Mathematics Test

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Mathematics Test - 20

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Question 1/10
1 / -0

The value of K, for which the equation (K–2)x² + 8x + K + 4 = 0 has both the roots real distinct and negative is:

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A
0

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B
2

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C
3

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D
–4

Solutions

$(K - 2) x^{2} + 8 x + K + 4 = 0$
$x^{2} + \frac{8}{(K - 2)} x + \frac{K + 4}{(K - 2)} = 0$
Let $f (x) = x^{2} + \frac{8}{(K - 2)} x + \frac{K + 4}{(K - 2)}$
If the equation $f (x) = 0,$ has real distinct and negative roots then:
$D > 0 & f (0) > 0 & - \frac{B}{2 A} < 0$
(i) $D > 0 \frac{8^{2}}{(K - 2)^{2}} - 4 \frac{(K + 4)}{K - 2} > 0$
$64 - 4 (k^{2} + 2 K - 8) > 0$
$- k^{2} - 2 K + 24 < 0 K^{2} + 2 K - 24 < 0$
$(K + 6) (K - 4) < 0$
(ii) $(f (0) > 0 \frac{(K + 4)}{(K - 2)} > 0) .$
$\Rightarrow K < - 4$ or $K > 2$
(iii) $- \frac{B}{2 Λ} < 0 - \frac{8}{2 (K - 2)} < 0 \frac{4}{(K - 2)} > 0$
$K - 2 > 0$
$K > 2$
$∴ D > 0 & f (0) > 0 &$
$- \frac{B}{2 A} < 0$
$- 6 < K < 4 & (K < - 4$ or $K > 2) & K > 2$
$- 6 < K < 48 k < - 48 K > 2)$ or $(K > 28 K > 2$
-62)
$2 < K < 4$
$K \in (2, 4)$
Hence, the correct option is C.
Question 2/10
1 / -0

If A and B are two square matrices such that B = –A^–1 BA, then (A+B)² is equal to

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A
O

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B
A² + B²

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C
A² + 2AB + B²

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D
A+B

Solutions

$B = - A^{- 1} BA$
$AB = A (- A^{- 1} BA)$
$AB = - {AA}^{- 1} ∣ BA)$
$AB = - BA$
$AB + BA = 0$
Now, $(A + B)^{2} = A^{2} + B^{2} + AB + BA$
$(A + B)^{2} = A^{2} + B^{2}, (∵ AB + BA = 0)$
Hence, the correct option is B.
Question 3/10
1 / -0

For $n \geq 2$ the product ${1 + α, {} + α^{2}} {1 + α^{2^{2}}}, \dots, {1 + α^{2^{n}}},$ where $α = (\frac{1 + i}{2}),$ is equal to...

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A
$(1 + i) (1 - 2^{- 2^{n}})$

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B
$(1 - i) (1 - 2^{- 2^{n}})$

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C
$(1 + i) (1 + 2^{- 2^{n}})$

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D
none of these

Solutions

Let $\frac{1 + i}{2} = x$ Hence the given series $S$ is $S = (1 + x / 1 + x^{2}) (1 + x^{4}) \cdot (1 + x^{2^{n}})$
$S (1 - x) = 1 - x^{2^{n} - 1}$
$S = \frac{1 - x^{2} n + 1}{1 - x} = \frac{1 - x^{2 + 1}}{1 - \frac{1 + i}{2}}$
$= \frac{1 - {(\frac{1 + i}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2})}^{2^{n + 1}}}{(\frac{1 - i}{2})} = \frac{1 - {(\frac{1 + i}{\sqrt{2}})}^{2^{n + 1}} {(\frac{\sqrt{2}}{2})}^{2^{n + 1}}}{(\frac{1 - i}{2})}$
$= \frac{2 (1 + i {1 - (- 1)^{2^{n - 1}} 2^{- 2^{n}}}}{2}$
$= (1 + i (1 - 2^{- 2^{n}}) (∵ n \geq 2 \Rightarrow (- 1)^{2^{n - 1}} = 1)$
Hence, the correct option is A.
Question 4/10
1 / -0

All the roots of $a_{1} z^{3} + a_{2} z^{2} + a_{3} z + a_{4} = 3$ , where $| a_{i} | \leq 1 (i = 1, 2, 3, 4)$

