br>Sum of real values of x =... |
The number of all possible positive integral values of $$\alpha $$ for which the roots of the quadratic equation, 6x<sup>2</sup> $$-$$ 11x + $$\alpha $$ = 0 are rational numbers is :
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier... | ["A"]
Explanation:
For rational D must be perfect square
<br><br>D = 121 $$-$$ 24$$\alpha $$
<br><br>for 121 $$-$$ 24$$\alpha $$ to be perfect square a must be 3, 4, 5
<br><br>So, ans $$\alpha $$ = 3 |
The number of integral values of m for which the
equation
<br/><br/>(1 + m<sup>2</sup>
)x<sup>2</sup>
– 2(1 + 3m)x + (1 + 8m) = 0
has no real root is :
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "infinitely many"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": ... | ["B"]
Explanation:
(1 + m<sup>2</sup>
)x<sup>2</sup>
– 2(1 + 3m)x + (1 + 8m) = 0
<br><br>Given equation has no real solution,
<br><br>$$ \therefore $$ Discriminant (D) < 0
<br><br>$$ \Rightarrow $$ 4(1 + 3m)<sup>2</sup> - 4(1 + m<sup>2</sup>)(1 + 8m) < 0
<br><br>$$ \Rightarrow $$ 4[9m<sup>2</sup> + 6m + 1 - 8m ... |
If m is chosen in the quadratic equation
<br/><br/>(m<sup>2</sup> + 1)
x<sup>2</sup> – 3x + (m<sup>2</sup> + 1)<sup>2</sup> = 0
<br/><br/>such that the sum of its
roots is greatest, then the absolute difference of
the cubes of its roots is :-
Options:
[{"identifier": "A", "content": "$$4\\sqrt 3 $$"}, {"identifier":... | ["C"]
Explanation:
Given quadratic equation
<br><br>(m<sup>2</sup> + 1)
x<sup>2</sup> – 3x + (m<sup>2</sup> + 1)<sup>2</sup> = 0
<br><br>Let roots of the equation $$\alpha $$ and $$\beta $$.
<br><br>$$ \therefore $$ Sum of roots = $$\alpha $$ + $$\beta $$ = $${3 \over {{m^2} + 1}}$$
<br><br>Product of roots = $$\alp... |
The least positive value of 'a' for which the
equation <br/><br/>2x<sup>2</sup> + (a – 10)x + $${{33} \over 2}$$
= 2a has real
roots is
Options:
[] | 8
Explanation:
For real roots Discriminate $$ \ge $$ 0.
<br><br>(a – 10)<sup>2</sup>
– 4$$\left( {{{33} \over 2} - 2a} \right).2$$ $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a<sup>2</sup>
+ 100 – 20a – 132 + 16a $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a
<sup>2</sup>
– 4a – 32 $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ (a – 8) (a... |
Let f(x) be a quadratic polynomial such that
<br/>f(–1) + f(2) = 0. If one of the roots of f(x) = 0
<br/>is 3, then its other root lies in :
Options:
[{"identifier": "A", "content": "(\u20133, \u20131)"}, {"identifier": "B", "content": "(1, 3)"}, {"identifier": "C", "content": "(\u20131, 0)"}, {"identifier": "D", "con... | ["C"]
Explanation:
Let the other root is $$\alpha $$.
<br><br>$$ \therefore $$ f(x)
=
a(x
–
3)
(x
–
$$\alpha $$)
<br><br>f(2) = a($$\alpha $$– 2)
<br><br>f(–1) = 4a(1 + $$\alpha $$)
<br><br>Given f(–1) + f(2) = 0
<br><br>$$ \Rightarrow $$a($$\alpha $$ – 2 + 4 + 4$$\alpha $$) = 0
<br><br>$$ \Rightarrow $$ 5$$\... |
The sum of all integral values of k (k $$\ne$$ 0) for which the equation $${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$ in x has no real roots, is ____________.
Options:
[] | 66
Explanation:
$${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$<br><br>$$x \in R - \{ 1,2\} $$<br><br>$$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$$<br><br>$$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$$<br><br>for x $$\ne$$ 3, $$k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$$<br><br>$$x - 3 + {2 \... |
The set of all values of K > $$-$$1, for which the equation $${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$ has real roots, is :
Options:
[{"identifier": "A", "content": "$$\\left( {1,{5 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "[2, 3)"}, {"identifie... | ["A"]
Explanation:
$${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$<br><br>Let $$3{x^2} + 4x + 3 = a$$<br><br>and $$3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$$<br><br>Given equation becomes<br><br>$$ \Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$$<br><br>$$ \Rightarrow a(... |
The numbers of pairs (a, b) of real numbers, such that whenever $$\alpha$$ is a root of the equation x<sup>2</sup> + ax + b = 0, $$\alpha$$<sup>2</sup> $$-$$ 2 is also a root of this equation, is :
Options:
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, ... | ["A"]
Explanation:
Consider the equation x<sup>2</sup> + ax + b = 0<br><br>If has two roots (not necessarily real $$\alpha$$ & $$\beta$$)<br><br>Either $$\alpha$$ = $$\beta$$ or $$\alpha$$ $$\ne$$ $$\beta$$<br><br>Case (1) If $$\alpha$$ = $$\beta$$, then it is repeated root. Given that $$\alpha$$<sup>2</sup> $$-$$... |
<p>The number of real solutions of the equation $${e^{4x}} + 4{e^{3x}} - 58{e^{2x}} + 4{e^x} + 1 = 0$$ is ___________.</p>
Options:
[] | 2
Explanation:
<p>Dividing by e<sup>2x</sup></p>
<p>$${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$$</p>
<p>$$ \Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$$</p>
<p>Let $${e^x} + {e^{ - x}} = t \in [2,\infty )$$</p>
<p>$$ \Rightarrow {t^2} + 4t - 60 = 0$$</p>
<p>$$ \Rightarrow t = 6$... |
<p>The number of distinct real roots of x<sup>4</sup> $$-$$ 4x + 1 = 0 is :</p>
Options:
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["B"]
Explanation:
<p>$$f(x) = {x^4} - 4x + 1 = 0$$</p>
<p>$$f'(x) = 4{x^3} - 4$$</p>
<p>$$ = 4(x - 1)({x^2} + 1 + x)$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9sp3c/20ce78f3-fabd-4839-b609-84bc56bb070f/63f06880-0655-11ed-903e-c9687588b3f3/file-1l5q9sp3d.png?format=png" data-orsrc="... |
<p>Let p and q be two real numbers such that p + q = 3 and p<sup>4</sup> + q<sup>4</sup> = 369. Then $${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}}$$ is equal to _________.</p>
Options:
[] | 4
Explanation:
<p>$$\because$$ $$p + q = 3$$ ...... (i)</p>
<p>and $${p^4} + {q^4} = 369$$ ...... (ii)</p>
<p>$${\{ {(p + q)^2} - 2pq\} ^2} - 2{p^2}{q^2} = 369$$</p>
<p>or $${(9 - 2pq)^2} - 2{(pq)^2} = 369$$</p>
<p>or $${(pq)^2} - 18pq - 144 = 0$$</p>
<p>$$\therefore$$ $$pq = - 6$$ or 24</p>
<p>But $$pq = 24$$ is not... |
<p>The sum of all the real roots of the equation <br/><br/>$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$ is</p>
Options:
[{"identifier": "A", "content": "$${\\log _e}3$$"}, {"identifier": "B", "content": "$$ - {\\log _e}3$$"}, {"identifier": "C", "content": "$${\\log _e}6$$"}, {"identifier": "D", "content": "$$ - {\\... | ["B"]
Explanation:
<p>$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$</p>
<p>Let $${e^x} = t$$</p>
<p>$$\therefore$$ $$({t^2} - 4)(6{t^2} - 5t + 1) = 0$$</p>
<p>$$ \Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0$$</p>
<p>$$\therefore$$ t = 2, $$-$$2, $${1 \over 2}$$, $${1 \over 3}$$</p>
<p>$$\therefore$$ $${e^x} = 2 \Right... |