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A
lie outside or on the circle |z| = 2/3

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B
lie on the circle |z| = 2/3

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C
lie inside or on the circle |z| = 2/3

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D
none of these

Solutions

If possible let $z =$ re $i θ$ be a root which lies inside $| z | = 2 / 3 .$ Hence $0 \leq r < 2 / 3$ Hence $a_{1} {({re}^{i θ})}^{3} + a_{2} {({re}^{i θ})}^{2} + a_{3} ({re}^{i θ}) + a_{4} = 3$
Comparing real parts we get $a_{1} t^{3} \cos 3 θ + a_{2} r^{2} \cos 2 θ + a_{3} r \cos θ + a_{4} = 3$
but $(r^{3} a_{1} \cos 3 θ + r^{2} a_{2} \cos 2 θ + {ra}_{3} \cos θ + a_{4})$
$\leq r^{3} + r^{2} + r + 1$
(Qlail $\leq 1 &$ so also $\cos 3 θ, \cos 2 θ & \cos θ)$
$3 \leq r^{3} + r^{2} + r + 1$
$\frac{r^{4} - 1}{r - 1} \geq 3$
$\frac{1 - r^{4}}{1 - r} \geq 3$
$\frac{1}{1 - r} \geq 3$
$1 - r \leq \frac{1}{3}$
$1 - \frac{1}{3} \leq$
$r \geq 2 / 3$
Which is a contradiction to the assumption $0 \leq x < 2 / 3$ . Hence no root of the given equation can lie inside the circle. $| z | = \frac{2}{3}$
Hence, the correct option is A.
Question 5/10
1 / -0

The coefficient of the term independent of x in the expansion of $(1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is:

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A
1/3

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B
19/54

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C
17/54

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D
1/4

Solutions

The general term of the expansion of ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ is
$T_{r + 1} =^{9} C_{r} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r}$
$=^{9} C_{r} {(\frac{3}{2})}^{9 - t} {(- \frac{1}{3 x})}^{r}$
The coefficient of independent of $x$ in the expansion of $(1 + x + 2 x^{3})$ ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} =$ sum of coefficient of $x^{0}, x^{- 1} 2 x^{- 3}$ in expansion of ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$
(i) For coefficient of $x^{0} : 18 - 3 r = 0,,$ according to (1) $: r = 6 .$ Put in (i):
$T_{7} =^{9} C_{6} {(\frac{3}{2})}^{9 - 6} {(- \frac{1}{3})}^{6} =^{9} C_{3} {(\frac{3}{2})}^{3} \cdot {(\frac{1}{3})}^{6}$
$=^{9} C_{3} \frac{1}{6^{3}} = \frac{7}{18}$
(ii) For Coefficient of $x^{- 3} : 18 - 3 r = - 1$ or, $r = 19 / 3 =$ fraction. No such term exists.
(iii) For coefficient of $2 x^{- 3} : 18 - 3 r = - 3$ or, $r = 7 .$ Put in (i):
$T_{8} = 2 [9 C_{7} {(\frac{3}{2})}^{9 - 7} {(- \frac{1}{3})}^{7}]$
$= (9 C_{2}) (2 {(\frac{3}{2})}^{2} (- 1) {(\frac{1}{3})}^{7} = - \frac{2}{27}$
Sum of coefficients $= \frac{7}{18} - \frac{2}{27} = \frac{17}{54}$
Hence, the correct option is C.
Question 6/10
1 / -0

a, b, c are positive numbers and abc² has the greatest value 1/ 64. Then...

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A
$a = b = \frac{1}{2}, c = \frac{1}{4}$

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B
$a = b = \frac{1}{4}, c = \frac{1}{2}$

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C
$a = b = c = \frac{1}{3}$

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D
None of these

Solutions

We have,
$\frac{a + b + \frac{c}{2} + \frac{c}{2}}{4} \geq \sqrt[4]{a b \cdot \frac{c}{2} \cdot \frac{c}{2}}$
or, $\frac{a + b + c}{4} \geq \sqrt[4]{\frac{a b c^{2}}{4}}$
$∴ {(\frac{a + b + c}{4})}^{4} \geq \frac{a b c^{2}}{4}$
or $a b c^{2} \leq \frac{1}{64} (a + b + c)^{4}$
$∴$ the greatest value of $a b c^{2} = \frac{1}{64} (a + b + c)^{4}$
Also for the greatest value of abc $^{2}$ the numbers have to be equal, i.e $a = b = c / 2$ Also given that maximum value $= 1 / 64$ $s 0, a + b + c = 1$
i.e. $a = b = 1 / 4, c = 1 / 2$
Hence, the correct option is B.
Question 7/10
1 / -0

If $f (x) = x + \frac{1}{x} \nabla x \in R - (0)$ then,

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A
$f^{3} (x) = f (x^{3}) + 3 f (\frac{1}{x})$

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B
$f^{3} (x) = 3 f (x^{3}) + f (\frac{1}{x})$

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C
$f^{3} (x) + f (x^{3}) + f (x)$

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D
None of these

Solutions

$f (x) = x + \frac{1}{x} f (x^{3}) = x^{3} + \frac{1}{x^{3}}$
$f (x) = x + \frac{1}{x} = f (\frac{1}{x}) = \frac{1}{x} + x f (x) = f (\frac{1}{x})$
$∴ f^{3} (x) = {(x + \frac{1}{x})}^{3}$
$= (x^{3} + \frac{1}{x^{3}}) + 3 x \cdot \frac{1}{x} (x + \frac{1}{x})$
$= (x^{3} + \frac{1}{x^{3}}) + 3 (x + \frac{1}{x})$
$= f (x^{3}) + 3 f (x)$
$= f (x^{3}) + 3 f (\frac{1}{x}) \cdot [∵ f (x) = f (\frac{1}{x})]$
Hence, the correct option is A.
Question 8/10
1 / -0

The value of $lim_{x \to 1} \frac{3^{x + 1} - 9}{4^{2 x + 1} - 64}$ is...

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A
$\frac{\ln 3}{128 \ln 4}$

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B
$\frac{9 \ln 3}{128 \ln 4}$

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C
$\frac{9 \ln 3}{\ln 4}$

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D
$\frac{9 \ln 3}{64 \ln 4}$

Solutions

Substitute $(x - 1) = t .$
$x \to 1 (x - 1) \to 0 t \to 0$
Here given limit is:
$lim_{x \to 1} \frac{3^{x + 1} - 9}{4^{2 x + 1} - 64} = lim_{t \to 0} \frac{3^{t + 2} - 9}{4^{2 (t + 1) + 1} - 64}$
$= lim_{t \to 0} \frac{3^{t} \cdot 3^{2} - 3^{2}}{4^{2 t} \cdot 4^{3} - 4^{3}} = lim_{t \to 0} \frac{9 (3^{t} - 1)}{64 (4^{2 t} - 1)}$
$= lim_{t \to 0} \frac{9}{64} \frac{(\frac{3^{t} - 1}{t}) \cdot t}{(\frac{4^{2 t} - 1}{2 t}) \cdot 2 t} = lim_{t \to 0} \frac{9}{64} \cdot \frac{\ln 3}{\ln 4} \cdot \frac{1}{2} = \frac{9 \ln 3}{128 \ln 4}$

Hence, the correct option is B.
Question 9/10
1 / -0

The altitude of a cone is 20cm and its semi-vertical angle is 30 °. If the semi-vertical angle is increasing at the rate of 2 ° per second, then the radius of the base is increasing at the rate of.....

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A
$\frac{π}{27} cm / \sec$

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B
$\frac{8 π}{27} cm / \sec$

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C
$\frac{8 π}{9} cm / \sec$

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D
None of these

Solutions

$\frac{d θ}{d t} = 2^{\circ}$ per second
$= 2 \times \frac{π}{180} rad / \sec$
$= \frac{π}{90} rad / \sec$
$θ = 30^{\circ} = \frac{π}{6} radian$
Let \thetabe the semi-vertical angle and r be the base radius of the cone at time t. Then, $r = 20 \tan θ$
$\frac{dr}{dt} = 20 \sec^{2} θ \frac{d θ}{dt}$
${\frac{d r}{d t}]}_{θ = \frac{π}{6}} = (20 \sec^{2} \frac{π}{6}) \times \frac{π}{90}$
${\frac{d r}{d t}]}_{θ = \frac{π}{6}} = 20 \times \frac{4}{3} \times \frac{π}{90} = \frac{8 π}{27} cm / \sec$
Hence, the correct option is B.
Question 10/10
1 / -0