<p>The number of distinct real roots of the equation <br/><br/>x<sup>7</sup> $$-$$ 7x $$-$$ 2 = 0 is</p>
Options:
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"]
Explanation:
<p>Given equation $${x^7} - 7x - 2 = 0$$</p>
<p>Let $$f(x) = {x^7} - 7x - 2$$</p>
<p>$$f'(x) = 7{x^6} - 7 = 7({x^6} - 1)$$</p>
<p>and $$f'(x) = 0 \Rightarrow x = \, + \,1$$</p>
<p>and $$f( - 1) = - 1 + 7 - 2 = 5 > 0$$</p>
<p>$$f(1) = 1 - 7 - 2 = - 8 < 0$$</p>
<p>So, roughly sketch of f(x) wi... |
<p>Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$ is minimum. Then S :</p>
Options:
[{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "is a singleton"}, {"ide... | ["A"]
Explanation:
<p>Given quadratic equation,</p>
<p>$$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$</p>
<p>Let, a and b are the roots of the equation,</p>
<p>$$\therefore$$ $$a + b = - {{\alpha - 6} \over 3}$$</p>
<p>and $$ab = {{\alpha + 3} \over 3}$$</p>
<p>For real roots,</p>
<p>$$D \ge 0$$</p>
<p>$$ \Rightar... |
<p>The number of distinct real roots of the equation $$x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$$ is ______________.</p>
Options:
[] | 3
Explanation:
<p>$${x^8} - {x^7} - {x^6} + {x^5} + 3{x^4} - 4{x^3} - 2{x^2} + 4x - 1 = 0$$</p>
<p>$$ \Rightarrow {x^7}(x - 1) - {x^5}(x - 1) + 3{x^3}(x - 1) - x({x^2} - 1) + 2x(1 - x) + (x - 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - x(x + 1) - 2x + 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^7} - ... |
<p>The sum of all real values of $$x$$ for which $$\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$$ is equal to __________.</p>
Options:
[] | 6
Explanation:
<p>$${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$$</p>
<p>$$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$$</p>
<p>$${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over ... |
The equation $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ has :
Options:
[{"identifier": "A", "content": "two solutions and both are negative"}, {"identifier": "B", "content": "two solutions and only one of them is negative"}, {"identifier": "C", "content": "four sol... | ["A"]
Explanation:
$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$
<br/><br/>Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$
<br/><br/>Now, $\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0$
<br/><br/>Dividing equation by $\mathrm{t}^{2}$
<br/><br/>$$
\begin{aligned}
& t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}... |
<p>The number of real roots of the equation $$\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+6}$$, is :</p>
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["B"]
Explanation:
$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$
<br/><br/>$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$
<br/><br/>$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain
<br/><br/>or
<br/><br/>$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$
<br/>... |
<p>The number of real solutions of the equation $$3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$$, is</p>
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["C"]
Explanation:
$3\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0$
<br/><br/>
$3\left[\left(x+\frac{1}{x}\right)^{2}-2\right]-2\left(x+\frac{1}{x}\right)+5=0$
<br/><br/>
Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$
<br/><br/>
$$
\begin{aligned}
& 3 t^{2}-2 t-1=0 \\\\
& 3 t^... |
<p> Let m and $$\mathrm{n}$$ be the numbers of real roots of the quadratic equations $$x^{2}-12 x+[x]+31=0$$ and $$x^{2}-5|x+2|-4=0$$ respectively, where $$[x]$$ denotes the greatest integer $$\leq x$$. Then $$\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$$ is equal to __________.</p>
Options:
[] | 9
Explanation:
The givne eqn is : $x^2-12 x+[x]+31=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\{x\}-x=x^2-12 x+31 \\\\
& \Rightarrow\{x\}=x^2-11 x+31
\end{aligned}
$$
<br/><br/>So, $0 \leq x^2-11 x+31<1$
<br/><br/>$$
\begin{aligned}
& \Rightarrow x^2-11 x+30 \leq 0 \\\\
& \Rightarrow(x-5)(x-6)<0 \\\\
& \Rightarrow x... |
If $$x$$ is real, the maximum value of $${{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ is
Options:
[{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$$41$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$${17 \\over 7}$$ "}] | ["B"]
Explanation:
$$y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$
<br><br>$$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$
<br><br>$$D \ge 0$$ as $$x$$ is real
<br><br>$$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$$
<br><br>$$ \Rightarrow... |
The sum of all real values of $$x$$ satisfying the equation $${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}}\, = 1$$ is :
Options:
[{"identifier": "A", "content": "$$6$$"}, {"identifier": "B", "content": "$$5$$ "}, {"identifier": "C", "content": "$$3$$"}, {"identifier": "D", "content": "$$-4$$ "}] | ["C"]
Explanation:
Given equation,
$${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$$
<br><br><b>Case 1 : </b>When x<sup>2</sup> - 5x + 5 = 1 and x<sup>2</sup> + 4x - 60 is any real no then this equation satisfy.
<br><b>Note :</b> When we put any real number as a power of 1 the value stays always 1 (1<sup> an... |
The number of integral values of m for which the quadratic expression, (1 + 2m)x<sup>2</sup> – 2(1 + 3m)x + 4(1 + m), x $$ \in $$ R, is always positive, is :
Options:
[{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["A"]
Explanation:
Expression is always positive it
<br><br>2m + 1 > 0 $$ \Rightarrow $$ m > $$-$$ $${1 \over 2}$$ &
<br><br>D < 0 $$ \Rightarrow $$ m<sup>2</sup> $$-$$ 6m $$-$$ 3 < 0
<br><br> 3 $$-$$ $$\sqrt {12} $$ < m < 3 + $$\sqrt {12} $$ . . . . (iii)
<b... |
<p>Let $$a, b, c$$ be the lengths of three sides of a triangle satistying the condition $$\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$$. If the set of all possible values of $$x$$ is the interval $$(\alpha, \beta)$$, then $$12\left(\alpha^2+\beta^2\right)$$ is equal to __________.</p>
Options:
[] | 36
Explanation:
<p>$$\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\
& \Rightarrow(a x-b)^2+(b x-c)^2=0 \\
& \Rightarrow a x-b=0, \quad b x-c=0 \\
& \Rightarrow a+b>c \quad b+c>a \quad c+a>b
\end{aligned}$$</p>
<p>$$\begin{array}{l|l|l... |
Difference between the corresponding roots of $${x^2} + ax + b = 0$$ and $${x^2} + bx + a = 0$$ is same and $$a \ne b,$$ then
Options:
[{"identifier": "A", "content": "$$a + b + 4 = 0$$ "}, {"identifier": "B", "content": "$$a + b - 4 = 0$$ "}, {"identifier": "C", "content": "$$a - b - 4 = 0$$ "}, {"identifier": "D",... | ["A"]
Explanation:
Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations $${x^2} + ax + b = 0$$
<br><br>and $${x^2} + bx + a = 0$$ respectively.
<br><br>$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$
<br><br>and $$\gamma + \delta = - b,\gamma \delta = a.$$
<br><br>Given $$\l... |
If $$p$$ and $$q$$ are the roots of the equation $${x^2} + px + q = 0,$$ then
Options:
[{"identifier": "A", "content": "$$p = 1,\\,\\,q = - 2$$ "}, {"identifier": "B", "content": "$$p = 0,\\,\\,q = 1$$ "}, {"identifier": "C", "content": "$$p = - 2,\\,\\,q = 0$$ "}, {"identifier": "D", "content": "$$p = - 2,\\,\\,q... | ["A"]
Explanation:
$$p + q = - p$$ and $$pq = q \Rightarrow q\left( {p - 1} \right) = 0$$
<br><br>$$ \Rightarrow q = 0$$ or $$p=1.$$
<br><br>If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$
<br><br>$$\therefore$$ $$p=1$$ and $$q=-2.$$ |
If $$\alpha \ne \beta $$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta $$ and $$\beta /\alpha \,\,$$ as its roots is
Options:
[{"identifier": "A", "content": "$$3{x^2} - 19x + 3 = 0$$"}, {"identifier": "B", "content": "$$3{x^2} + 19x - 3 = 0$$ "}, {"iden... | ["A"]
Explanation:
We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$
<br><br>$$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of
<br><br>equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$
<br><br>$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$
<br><br>Thus, the equ... |
If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in
Options:
[{"identifier": "A", "content": "Arithmetic - Geometric Progression "}, {"identifier": "B", "content": "Arithmetic P... | ["D"]
Explanation:
$$a{x^2} + bx + c = 0,$$ $$\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$$
<br><br>As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$
<br><br>$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$... |
The value of '$$a$$' for which one root of the quadratic equation
$$$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$$
<br/>is twice as large as the other is
Options:
[{"identifier": "A", "content": "$$ - {1 \\over 3}$$ "}, {"identifier": "B", "content": "$$ {2 \\over 3}$$"}, {"identifier":... | ["B"]
Explanation:
Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then
<br><br>$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$
<br><br>and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$
<br><br>$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a +... |
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation
Options:
[{"identifier": "A", "content": "$${x^2} - 18x - 16 = 0$$ "}, {"identifier": "B", "content": "$${x^2} - 18x + 16 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 18x - 16 = 0$$"}, {"ident... | ["B"]
Explanation:
Let two numbers be a and b then $${{a + b} \over 2} = 9$$
<br><br>and $$\sqrt {ab} = 4$$
<br><br>$$\therefore$$ Equation with roots $$a$$ and $$b$$ is
<br><br>$${x^2} - \left( {a + b} \right)x + ab = 0$$
<br><br>$$ \Rightarrow {x^2} - 18x + 16 = 0$$ |
If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are
Options:
[{"identifier": "A", "content": "$$ - 1,2$$ "}, {"identifier": "B", "content": "$$ - 1,1$$"}, {"identifier": "C", "content": "$$ 0,-1$$"}, {"identifier": "D", "content": "$$0,1$$ "}... | ["C"]
Explanation:
Let the second root be $$\alpha .$$
<br><br>Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$
<br><br>Also $$\alpha .\left( {1 - p} \right) = 1 - p$$
<br><br>$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$
<br><br>$$ \Rightarrow p = 1$$ [as $$\alpha ... |
In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$ \tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then
Options:
[{"identifier": "A", "content": "$$a = b + c$$ "}, {"identifier": "B", "content": "$$c = a + b$$ "}, {"identi... | ["B"]
Explanation:
$$\angle $$R = 90<sup>o</sup> $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90<sup>o</sup>
<br><br>$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$<sup>o</sup>
<br><br>$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0... |
The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
assume the least value is :
Options:
[{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "co... | ["A"]
Explanation:
$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
<br><br>$$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$
<br><br>$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$
<br><br>$$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$... |
The value of $$a$$ for which the sum of the squares of the roots of the equation <br/>$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is
Options:
[{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "c... | ["A"]
Explanation:
Given quadratic equation,
<br><br>$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
<br><br>Let $$\alpha $$ and $$\beta $$ are the roots of the equation.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$a - 2$$
<br><br>and $$\alpha $$$$\beta $$ = $$ - a - 1$$
<br><br>Now $${\alpha ^2} + {\beta ^2... |
If the roots of the quadratic equation $${x^2} + px + q = 0$$ are $$\tan {30^ \circ }$$ and $$\tan {15^ \circ }$$, respectively, then the value of $$2 + q - p$$ is
Options:
[{"identifier": "A", "content": "2 "}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0 "}, {"identifier": "D", "content":... | ["B"]
Explanation:
$${x^2} + px + q = 0$$
<br><br>Sum of roots $$ = \tan {30^ \circ } + \tan {15^ \circ } = - p$$
<br><br>Products of roots $$ = \tan {30^ \circ }.\tan {15^ \circ } = q$$
<br><br>$$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$$
... |
If the difference between the roots of the equation $${x^2} + ax + 1 = 0$$ is less than $$\sqrt 5 ,$$ then the set of possible values of $$a$$ is
Options:
[{"identifier": "A", "content": "$$\\left( {3,\\infty } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\infty , - 3} \\right)$$ "}, {"identifier": "C... | ["C"]
Explanation:
Let $$\alpha $$ and $$\beta $$ are roots of the equation $${x^2} + ax + 1 = 0$$
<br><br>So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$
<br><br>given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$
<br><br>$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } ... |
If $$\alpha $$ and $$\beta $$ are the roots of the equation $${x^2} - x + 1 = 0,$$ then $${\alpha ^{2009}} + {\beta ^{2009}} = $$
Options:
[{"identifier": "A", "content": "$$\\, - 1$$ "}, {"identifier": "B", "content": "$$\\, 1$$"}, {"identifier": "C", "content": "$$\\, 2$$ "}, {"identifier": "D", "content": "$$\\... | ["B"]
Explanation:
$${x^2} - x + 1 = 0$$
<br><br>$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$
<br><br>$$x = {{1 \pm \sqrt 3 i} \over 2}$$
<br><br>$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$
<br><br>$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$
<br><br>$${\alpha ^{2009}} ... |
Let $$\alpha $$ and $$\beta $$ be the roots of equation $$p{x^2} + qx + r = 0,$$ $$p \ne 0.$$ If $$p,\,q,\,r$$ in A.P. and $${1 \over \alpha } + {1 \over \beta } = 4,$$ then the value of $$\left| {\alpha - \beta } \right|$$ is :
Options:
[{"identifier": "A", "content": "$${{\\sqrt {34} } \\over 9}$$ "}, {"identifier... | ["B"]
Explanation:
Let $$p,q,r$$ are in $$AP$$
<br><br>$$ \Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Given $${1 \over \alpha } + {1 \over \beta } = 4 \Rightarrow {{\alpha + \beta } \over {\alpha \beta }} = 4$$
<br><br>We have $$\alpha + \beta = - q/p$$ and $$\alpha \beta = {r \ov... |
Let $$\alpha $$ and $$\beta $$ be the roots of equation $${x^2} - 6x - 2 = 0$$. If $${a_n} = {\alpha ^n} - {\beta ^n},$$ for $$n \ge 1,$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is equal to :
Options:
[{"identifier": "A", "content": "$$3$$"}, {"identifier": "B", "content": "$$ - 3$$ "}, {"identifi... | ["A"]
Explanation:
Given equation, x<sup>2</sup> - 6x - 2 = 0
<br><br> Roots are $$\alpha $$ and $$\beta $$.