The set of all values of the parameter a for which the points of minimum of the function $y = 1 + a^{2} x - x^{3}$ Satisfy the inequality $\frac{x^{2} + x + 2}{x^{2} + 5 x + 6} \leq 0$ is,

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A
an empty set

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B
$(- 3 \sqrt{3}, - 2 \sqrt{3})$

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C
$(2 \sqrt{3}, 3 \sqrt{3})$

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D
$(- 3 \sqrt{3}, - 2 \sqrt{3}) \cup (2 \sqrt{3}, 3 \sqrt{3})$

Solutions

$y = 1 + a^{2} x - x^{3}$
$y^{'} = a^{2} - 3 x^{2}$
$y^{''} = - 6 x$
$y^{'} = 0 a^{2} - 3 x^{2} = 0 x = \pm \frac{a}{\sqrt{3}}$
(i) Let $a > 0 :$ Hence $y^{/ /} > 0$ for $x = - \frac{a}{\sqrt{3}} .$ Hence $y$ has minima at $x = - \frac{a}{\sqrt{3}}$ for $a > 0 .$ $∴ - 3 < - \frac{a}{\sqrt{3}} < - 2 2 \sqrt{3} < 2 < 3 \sqrt{3}$
(ii) Let $a < 0$ :
Hence $y^{/ /} > 0$ for $x = \frac{a}{\sqrt{3}} .$ Hence $y$ has minima at $x = \frac{a}{\sqrt{3}}$ for $a < 0$ . $∴ - 3 < \frac{a}{\sqrt{3}} < - 2 - 3 \sqrt{3} < 2 < - 2 \sqrt{3}$
From
(i) $8 ($ ii $)$ a $\in (- 3 \sqrt{3}, - 2 \sqrt{3}) \cup (2 \sqrt{3}, 3 \sqrt{3})$
Hence, the correct option is $D$ .

Question 1/10

0

2

3

–4

Question 2/10

O

A2 + B2

A2 + 2AB + B2

A+B

Question 3/10

(1+i)(1−2−2n)

(1−i)(1−2−2n)

(1+i)(1+2−2n)

none of these

Question 4/10

lie outside or on the circle |z| = 2/3

lie on the circle |z| = 2/3

lie inside or on the circle |z| = 2/3

none of these

Question 5/10

1/3

19/54

17/54

1/4

Question 6/10

a=b=12,c=14

a=b=14,c=12

a=b=c=13

None of these

Question 7/10

f3(x)=f(x3)+3f(1x)

f3(x)=3f(x3)+f(1x)

f3(x)+f(x3)+f(x)

None of these

Question 8/10

ln⁡3128ln⁡4

9ln⁡3128ln⁡4

9ln⁡3ln⁡4

9ln⁡364ln⁡4

Question 9/10

π27cm/sec

8π27cm/sec

8π9cm/sec

None of these

Question 10/10

an empty set

(−33,−23)

(23,33)

(−33,−23)∪(23,33)

A² + B²

A² + 2AB + B²

$(1 + i) (1 - 2^{- 2^{n}})$

$(1 - i) (1 - 2^{- 2^{n}})$

$(1 + i) (1 + 2^{- 2^{n}})$

$a = b = \frac{1}{2}, c = \frac{1}{4}$

$a = b = \frac{1}{4}, c = \frac{1}{2}$

$a = b = c = \frac{1}{3}$

$f^{3} (x) = f (x^{3}) + 3 f (\frac{1}{x})$

$f^{3} (x) = 3 f (x^{3}) + f (\frac{1}{x})$

$f^{3} (x) + f (x^{3}) + f (x)$

$\frac{\ln 3}{128 \ln 4}$

$\frac{9 \ln 3}{128 \ln 4}$

$\frac{9 \ln 3}{\ln 4}$

$\frac{9 \ln 3}{64 \ln 4}$

$\frac{π}{27} cm / \sec$

$\frac{8 π}{27} cm / \sec$

$\frac{8 π}{9} cm / \sec$

$(- 3 \sqrt{3}, - 2 \sqrt{3})$

$(2 \sqrt{3}, 3 \sqrt{3})$

$(- 3 \sqrt{3}, - 2 \sqrt{3}) \cup (2 \sqrt{3}, 3 \sqrt{3})$