<br><br>So, $$\alpha + \beta = 6$$ and $$\alpha \beta = - 2$$
<br><br>In the question given, $${a_n} = {\alpha ^n} - {\beta ^n}$$
<br><br>$$\therefore$$ $${a_8} = {\alpha ^8} - {\beta ^8}$$
<br><br>and $${a... |
If x is a solution of the equation, $$\sqrt {2x + 1} $$ $$ - \sqrt {2x - 1} = 1,$$ $$\,\,\left( {x \ge {1 \over 2}} \right),$$ then $$\sqrt {4{x^2} - 1} $$ is equal to :
Options:
[{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "... | ["A"]
Explanation:
Given,
<br><br>$$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$$
<br><br>$$ \Rightarrow $$ $$\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $$
<br><br>Squaring both sides, we get
<br><br>2x + 1 $$=$$ 1 + 2x $$-$$ 1 + 2$$\sqrt {2x - 1} $$
<br><br>$$ \Rightarrow $$ 1 $$=$$ 2$$\sqrt {2... |
If for a positive integer n, the quadratic equation
<br/><br/>$$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)$$$$ + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)$$$$ = 10n$$
<br/><br/>has two consecutive integral solutions, then n is equal to :
Options:
[{"identifier": "... | ["C"]
Explanation:
$$\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right... |
Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$$-$$ 1 and it leaves remainder 6 when divided by x + 1; then :
Options:
[{"identifier": "A", "content": "p(2) = 11"}, {"identifier": "B", "content": "p(2) = 19"}, {"identifier": "C", "content": "p($$-$$ 2) = 19"}, {"iden... | ["C"]
Explanation:
Let, P(x) = ax<sup>2</sup> + bx + c
<br><br>As, P(0) = 1,
<br><br>$$\therefore\,\,\,$$ a(0)<sup>2</sup> + b(0) + c = 1
<br><br>$$ \Rightarrow $$$$\,\,\,$$ c = 1
<br><br>$$\therefore\,\,\,$$ P(x) = ax<sup>2</sup> + bx + 1
<br><br>If P(x) is divided by x $$-$$ 1, remainder = 4
<br><br>$$ ... |
If tanA and tanB are the roots of the quadratic equation, 3x<sup>2</sup> $$-$$ 10x $$-$$ 25 = 0, then the value of 3 sin<sup>2</sup>(A + B) $$-$$ 10 sin(A + B).cos(A + B) $$-$$ 25 cos<sup>2</sup>(A + B) is :
Options:
[{"identifier": "A", "content": "$$-$$ 10"}, {"identifier": "B", "content": "10"}, {"identifier": "C"... | ["C"]
Explanation:
As tan A and tan B are the roots of 3x<sup>2</sup> $$-$$ 10x $$-$$ 25 = 0,
<br><br>So, tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$
<br><br>= $${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$$ = $${{10/3} \over {28/3}}$$ = $${5 \over {14}}$$
<br><br>Now, cos<sup>2</sup> (A + B) = $... |
If an angle A of a $$\Delta $$ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x<sup>2</sup> + 27x + 20 = 0 are :
Options:
[{"identifier": "A", "content": "secA, cotA"}, {"identifier": "B", "content": "sinA, secA"}, {"identifier": "C", "content": "secA, tanA"}, {"identifier": "D", "content":... | ["C"]
Explanation:
Here, 9x<sup>2</sup> + 27x + 20 = 0
<br><br>$$\therefore\,\,\,$$ x = $${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$ $${4 \over 3}$$, $... |
Let p, q and r be real numbers (p $$ \ne $$ q, r $$ \ne $$ 0), such that the roots of the equation $${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :
Options:
[{"identifier": "A", "content": "$${{{p^2} + {q^2}} \\... | ["B"]
Explanation:
Given,
<br><br>$${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ (2x + p + q) r = x<sup>2</sup> + px + qx + pq
<br><br>$$ \Rightarro... |
If f(x) is a quadratic expression such that f (1) + f (2) = 0, and $$-$$ 1 is a root of f (x) = 0, then the other root of f(x) = 0 is :
Options:
[{"identifier": "A", "content": "$$-$$ $${5 \\over 8}$$"}, {"identifier": "B", "content": "$$-$$ $${8 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 8}$$"}, {"id... | ["D"]
Explanation:
Let $$\alpha $$ and $$\beta $$ = - 1 are the roots of the polynomial, then we get
<br><br>f(x) = x<sup>2</sup> + (1 - $$\alpha $$)x - $$\alpha $$
<br><br>$$ \therefore $$ f(1) = 2 - 2$$\alpha $$
<br><br>and f(2) = 6 - 3$$\alpha $$
<br><br>Also given,
<br><br> f (1) + f (2) = 0
<br><br>$$ \therefore ... |
If $$\lambda $$ $$ \in $$ <b>R</b> is such that the sum of the cubes of the roots of the equation,
<br/>x<sup>2</sup> + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :
Options:
[{"identifier": "A", "content": "$$4\\sqrt 2 $$"... | ["B"]
Explanation:
Let $$\alpha $$, $$\beta $$ are the roots of the equation,
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 2 and $$\alpha $$$$\beta $$ = 10 $$-$$ $$\lambda $$
<br><br>$${\alpha ^3} + {\beta ^3}$$ = ($$\alpha $$ + $$\beta $$)<sup>3</sup> $$-$$ 3$$\alpha $$$$\beta $$ ($$\alpha ... |
The value of $$\lambda $$ such that sum of the squares of the roots of the quadratic equation, x<sup>2</sup> + (3 – $$\lambda $$)x + 2 = $$\lambda $$ has the least value is -
Options:
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${{15} \\over 8}$$"}, {"ide... | ["B"]
Explanation:
$$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 3
<br><br>$$\alpha $$$$\beta $$ = 2 $$-$$ $$\lambda $$
<br><br>$$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup> = ($$\alpha $$ + $$\beta $$)<sup>2</sup> $$-$$ 2$$\alpha $$$$\beta $$ = ($$\lambda $$ $$-$$ 3)<sup>2</sup> $$-$$ 2$$\left( {2 - \lambda } \rig... |
If one real root of the quadratic equation 81x<sup>2</sup> + kx + 256 = 0 is cube of the other root, then a value of k is
Options:
[{"identifier": "A", "content": "$$-$$ 81"}, {"identifier": "B", "content": "$$-$$ 300"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "144"}] | ["B"]
Explanation:
81x<sup>2</sup> + kx + 256 = 0 ; x = $$\alpha $$, $$\alpha $$<sup>3</sup>
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>4</sup> = $${{256} \over {81}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$ \pm $$ $${{4} \over {3}}$$
<br><br>Now $$-$$ $${k \over {81}}$$ = $$\alpha $$ ... |
Let $$\alpha $$ and $$\beta $$ be the roots of the quadratic equation x<sup>2</sup>
sin $$\theta $$ – x(sin $$\theta $$ cos $$\theta $$ + 1) + cos $$\theta $$ = 0 (0 < $$\theta $$ < 45<sup>o</sup>), and $$\alpha $$ < $$\beta $$. Then $$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \righ... | ["B"]
Explanation:
D = (1 + sin$$\theta $$ cos$$\theta $$)<sup>2</sup> $$-$$ 4sin$$\theta $$cos$$\theta $$ = (1 $$-$$ sin$$\theta $$ cos$$\theta $$)<sup>2</sup>
<br><br>$$ \Rightarrow $$ roots are $$\beta $$ = cosec$$\theta $$ and $$\alpha $$ = cos$$\theta $$
<br><br>$$\sum\limits_{n = 0}^\infty {\left( {{... |
If $$\lambda $$ be the ratio of the roots of the quadratic equation in x, 3m<sup>2</sup>x<sup>2</sup> + m(m – 4)x + 2 = 0, then the least value of m for which $$\lambda + {1 \over \lambda } = 1,$$ is
Options:
[{"identifier": "A", "content": "$$ - 2 + \\sqrt 2 $$"}, {"identifier": "B", "content": "4$$-$$3$$\\sqrt 2 $... | ["B"]
Explanation:
3m<sup>2</sup>x<sup>2</sup> + m(m $$-$$ 4) x + 2 = 0
<br><br>$$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta $$
<br><br>($$\alpha $$ + $$\beta $$)<sup>2</sup> = 3$$\alpha $$$$\beta $$
<br><br>$${\left( { - {{m\left( {m -... |
Let p, q $$ \in $$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic
equation, x<sup>2</sup> + px + q = 0, then :
Options:
[{"identifier": "A", "content": "p<sup>2</sup> \u2013 4q \u2013 12 = 0"}, {"identifier": "B", "content": "q<sup>2</sup> \u2013 4p \u2013 16 = 0"}, {"identifier": "C", "content": "q<sup>2</sup> + 4p... | ["A"]
Explanation:
If a quadratic equation with rational coefficient has one irrational root then other root will be the conjugate of the irrational root.
<br><br>Here x<sup>2</sup> + px + q = 0 has one root 2 - $$\sqrt 3$$.
<br><br>$$ \therefore $$ Other root will be 2 + $$\sqrt 3$$.
<br><br>Sum of the roots = -p = 4... |
If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
<br/>x<sup>2</sup> + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
<br/>$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta ... | ["C"]
Explanation:
Given $$\alpha + \beta = - \sin \theta $$ and$$\alpha \beta = - 2\sin \theta $$<br><br>
$${{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta }... |
If $$\alpha $$ and $$\beta $$ be two roots of the equation<br/> x<sup>2</sup> – 64x + 256 = 0. Then the value of
<br/>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ is :
Options:
[{"identifier": "A", "content": "1"}, {"identifier": "B", "co... | ["C"]
Explanation:
x<sup>2</sup> – 64x + 256 = 0
<br><br>$$\alpha $$ + $$\beta $$ = 64, $$\alpha $$$$\beta $$ = 256
<br><br>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$
<br><br>= $${{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}... |
If $$\alpha $$ and $$\beta $$ are the roots of the equation
<br/>2x(2x + 1) = 1, then $$\beta $$ is equal to :
Options:
[{"identifier": "A", "content": "$$ - 2\\alpha \\left( {\\alpha + 1} \\right)$$"}, {"identifier": "B", "content": "$$ 2\\alpha \\left( {\\alpha + 1} \\right)$$"}, {"identifier": "C", "content": "$... | ["A"]
Explanation:
$$\alpha $$ and
$$\beta $$ are the roots of the equation
<br>4x<sup>2</sup> + 2x – 1 = 0.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$ - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ -1 = 2$$\alpha $$ + 2$$\beta $$
<br><br>and 4$$\alpha $$<sup>2</sup> + 2$$\alpha $$ - 1 = 0
<br><br>$$ \Rightarro... |
If $$\alpha $$ and $$\beta $$ are the roots of the equation,
<br/>7x<sup>2</sup> – 3x – 2 = 0, then the value of
<br/>$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$ is equal to :
Options:
[{"identifier": "A", "content": "$${1 \\over {24}}$$"}, {"identifier": "B", "content": "$${{27} \\over {32... | ["C"]
Explanation:
Given, 7x<sup>2</sup> – 3x – 2 = 0
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $${3 \over 7}$$
<br><br>$$\alpha $$$$\beta $$ = - $${2 \over 7}$$
<br><br>$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$
<br><br>= $${{\alpha + \beta - \alpha \beta \left( {\alpha + \be... |
Let $$\lambda \ne 0$$ be in R. If $$\alpha $$ and $$\beta $$ are the roots of the <br/>equation, x<sup>2</sup> - x + 2$$\lambda $$ = 0 and $$\alpha $$ and $$\gamma $$ are the roots of <br/>the equation, $$3{x^2} - 10x + 27\lambda = 0$$, then $${{\beta \gamma } \over \lambda }$$ is equal to:
Options:
[{"identifier... | ["D"]
Explanation:
$$\alpha $$ and $$\beta $$ are the roots of the <br>equation x<sup>2</sup> - x + 2$$\lambda $$ = 0 .....(1)
<br><br>$$ \therefore $$ $$\alpha + \beta = 1,\,\alpha \beta = 2\lambda $$
<br><br>$$\alpha $$ and $$\gamma $$ are the roots of <br>the equation, $$3{x^2} - 10x + 27\lambda = 0$$ ......(2)... |
Let $$\alpha $$ and $$\beta $$ be the roots of x<sup>2</sup> - 3x + p=0 and $$\gamma $$ and $$\delta $$ be the roots of x<sup>2</sup> - 6x + q = 0. If $$\alpha, \beta, \gamma, \delta $$
form a geometric progression.Then ratio (2q + p) : (2q - p) is:
Options:
[{"identifier": "A", "content": "9 : 7"}, {"identifier": "B"... | ["A"]
Explanation:
$$\alpha $$ and $$\beta $$ are the roots of x<sup>2</sup> $$-$$ 3x + p = 0<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 3 and $$\alpha \beta $$ = p<br><br>$$\gamma $$ and $$\delta $$ are the roots of x<sup>2</sup> $$-$$ 6x + q = 0<br><br>$$ \therefore $$ $$\gamma $$ + $$\delta $$ = 6 and $$\ga... |
Let
$$\alpha $$ and
$$\beta $$ be the roots of the equation
<br/>5x<sup>2</sup> + 6x – 2 = 0. If S<sub>n</sub>
=
$$\alpha $$<sup>n</sup> +
$$\beta $$<sup>n</sup>, n = 1, 2, 3....,
then :
Options:
[{"identifier": "A", "content": "5S<sub>6</sub>\n + 6S<sub>5</sub>\n = 2S<sub>4</sub>"}, {"identifier": "B", "content": "5... | ["A"]
Explanation:
$$\alpha $$ and
$$\beta $$ be the roots of the equation
<br>5x<sup>2</sup> + 6x – 2 = 0.
<br><br>$$ \Rightarrow $$ 5$$\alpha $$<sup>2</sup> + 6$$\alpha $$ - 2 = 0
<br><br>$$ \Rightarrow $$ 5$$\alpha $$<sup>n + 2</sup> + 6$$\alpha $$<sup>n + 2</sup> - 2$$\alpha $$<sup>n</sup> = 0 ......(1)
<br><br>(... |
If $$\alpha $$ and $$\beta $$ are the roots of the equation
<br/>x<sup>2</sup>
+ px + 2 = 0 and $${1 \over \alpha }$$ and $${1 \over \beta }$$ are the<br/> roots of
the equation 2x<sup>2</sup>
+ 2qx + 1 = 0, then
<br/>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alp... | ["C"]
Explanation:
$$\alpha $$ and $$\beta $$ are the roots of the <br><br>equation
x<sup>2</sup>
+ px + 2 = 0
<br><br>$$ \therefore $$ $$\alpha + \beta = - p,\,\alpha \beta = 2$$
<br><br>$${1 \over \alpha }$$ and $${1 \over \beta }$$ are the roots of
the <br><br>equation 2x<sup>2</sup>
+ 2qx + 1 = 0
<br><br>$$ ... |
Let $$\alpha $$ and $$\beta $$ be two real roots of the equation <br/>(k + 1)tan<sup>2</sup>x - $$\sqrt 2 $$ . $$\lambda $$tanx = (1 - k), where k($$ \ne $$ - 1)
and $$\lambda $$ are real numbers. if tan<sup>2</sup> ($$\alpha $$ + $$\beta $$) = 50, then a value of $$\lambda $$ is:
Options:
[{"identifier": "A", "conte... | ["B"]
Explanation:
Let tan$$\alpha $$ and tan$$\beta $$ are the roots of
<br><br>(k + 1)tan<sup>2</sup>x - $$\sqrt 2 $$ . $$\lambda $$tanx - (1 - k) = 0
<br><br>$$ \therefore $$ tan$$\alpha $$ + tan$$\beta $$ = $${{\sqrt 2 \lambda } \over {k + 1}}$$
<br><br>and an$$\alpha $$.tan$$\beta $$ = $${{k - 1} \over {k + 1}}$$... |
Let $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$. <br/>If $$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$ and<br/> $$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $$, then a and b are the roots of the quadratic equation :
Options:
[{"identifier": "A", "content": "x<sup>2</sup> + 101x + 1... | ["C"]
Explanation:
$$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$
<br><br>$$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$
<br><br>= $$\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}} $$
<br><br>= $$\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\om... |
Let $$\alpha $$ and $$\beta $$ be the roots of the equation x<sup>2</sup>
- x - 1 = 0. <br/>If p<sub>k</sub>
= $${\left( \alpha \right)^k} + {\left( \beta \right)^k}$$
, k $$ \ge $$ 1, then which one
of the following statements is not true?
Options:
[{"identifier": "A", "content": "(p<sub>1</sub> + p<sub>2</sub> +... | ["D"]
Explanation:
x<sup>2</sup>
- x - 1 = 0
<br><br>$$ \therefore $$ $$\alpha $$<sup>2</sup>
- $$\alpha $$ - 1 = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>2</sup> = $$\alpha $$ + 1
<br><br>$$ \therefore $$ $$\alpha $$<sup>3</sup> = $$\alpha $$<sup>2</sup> + $$\alpha $$
<br><br>= $$\alpha $$ + 1 + $$\alpha $$
<br>... |
Let $$\alpha$$ and $$\beta$$ be the roots of x<sup>2</sup> $$-$$ 6x $$-$$ 2 = 0. If a<sub>n</sub> = $$\alpha$$<sup>n</sup> $$-$$ $$\beta$$<sup>n</sup> for n $$ \ge $$ 1, then the value of $${{{a_{10}} - 2{a_8}} \over {3{a_9}}}$$ is :
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {... | ["B"]
Explanation:
Given, $$\alpha$$ and $$\beta$$ be the roots of $${x^2} - 6x - 2 = 0$$<br><br>$$\matrix{
{\alpha + \beta = 6} \cr
{\alpha \beta = - 2} \cr
} $$<br><br>and $${\alpha ^2} - 6\alpha - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha $$<br><br>$${\beta ^2} - 6\beta - 2 = 0 \Rightarrow {\bet... |
Let $$\alpha$$ and $$\beta$$ be two real numbers such that $$\alpha$$ + $$\beta$$ = 1 and $$\alpha$$$$\beta$$ = $$-$$1. Let p<sub>n</sub> = ($$\alpha$$)<sup>n</sup> + ($$\beta$$)<sup>n</sup>, p<sub>n$$-$$1</sub> = 11 and p<sub>n+1</sub> = 29 for some integer n $$ \ge $$ 1. Then, the value of p$$_n^2$$ is ___________.
... | 324
Explanation:
Given, $$\alpha$$ + $$\beta$$ = 1, $$\alpha$$$$\beta$$ = $$-$$ 1<br><br>$$ \therefore $$ Quadratic equation with roots $$\alpha$$, $$\beta$$ is x<sup>2</sup> $$-$$ x $$-$$ 1 = 0<br><br>$$ \Rightarrow $$ $$\alpha$$<sup>2</sup> = $$\alpha$$ + 1<br><br>Multiplying both sides by $$\alpha$$<sup>n$$-$$1</su... |
The value of $$4 + {1 \over {5 + {1 \over {4 + {1 \over {5 + {1 \over {4 + ......\infty }}}}}}}}$$ is :
Options:
[{"identifier": "A", "content": "2 + $${2 \\over 5}\\sqrt {30} $$"}, {"identifier": "B", "content": "2 + $${4 \\over {\\sqrt 5 }}\\sqrt {30} $$"}, {"identifier": "C", "content": "5 + $${2 \\over 5}\\sqrt {3... | ["A"]
Explanation:
$$y = 4 + {1 \over {5 + {1 \over y}}}$$<br><br>$$ \Rightarrow y = 4 + {y \over {5y + 1}}$$<br><br>$$ \Rightarrow 5{y^2} - 20y - 4 = 0$$<br><br>$$ \Rightarrow y = {{20 \pm \sqrt {400 + 80} } \over {10}}$$<br><br>$$ \Rightarrow y = {{20 \pm 4\sqrt {30} } \over {10}},y > 0$$<br><br>$$ \therefore $$ ... |
The value of $$3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$ is equal to
Options:
[{"identifier": "A", "content": "1.5 + $$\\sqrt 3 $$"}, {"identifier": "B", "content": "2 + $$\\sqrt 3 $$"}, {"identifier": "C", "content": "3 + 2$$\\sqrt 3 $$"}, {"identifier": "D", "content": "4 + $... | ["A"]
Explanation:
Let $$x = 3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$<br><br>So, $$x = 3 + {1 \over {4 + {1 \over x}}} = 3 + {1 \over {{{4x + 1} \over x}}}$$<br><br>$$ \Rightarrow (x - 3) = {x \over {(4x + 1)}}$$<br><br>$$ \Rightarrow (4x + 1)(x - 3) = x$$<br><br>$$ \Rightarrow... |
If $$\alpha$$ and $$\beta$$ are the distinct roots of the equation $${x^2} + {(3)^{1/4}}x + {3^{1/2}} = 0$$, then the value of $${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)$$ is equal to :
Options:
[{"identifier": "A", "content": "56 $$\\times$$ 3<sup>25</sup>"}, {"identifier": "B", "content"... | ["C"]
Explanation:
As, $$({\alpha ^2} + \sqrt 3 ) = - {(3)^{1/4}}.\alpha $$<br><br>$$ \Rightarrow ({\alpha ^4} + 2\sqrt 3 {\alpha ^2} + 3) = \sqrt 3 {\alpha ^2}$$ (On squaring)<br><br>$$\therefore$$ $$({\alpha ^4} + 3) = ( - )\sqrt 3 {\alpha ^2}$$<br><br>$$ \Rightarrow {\alpha ^8} + 6{\alpha ^4} + 9 = 3{\alpha ^4}$$ ... |
If $$\alpha$$, $$\beta$$ are roots of the equation $${x^2} + 5(\sqrt 2 )x + 10 = 0$$, $$\alpha$$ > $$\beta$$ and $${P_n} = {\alpha ^n} - {\beta ^n}$$ for each positive integer n, then the value of $$\left( {{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}} \right)$$ is eq... | 1
Explanation:
$${x^2} + 5\sqrt 2 x + 10 = 0$$<br><br>& $${P_n} = {\alpha ^n} - {\beta ^n}$$ (Given)<br><br>Now, $${{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}$$ = $${{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}$$<br>... |
Let $$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$ and $$\beta = \mathop {\min }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$. If $$8{x^2} + bx + c = 0$$ is a quadratic equation whose roots are $$\alpha$$<sup>1/5</sup> and $$\beta$$<sup>1/5</sup>, then the value of c $$-$$... | ["A"]
Explanation:
$$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$<br><br>$$ = \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\} $$<br><br>$$ = max\{ {2^{6\sin 3x + 8\cos 3x}}\} $$<br><br>and $$\beta = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} = \min \{ {2^{6\sin 3x + 8\cos 3x}}\} $$<br><br>No... |
cosec18$$^\circ$$ is a root of the equation :
Options:
[{"identifier": "A", "content": "x<sup>2</sup> + 2x $$-$$ 4 = 0"}, {"identifier": "B", "content": "4x<sup>2</sup> + 2x $$-$$ 1 = 0"}, {"identifier": "C", "content": "x<sup>2</sup> $$-$$ 2x + 4 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ 2x $$-$$ 4 =... | ["D"]
Explanation:
$$\cos ec18^\circ = {1 \over {\sin 18^\circ }} = {4 \over {\sqrt 5 - 1}} = \sqrt 5 + 1$$<br><br>Let $$\cos ec18^\circ = x = \sqrt 5 + 1$$<br><br>$$ \Rightarrow x - 1 = \sqrt 5 $$<br><br>Squaring both sides, we get<br><br>$${x^2} - 2x + 1 = 5$$<br><br>$$ \Rightarrow {x^2} - 2x - 4 = 0$$ |
<p>Let f(x) be a quadratic polynomial such that f($$-$$2) + f(3) = 0. If one of the roots of f(x) = 0 is $$-$$1, then the sum of the roots of f(x) = 0 is equal to :</p>
Options:
[{"identifier": "A", "content": "$${{11} \\over 3}$$"}, {"identifier": "B", "content": "$${{7} \\over 3}$$"}, {"identifier": "C", "content": ... | ["A"]
Explanation:
<p>$$\because$$ x = $$-$$1 be the roots of f(x) = 0</p>
<p>$$\therefore$$ Let $$f(x) = A(x + 1)(x - 1)$$ ...... (i)</p>
<p>Now, $$f( - 2) + f(3) = 0$$</p>
<p>$$ \Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0$$</p>
<p>$$b = {{14} \over 3}$$</p>
<p>$$\therefore$$ Second root of f(x) = 0 will be $${{14} ... |
<p>Let $$\alpha$$, $$\beta$$ be the roots of the equation $${x^2} - 4\lambda x + 5 = 0$$ and $$\alpha$$, $$\gamma$$ be the roots of the equation $${x^2} - \left( {3\sqrt 2 + 2\sqrt 3 } \right)x + 7 + 3\lambda \sqrt 3 = 0$$, $$\lambda$$ > 0. If $$\beta + \gamma = 3\sqrt 2 $$, then $${(\alpha + 2\beta + \gamma )... | 98
Explanation:
<p>$$\because$$ $$\alpha$$, $$\beta$$ are roots of x<sup>2</sup> $$-$$ 4$$\lambda$$x + 5 = 0</p>
<p>$$\therefore$$ $$\alpha$$ + $$\beta$$ = 4$$\lambda$$ and $$\alpha$$$$\beta$$ = 5</p>
<p>Also, $$\alpha$$, $$\gamma$$ are roots of</p>
<p>$${x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\... |
<p>The sum of the cubes of all the roots of the equation <br/><br/>$${x^4} - 3{x^3} - 2{x^2} + 3x + 1 = 0$$ is _________.</p>
Options:
[] | 36
Explanation:
<p>$${x^4} - 3{x^3} - {x^2} - {x^2} + 3x + 1 = 0$$</p>
<p>$$({x^2} - 1)({x^2} - 3x - 1) = 0$$</p>
<p>Let the root of $${x^2} - 3x - 1 = 0$$ be $$\alpha$$ and $$\beta$$ and other two roots of given equation are 1 and $$-$$1</p>
<p>So sum of cubes of roots</p>
<p>$$ = {1^3} + {( - 1)^3} + {\alpha ^3} + {... |
<p>If the sum of the squares of the reciprocals of the roots $$\alpha$$ and $$\beta$$ of <br/><br/>the equation 3x<sup>2</sup> + $$\lambda$$x $$-$$ 1 = 0 is 15, then 6($$\alpha$$<sup>3</sup> + $$\beta$$<sup>3</sup>)<sup>2</sup> is equal to :</p>
Options:
[{"identifier": "A", "content": "18"}, {"identifier": "B", "cont... | ["B"]
Explanation:
<p>$$3{x^2} + \lambda x - 1 = 0$$</p>
<p>Given, two roots are $$\alpha$$ and $$\beta$$.</p>
<p>$$\therefore$$ Sum of roots $$ = \alpha + \beta = {-\lambda \over 3}$$</p>
<p>And product of roots $$ = \alpha \beta = {-1 \over 3}$$</p>
<p>Given that,</p>
<p>Sum of square of reciprocal of roots $$\... |
<p>The minimum value of the sum of the squares of the roots of $$x^{2}+(3-a) x+1=2 a$$ is:</p>
Options:
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11ed-a18d-5933e4fde865/file-1l7nny14b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11... |
<p>If $$\alpha, \beta$$ are the roots of the equation</p>
<p>$$
x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0
$$,</p>
<p>then the equation, whose roots are $$\alpha+\frac{1}{\beta}$$ and $$\beta... | ["B"]
Explanation:
<p>$${3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}$$</p>
<p>$${3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log... |
<p>Let $$\alpha$$, $$\beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+\sqrt{6}=0$$ and $$\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$$ be the roots of the equation $$x^{2}+a x+b=0$$. Then the roots of the equation $$x^{2}-(a+b-2) x+(a+b+2)=0$$ are :</p>
Options:
[{"identifier": "A", "content": "non-real compl... | ["B"]
Explanation:
<p>$$\alpha + \beta = \sqrt 2 $$, $$\alpha \beta = \sqrt 6 $$</p>
<p>$${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$$</p>
<p>$$ = 2 + {{2 - 2\sqrt 6 } \over 6} = - a$$</p>
<p>$$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\... |
<p>If $$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$$, then the maximum value of $$\mathrm{a}$$ is :</p>
Options:
[{"identifier": "A", "content": "198"}, {"identifier": "B", "content": "202"}, {"identifier": "C", "content": "212"}, {"identifier": "D", "content": "218"}] | ["C"]
Explanation:
<p>$${1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$</p>
<p>$$ \Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \ove... |
<p>Let $$\alpha, \beta(\alpha>\beta)$$ be the roots of the quadratic equation $$x^{2}-x-4=0 .$$ If $$P_{n}=\alpha^{n}-\beta^{n}$$, $$n \in \mathrm{N}$$, then $$\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2}+P_{14} P_{15}}{P_{13} P_{14}}$$ is equal to __________.</p>
Options:
[] | 16
Explanation:
<p>$$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $${x^2} - x - 4 = 0$$.</p>
<p>$$\therefore$$ $$\alpha$$ and $$\beta$$ are satisfy the given equation.</p>
<p>$${\alpha ^2} - \alpha - 4 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$$ ...... (1)<... |
<p>Let $$S = \left\{ {x:x \in \mathbb{R}\,\mathrm{and}\,{{(\sqrt 3 + \sqrt 2 )}^{{x^2} - 4}} + {{(\sqrt 3 - \sqrt 2 )}^{{x^2} - 4}} = 10} \right\}$$. Then $$n(S)$$ is equal to</p>
Options:
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "... | ["B"]
Explanation:
Let $(\sqrt{3}+\sqrt{2})^{x^{2}-4}=t$
<br/><br/>$$
\begin{aligned}
& t+\frac{1}{t}=10 \\\\
\Rightarrow & t^{2}-10 t+1=0 \\\\
\Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6}
\end{aligned}
$$
<br/><br/><b>Case-I</b>
<br/><br/>$$
\begin{aligned}
& t=5+2 \sqrt{6} = (\sqrt{3}+\sqrt{2})^... |
<p>Let $$\alpha_1,\alpha_2,....,\alpha_7$$ be the roots of the equation $${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$ and $$|{\alpha _1}| \ge |{\alpha _2}| \ge \,...\, \ge \,|{\alpha _7}|$$. Then $$\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6$$ is equal to _________.</p>
Options:
[] | 9
Explanation:
<p>$${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$</p>
<p>$$x({x^6} + 3{x^4} - 13{x^2} - 15) = 0$$</p>
<p>$$x = 0 = {\alpha _7}$$</p>
<p>Let $${x^2} = t$$</p>
<p>$${t^3} + 3{t^2} - 13t - 15 = 0$$</p>
<p>$$(t + 1)(t + 5)(t - 3) = 0$$</p>
<p>$$t = {x^2} = - 1, - 5,3$$</p>
<p>$$x\, = \, \pm \,i, \pm \,\sqrt 5 i, \... |
<p>Let $$\lambda \ne 0$$ be a real number. Let $$\alpha,\beta$$ be the roots of the equation $$14{x^2} - 31x + 3\lambda = 0$$ and $$\alpha,\gamma$$ be the roots of the equation $$35{x^2} - 53x + 4\lambda = 0$$. Then $${{3\alpha } \over \beta }$$ and $${{4\alpha } \over \gamma }$$ are the roots of the equation</p>
O... | ["B"]
Explanation:
$14 x^{2}-31 x+3 \lambda=0$
<br/><br/>
$$
\begin{aligned}
& \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\
& 35 x^{2}-53 x+4 \lambda=0 \\\\
& \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) ... |
<p>Let $$\alpha \in\mathbb{R}$$ and let $$\alpha,\beta$$ be the roots of the equation $${x^2} + {60^{{1 \over 4}}}x + a = 0$$. If $${\alpha ^4} + {\beta ^4} = - 30$$, then the product of all possible values of $$a$$ is ____________.</p>
Options:
[] | 45
Explanation:
$x^{2}+60^{\frac{1}{4}} x+a=0$
<br/><br/>
$$
\therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a
$$
<br/><br/>
Now $\alpha^{4}+\beta^{4}=-30$
<br/><br/>
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
<br/><br/>
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
<br/>... |
<p>Let $$\alpha, \beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+2=0$$. Then $$\alpha^{14}+\beta^{14}$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$-64$$"}, {"identifier": "B", "content": "$$-64 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$-128 \\sqrt{2}$$"}, {"identifier": "D", "content": ... | ["D"]
Explanation:
<ol>
<li><p><strong>Find the roots of the quadratic equation:</strong></p>
<p>The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. For the quadratic equation $$x^2 - \sqrt{2}x + 2 = 0$$, we have $$a = 1, b = -\sqrt{2}, c = 2$$. Plugging these into the quadratic formula gives:
<br/><b... |
<p>Let $$\alpha, \beta$$ be the roots of the quadratic equation $$x^{2}+\sqrt{6} x+3=0$$. Then $$\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "9"}, {"identifier": "C... | ["D"]
Explanation:
Given quadratic equation: $$x^{2}+\sqrt{6} x+3=0$$
<br/><br/>We can find the roots using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
<br/><br/>Here, $$a = 1$$, $$b = \sqrt{6}$$, and $$c = 3$$.
<br/><br/>Substituting these values into the quadratic formula, we have:
<br/><br... |
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